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devel / comp.theory / Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

SubjectAuthor
* Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
|+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
|||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
|||  `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
|||   `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
|+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
|||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| | `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |  `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |   `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |    `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |     `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |      `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |       `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |        `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |         +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |         |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |         | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||| |         | |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |         | | `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||| |         | |  `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |         | |   `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||| |         | |    `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
||| |         | |     `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||| |         | |      `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFColcott
||| |         | |       +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior YES/NOolcott
||| |         | |       |`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior YES/NORichard Damon
||| |         | |       +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFCRichard Damon
||| |         | |       |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFColcott
||| |         | |       | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFCRichard Damon
||| |         | |       | |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFColcott
||| |         | |       | | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFCRichard Damon
||| |         | |       | | |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --aolcott
||| |         | |       | | ||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --aimmibis
||| |         | |       | | || `- H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct when reports on the actual behavior that it seesolcott
||| |         | |       | | |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --aolcott
||| |         | |       | | ||+* When H ⟨Ĥ⟩ ⟨Ĥ⟩ reports on the behavior it actually sees then it is correctolcott
||| |         | |       | | |||+* Re: When H ⟨Ĥ⟩ ⟨Ĥ⟩ reports on the behavior it actually sees then it is correctMikko
||| |         | |       | | ||||`- Re: When H ⟨Ĥ⟩ ⟨Ĥ⟩ reports on the behavior it actually sees then it is correctRichard Damon
||| |         | |       | | |||`* Re: When H ⟨Ĥ⟩ ⟨Ĥ⟩ reports on the behavior it actually sees then it is correctolcott
||| |         | |       | | ||| `- Re: When H ⟨Ĥ⟩ ⟨Ĥ⟩ reports on the behavior it actually sees then it is correctolcott
||| |         | |       | | ||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | || `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --lRichard Damon
||| |         | |       | | |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --aRichard Damon
||| |         | |       | | | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | | |`- H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct when reports on the actual behavior that it seesolcott
||| |         | |       | | | | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | | |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --wRichard Damon
||| |         | |       | | | | ||+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --Oimmibis
||| |         | |       | | | | ||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --ORichard Damon
||| |         | |       | | | | |+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --wimmibis
||| |         | |       | | | | |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | | | `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | | |  +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | | |  |+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | | |  |+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | | |  |`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | | |  `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | | `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | |||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --wolcott
||| |         | |       | | | ||||+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --Oolcott
||| |         | |       | | | ||||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --Oolcott
||| |         | |       | | | |||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --ORichard Damon
||| |         | |       | | | ||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | || `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | ||  |+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  |+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | ||  ||+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  ||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | ||  |||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  ||||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | ||  |||+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  |||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | ||  ||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  || `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | ||  ||  `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  ||+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | ||  |||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | ||  ||`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | ||  || +* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  || |`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | ||  || `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | ||  |+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  ||+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | | ||  ||+- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  ||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | ||  |`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | ||  `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nimmibis
||| |         | |       | | | |`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nMikko
||| |         | |       | | | +- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nolcott
||| |         | |       | | | `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --nRichard Damon
||| |         | |       | | `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFC --aRichard Damon
||| |         | |       | `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFCimmibis
||| |         | |       `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior ZFCimmibis
||| |         | `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
||| |         `- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||| `* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
||`- Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorRichard Damon
|`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott
+* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorimmibis
`* Re: Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behaviorolcott

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Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut1t4k$2c29l$3@dont-email.me>

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https://news.novabbs.org/devel/article-flat.php?id=55903&group=comp.theory#55903

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Fri, 15 Mar 2024 11:31:15 -0500
Organization: A noiseless patient Spider
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In-Reply-To: <ut181k$27qab$1@dont-email.me>
 by: olcott - Fri, 15 Mar 2024 16:31 UTC

On 3/15/2024 5:31 AM, Mikko wrote:
> On 2024-03-14 20:32:37 +0000, olcott said:
>
>> On 3/14/2024 7:19 AM, Mikko wrote:
>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>
>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>
>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>
>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as
>>>>>>>>>>>>>>>> unsound.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as
>>>>>>>>>>>>>>>> unsound
>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So you are saying that some Turing machines are not real
>>>>>>>>>>>>>>> Turing machines?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES otherwise.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> well the halting problem requires them to correctly say
>>>>>>>>>>>>>>> NO, so you haven't solved it
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> All decision problem instances of program/input such that
>>>>>>>>>>>>>> both
>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as
>>>>>>>>>>>>>> invalid.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> all decision problems are defined so that all instances are
>>>>>>>>>>>>> valid or else they are not defined properly
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>
>>>>>>>>>>> And now we are back to: Every Turing machine and input pair
>>>>>>>>>>> defines an execution sequence. Every sequence is either
>>>>>>>>>>> finite or infinite. Therefore it is well-defined and there is
>>>>>>>>>>> no paradox.
>>>>>>>>>>>
>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence of
>>>>>>>>>>> configurations that is not finite or infinite?
>>>>>>>>>>
>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>
>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>
>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>> or mentioned above.
>>>>>>>>>
>>>>>>>>
>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>
>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>> the wrong answer the other answer is right.
>>>>>>>
>>>>>>
>>>>>> *Your reasoning is just like this reasoning*
>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>
>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>> [Proving my 2004 claim that some decider/input pairs are incorrect
>>>>>> questions]
>>>>>
>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is right.
>>>>
>>>> *You have already affirmed otherwise*
>>>>
>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>  >> *Formalized*
>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>  >
>>>>  > Yes, this is one way to state Linz' conclusion.
>>>
>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>
>>
>> *The conventional spec is proved to be wrong for these instances*
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not halt
>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>
> Nice to see that you don't disagree with my observation about your mistake.
>

Best selling author of Theory of Computation textbooks:
*Introduction To The Theory Of Computation 3RD, by sipser*
https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/

Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H
correctly determines that its simulated D would never stop running
unless aborted then
(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

*When we apply the abort criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then H(D,D) is proven to meet this criteria*

This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.

*Lets stick with H(D,D)==0 is correct according to the above criteria*
until we have mutual agreement on that because:


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut1uhi$1vpl4$7@i2pn2.org>

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https://news.novabbs.org/devel/article-flat.php?id=55916&group=comp.theory#55916

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Path: i2pn2.org!.POSTED!not-for-mail
From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Fri, 15 Mar 2024 09:55:14 -0700
Organization: i2pn2 (i2pn.org)
Message-ID: <ut1uhi$1vpl4$7@i2pn2.org>
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 by: Richard Damon - Fri, 15 Mar 2024 16:55 UTC

On 3/15/24 9:31 AM, olcott wrote:
> On 3/15/2024 5:31 AM, Mikko wrote:
>> On 2024-03-14 20:32:37 +0000, olcott said:
>>
>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>
>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>
>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as
>>>>>>>>>>>>>>>>> unsound.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as
>>>>>>>>>>>>>>>>> unsound
>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not real
>>>>>>>>>>>>>>>> Turing machines?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES
>>>>>>>>>>>>>>>>> otherwise.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say
>>>>>>>>>>>>>>>> NO, so you haven't solved it
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> All decision problem instances of program/input such that
>>>>>>>>>>>>>>> both
>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as
>>>>>>>>>>>>>>> invalid.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> all decision problems are defined so that all instances
>>>>>>>>>>>>>> are valid or else they are not defined properly
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>
>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair
>>>>>>>>>>>> defines an execution sequence. Every sequence is either
>>>>>>>>>>>> finite or infinite. Therefore it is well-defined and there
>>>>>>>>>>>> is no paradox.
>>>>>>>>>>>>
>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence
>>>>>>>>>>>> of configurations that is not finite or infinite?
>>>>>>>>>>>
>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>
>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>
>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>> or mentioned above.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>
>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>
>>>>>>>
>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>
>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>> [Proving my 2004 claim that some decider/input pairs are
>>>>>>> incorrect questions]
>>>>>>
>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is right.
>>>>>
>>>>> *You have already affirmed otherwise*
>>>>>
>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>  >> *Formalized*
>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>  >
>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>
>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>
>>>
>>> *The conventional spec is proved to be wrong for these instances*
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not halt
>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>
>> Nice to see that you don't disagree with my observation about your
>> mistake.
>>
>
> Best selling author of Theory of Computation textbooks:
> *Introduction To The Theory Of Computation 3RD, by sipser*
> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>
> Date 10/13/2022 11:29:23 AM
> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
> (He has neither reviewed nor agreed to anything else in this paper)
> (a) If simulating halt decider H correctly simulates its input D until H
> correctly determines that its simulated D would never stop running
> unless aborted then
> (b) H can abort its simulation of D and correctly report that D
> specifies a non-halting sequence of configurations.
>
> *When we apply the abort criteria* (elaborated above)
> Will you halt if you never abort your simulation?
> *Then H(D,D) is proven to meet this criteria*
>
> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>
> *Lets stick with H(D,D)==0 is correct according to the above criteria*
> until we have mutual agreement on that because:
>
> I cannot afford to tolerate the [change the subject]
> form of rebuttal that wasted 15 years with Ben Bacarisse.
>
>


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut1ul7$2c29l$11@dont-email.me>

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https://news.novabbs.org/devel/article-flat.php?id=55919&group=comp.theory#55919

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Path: i2pn2.org!i2pn.org!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail
From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Fri, 15 Mar 2024 11:57:10 -0500
Organization: A noiseless patient Spider
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In-Reply-To: <ut1uhi$1vpl4$7@i2pn2.org>
 by: olcott - Fri, 15 Mar 2024 16:57 UTC

On 3/15/2024 11:55 AM, Richard Damon wrote:
> On 3/15/24 9:31 AM, olcott wrote:
>> On 3/15/2024 5:31 AM, Mikko wrote:
>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>
>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>
>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>
>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>
>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question
>>>>>>>>>>>>>>>>>> as unsound.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question
>>>>>>>>>>>>>>>>>> as unsound
>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not
>>>>>>>>>>>>>>>>> real Turing machines?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES
>>>>>>>>>>>>>>>>>> otherwise.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say
>>>>>>>>>>>>>>>>> NO, so you haven't solved it
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> All decision problem instances of program/input such
>>>>>>>>>>>>>>>> that both
>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as
>>>>>>>>>>>>>>>> invalid.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> all decision problems are defined so that all instances
>>>>>>>>>>>>>>> are valid or else they are not defined properly
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>
>>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair
>>>>>>>>>>>>> defines an execution sequence. Every sequence is either
>>>>>>>>>>>>> finite or infinite. Therefore it is well-defined and there
>>>>>>>>>>>>> is no paradox.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence
>>>>>>>>>>>>> of configurations that is not finite or infinite?
>>>>>>>>>>>>
>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>
>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>
>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>
>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>
>>>>>>>>
>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>
>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>> [Proving my 2004 claim that some decider/input pairs are
>>>>>>>> incorrect questions]
>>>>>>>
>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is
>>>>>>> right.
>>>>>>
>>>>>> *You have already affirmed otherwise*
>>>>>>
>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>  >> *Formalized*
>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>  >
>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>
>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>
>>>>
>>>> *The conventional spec is proved to be wrong for these instances*
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not
>>>> halt
>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>
>>> Nice to see that you don't disagree with my observation about your
>>> mistake.
>>>
>>
>> Best selling author of Theory of Computation textbooks:
>> *Introduction To The Theory Of Computation 3RD, by sipser*
>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>
>> Date 10/13/2022 11:29:23 AM
>> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
>> (He has neither reviewed nor agreed to anything else in this paper)
>> (a) If simulating halt decider H correctly simulates its input D until
>> H correctly determines that its simulated D would never stop running
>> unless aborted then
>> (b) H can abort its simulation of D and correctly report that D
>> specifies a non-halting sequence of configurations.
>>
>> *When we apply the abort criteria* (elaborated above)
>> Will you halt if you never abort your simulation?
>> *Then H(D,D) is proven to meet this criteria*
>>
>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>
>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>> until we have mutual agreement on that because:
>>
>> I cannot afford to tolerate the [change the subject]
>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>>
>>
>
> Except that it doesn't meet the requirements,


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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https://news.novabbs.org/devel/article-flat.php?id=56082&group=comp.theory#56082

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From: mikko.levanto@iki.fi (Mikko)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it sees
Date: Sat, 16 Mar 2024 13:51:05 +0200
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 by: Mikko - Sat, 16 Mar 2024 11:51 UTC

On 2024-03-15 16:31:15 +0000, olcott said:

> On 3/15/2024 5:31 AM, Mikko wrote:
>> On 2024-03-14 20:32:37 +0000, olcott said:
>>
>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>
>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>
>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound
>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not real Turing machines?
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES otherwise.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say NO, so you
>>>>>>>>>>>>>>>> haven't solved it
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> All decision problem instances of program/input such that both
>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as invalid.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> all decision problems are defined so that all instances are valid or
>>>>>>>>>>>>>> else they are not defined properly
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>
>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair defines an
>>>>>>>>>>>> execution sequence. Every sequence is either finite or infinite.
>>>>>>>>>>>> Therefore it is well-defined and there is no paradox.
>>>>>>>>>>>>
>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence of
>>>>>>>>>>>> configurations that is not finite or infinite?
>>>>>>>>>>>
>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>
>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>
>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>> or mentioned above.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>
>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>
>>>>>>>
>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>
>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>> [Proving my 2004 claim that some decider/input pairs are incorrect questions]
>>>>>>
>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is right.
>>>>>
>>>>> *You have already affirmed otherwise*
>>>>>
>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>  >> *Formalized*
>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>  >
>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>
>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>
>>>
>>> *The conventional spec is proved to be wrong for these instances*
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not halt
>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>
>> Nice to see that you don't disagree with my observation about your mistake.
>>
>
> Best selling author of Theory of Computation textbooks:
> *Introduction To The Theory Of Computation 3RD, by sipser*
> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>
> Date 10/13/2022 11:29:23 AM
> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
> (He has neither reviewed nor agreed to anything else in this paper)
> (a) If simulating halt decider H correctly simulates its input D until
> H correctly determines that its simulated D would never stop running
> unless aborted then
> (b) H can abort its simulation of D and correctly report that D
> specifies a non-halting sequence of configurations.
>
> *When we apply the abort criteria* (elaborated above)
> Will you halt if you never abort your simulation?
> *Then H(D,D) is proven to meet this criteria*
>
> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>
> *Lets stick with H(D,D)==0 is correct according to the above criteria*
> until we have mutual agreement on that because:
>
> I cannot afford to tolerate the [change the subject]
> form of rebuttal that wasted 15 years with Ben Bacarisse.


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut4bci$2uihj$2@dont-email.me>

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https://news.novabbs.org/devel/article-flat.php?id=56085&group=comp.theory#56085

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sat, 16 Mar 2024 09:46:42 -0500
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In-Reply-To: <ut4139$2sioj$1@dont-email.me>
 by: olcott - Sat, 16 Mar 2024 14:46 UTC

On 3/16/2024 6:51 AM, Mikko wrote:
> On 2024-03-15 16:31:15 +0000, olcott said:
>
>> On 3/15/2024 5:31 AM, Mikko wrote:
>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>
>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>
>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>
>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>
>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question
>>>>>>>>>>>>>>>>>> as unsound.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question
>>>>>>>>>>>>>>>>>> as unsound
>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not
>>>>>>>>>>>>>>>>> real Turing machines?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES
>>>>>>>>>>>>>>>>>> otherwise.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say
>>>>>>>>>>>>>>>>> NO, so you haven't solved it
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> All decision problem instances of program/input such
>>>>>>>>>>>>>>>> that both
>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as
>>>>>>>>>>>>>>>> invalid.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> all decision problems are defined so that all instances
>>>>>>>>>>>>>>> are valid or else they are not defined properly
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>
>>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair
>>>>>>>>>>>>> defines an execution sequence. Every sequence is either
>>>>>>>>>>>>> finite or infinite. Therefore it is well-defined and there
>>>>>>>>>>>>> is no paradox.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence
>>>>>>>>>>>>> of configurations that is not finite or infinite?
>>>>>>>>>>>>
>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>
>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>
>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>
>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>
>>>>>>>>
>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>
>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>> [Proving my 2004 claim that some decider/input pairs are
>>>>>>>> incorrect questions]
>>>>>>>
>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is
>>>>>>> right.
>>>>>>
>>>>>> *You have already affirmed otherwise*
>>>>>>
>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>  >> *Formalized*
>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>  >
>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>
>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>
>>>>
>>>> *The conventional spec is proved to be wrong for these instances*
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not
>>>> halt
>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>
>>> Nice to see that you don't disagree with my observation about your
>>> mistake.
>>>
>>
>> Best selling author of Theory of Computation textbooks:
>> *Introduction To The Theory Of Computation 3RD, by sipser*
>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>
>> Date 10/13/2022 11:29:23 AM
>> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
>> (He has neither reviewed nor agreed to anything else in this paper)
>> (a) If simulating halt decider H correctly simulates its input D until
>> H correctly determines that its simulated D would never stop running
>> unless aborted then
>> (b) H can abort its simulation of D and correctly report that D
>> specifies a non-halting sequence of configurations.
>>
>> *When we apply the abort criteria* (elaborated above)
>> Will you halt if you never abort your simulation?
>> *Then H(D,D) is proven to meet this criteria*
>>
>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>
>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>> until we have mutual agreement on that because:
>>
>> I cannot afford to tolerate the [change the subject]
>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>
> Anyway it is nice to see that you agree even if that may be
> a change of subject.
>


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut7471$3jp4a$1@dont-email.me>

  copy mid

https://news.novabbs.org/devel/article-flat.php?id=56270&group=comp.theory#56270

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From: mikko.levanto@iki.fi (Mikko)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it sees
Date: Sun, 17 Mar 2024 18:02:41 +0200
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 by: Mikko - Sun, 17 Mar 2024 16:02 UTC

On 2024-03-16 14:46:42 +0000, olcott said:

> On 3/16/2024 6:51 AM, Mikko wrote:
>> On 2024-03-15 16:31:15 +0000, olcott said:
>>
>>> On 3/15/2024 5:31 AM, Mikko wrote:
>>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>>
>>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>>
>>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound
>>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not real Turing machines?
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES otherwise.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say NO, so you
>>>>>>>>>>>>>>>>>> haven't solved it
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> All decision problem instances of program/input such that both
>>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as invalid.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> all decision problems are defined so that all instances are valid or
>>>>>>>>>>>>>>>> else they are not defined properly
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair defines an
>>>>>>>>>>>>>> execution sequence. Every sequence is either finite or infinite.
>>>>>>>>>>>>>> Therefore it is well-defined and there is no paradox.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence of
>>>>>>>>>>>>>> configurations that is not finite or infinite?
>>>>>>>>>>>>>
>>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>>
>>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>>
>>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>>
>>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>>
>>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>>> [Proving my 2004 claim that some decider/input pairs are incorrect questions]
>>>>>>>>
>>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is right.
>>>>>>>
>>>>>>> *You have already affirmed otherwise*
>>>>>>>
>>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>>  >> *Formalized*
>>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>  >
>>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>>
>>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>>
>>>>>
>>>>> *The conventional spec is proved to be wrong for these instances*
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>>
>>>> Nice to see that you don't disagree with my observation about your mistake.
>>>>
>>>
>>> Best selling author of Theory of Computation textbooks:
>>> *Introduction To The Theory Of Computation 3RD, by sipser*
>>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>>
>>> Date 10/13/2022 11:29:23 AM
>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
>>> (He has neither reviewed nor agreed to anything else in this paper)
>>> (a) If simulating halt decider H correctly simulates its input D until
>>> H correctly determines that its simulated D would never stop running
>>> unless aborted then
>>> (b) H can abort its simulation of D and correctly report that D
>>> specifies a non-halting sequence of configurations.
>>>
>>> *When we apply the abort criteria* (elaborated above)
>>> Will you halt if you never abort your simulation?
>>> *Then H(D,D) is proven to meet this criteria*
>>>
>>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>>
>>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>>> until we have mutual agreement on that because:
>>>
>>> I cannot afford to tolerate the [change the subject]
>>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>>
>> Anyway it is nice to see that you agree even if that may be
>> a change of subject.
>>
>
> I have no idea what you are saying.
>
> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
> that H correctly determined that it had to abort the simulation
> of its input to prevent the infinite execution of this input.


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut76a6$3jbbs$3@dont-email.me>

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https://news.novabbs.org/devel/article-flat.php?id=56275&group=comp.theory#56275

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 11:38:30 -0500
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 by: olcott - Sun, 17 Mar 2024 16:38 UTC

On 3/17/2024 11:02 AM, Mikko wrote:
> On 2024-03-16 14:46:42 +0000, olcott said:
>
>> On 3/16/2024 6:51 AM, Mikko wrote:
>>> On 2024-03-15 16:31:15 +0000, olcott said:
>>>
>>>> On 3/15/2024 5:31 AM, Mikko wrote:
>>>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>>>
>>>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>>>
>>>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>>>
>>>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question
>>>>>>>>>>>>>>>>>>>> as unsound.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not
>>>>>>>>>>>>>>>>>>> sound?
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question
>>>>>>>>>>>>>>>>>>>> as unsound
>>>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not
>>>>>>>>>>>>>>>>>>> real Turing machines?
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say
>>>>>>>>>>>>>>>>>>>> YES
>>>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES
>>>>>>>>>>>>>>>>>>>> otherwise.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly
>>>>>>>>>>>>>>>>>>> say NO, so you haven't solved it
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> All decision problem instances of program/input such
>>>>>>>>>>>>>>>>>> that both
>>>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as
>>>>>>>>>>>>>>>>>> invalid.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> all decision problems are defined so that all instances
>>>>>>>>>>>>>>>>> are valid or else they are not defined properly
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> And now we are back to: Every Turing machine and input
>>>>>>>>>>>>>>> pair defines an execution sequence. Every sequence is
>>>>>>>>>>>>>>> either finite or infinite. Therefore it is well-defined
>>>>>>>>>>>>>>> and there is no paradox.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Can you show me a Turing machine that specifies a
>>>>>>>>>>>>>>> sequence of configurations that is not finite or infinite?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is
>>>>>>>>>>>>> shown
>>>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>>>
>>>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>>>
>>>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>>>> [Proving my 2004 claim that some decider/input pairs are
>>>>>>>>>> incorrect questions]
>>>>>>>>>
>>>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is
>>>>>>>>> right.
>>>>>>>>
>>>>>>>> *You have already affirmed otherwise*
>>>>>>>>
>>>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>>>  >> *Formalized*
>>>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>  >
>>>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>>>
>>>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>>>
>>>>>>
>>>>>> *The conventional spec is proved to be wrong for these instances*
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does
>>>>>> not halt
>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>>>
>>>>> Nice to see that you don't disagree with my observation about your
>>>>> mistake.
>>>>>
>>>>
>>>> Best selling author of Theory of Computation textbooks:
>>>> *Introduction To The Theory Of Computation 3RD, by sipser*
>>>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>>>
>>>> Date 10/13/2022 11:29:23 AM
>>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is
>>>> correct*
>>>> (He has neither reviewed nor agreed to anything else in this paper)
>>>> (a) If simulating halt decider H correctly simulates its input D
>>>> until H correctly determines that its simulated D would never stop
>>>> running unless aborted then
>>>> (b) H can abort its simulation of D and correctly report that D
>>>> specifies a non-halting sequence of configurations.
>>>>
>>>> *When we apply the abort criteria* (elaborated above)
>>>> Will you halt if you never abort your simulation?
>>>> *Then H(D,D) is proven to meet this criteria*
>>>>
>>>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>>>
>>>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>>>> until we have mutual agreement on that because:
>>>>
>>>> I cannot afford to tolerate the [change the subject]
>>>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>>>
>>> Anyway it is nice to see that you agree even if that may be
>>> a change of subject.
>>>
>>
>> I have no idea what you are saying.
>>
>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>> that H correctly determined that it had to abort the simulation
>> of its input to prevent the infinite execution of this input.
>
> What would H return if it incorrecly detected that it had to
> abort the simulation of its input to prevent the infinite execution
> of this input?
>


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Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: news@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 18:35:36 +0100
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 by: immibis - Sun, 17 Mar 2024 17:35 UTC

On 17/03/24 17:38, olcott wrote:
> On 3/17/2024 11:02 AM, Mikko wrote:
>> On 2024-03-16 14:46:42 +0000, olcott said:
>>
>>> On 3/16/2024 6:51 AM, Mikko wrote:
>>>> On 2024-03-15 16:31:15 +0000, olcott said:
>>>>
>>>>> On 3/15/2024 5:31 AM, Mikko wrote:
>>>>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>>>>
>>>>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox
>>>>>>>>>>>>>>>>>>>>> question as unsound.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not
>>>>>>>>>>>>>>>>>>>> sound?
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide
>>>>>>>>>>>>>>>>>>>>>>> that:
>>>>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox
>>>>>>>>>>>>>>>>>>>>> question as unsound
>>>>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not
>>>>>>>>>>>>>>>>>>>> real Turing machines?
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly
>>>>>>>>>>>>>>>>>>>>> say YES
>>>>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES
>>>>>>>>>>>>>>>>>>>>> otherwise.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly
>>>>>>>>>>>>>>>>>>>> say NO, so you haven't solved it
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> All decision problem instances of program/input such
>>>>>>>>>>>>>>>>>>> that both
>>>>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as
>>>>>>>>>>>>>>>>>>> invalid.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> all decision problems are defined so that all
>>>>>>>>>>>>>>>>>> instances are valid or else they are not defined properly
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> And now we are back to: Every Turing machine and input
>>>>>>>>>>>>>>>> pair defines an execution sequence. Every sequence is
>>>>>>>>>>>>>>>> either finite or infinite. Therefore it is well-defined
>>>>>>>>>>>>>>>> and there is no paradox.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Can you show me a Turing machine that specifies a
>>>>>>>>>>>>>>>> sequence of configurations that is not finite or infinite?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is
>>>>>>>>>>>>>> shown
>>>>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>>>>
>>>>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>>>>
>>>>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>>>>> [Proving my 2004 claim that some decider/input pairs are
>>>>>>>>>>> incorrect questions]
>>>>>>>>>>
>>>>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is
>>>>>>>>>> right.
>>>>>>>>>
>>>>>>>>> *You have already affirmed otherwise*
>>>>>>>>>
>>>>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>>>>  >> *Formalized*
>>>>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>>  >
>>>>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>>>>
>>>>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>>>>
>>>>>>>
>>>>>>> *The conventional spec is proved to be wrong for these instances*
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does
>>>>>>> not halt
>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>>>>
>>>>>> Nice to see that you don't disagree with my observation about your
>>>>>> mistake.
>>>>>>
>>>>>
>>>>> Best selling author of Theory of Computation textbooks:
>>>>> *Introduction To The Theory Of Computation 3RD, by sipser*
>>>>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>>>>
>>>>> Date 10/13/2022 11:29:23 AM
>>>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is
>>>>> correct*
>>>>> (He has neither reviewed nor agreed to anything else in this paper)
>>>>> (a) If simulating halt decider H correctly simulates its input D
>>>>> until H correctly determines that its simulated D would never stop
>>>>> running unless aborted then
>>>>> (b) H can abort its simulation of D and correctly report that D
>>>>> specifies a non-halting sequence of configurations.
>>>>>
>>>>> *When we apply the abort criteria* (elaborated above)
>>>>> Will you halt if you never abort your simulation?
>>>>> *Then H(D,D) is proven to meet this criteria*
>>>>>
>>>>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>>>>
>>>>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>>>>> until we have mutual agreement on that because:
>>>>>
>>>>> I cannot afford to tolerate the [change the subject]
>>>>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>>>>
>>>> Anyway it is nice to see that you agree even if that may be
>>>> a change of subject.
>>>>
>>>
>>> I have no idea what you are saying.
>>>
>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>> that H correctly determined that it had to abort the simulation
>>> of its input to prevent the infinite execution of this input.
>>
>> What would H return if it incorrecly detected that it had to
>> abort the simulation of its input to prevent the infinite execution
>> of this input?
>>
>
> H(D,D)==0 means that it correctly detected that it must abort
> its simulation. H(D,D)==1 means that H(D,D) never returns and
> indeed this is a verified fact.
>


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

<ut7hkh$272r7$4@i2pn2.org>

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 12:51:44 -0700
Organization: i2pn2 (i2pn.org)
Message-ID: <ut7hkh$272r7$4@i2pn2.org>
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 by: Richard Damon - Sun, 17 Mar 2024 19:51 UTC

On 3/17/24 9:38 AM, olcott wrote:
> On 3/17/2024 11:02 AM, Mikko wrote:
>> On 2024-03-16 14:46:42 +0000, olcott said:
>>
>>> On 3
>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>> that H correctly determined that it had to abort the simulation
>>> of its input to prevent the infinite execution of this input.
>>
>> What would H return if it incorrecly detected that it had to
>> abort the simulation of its input to prevent the infinite execution
>> of this input?
>>
>
> H(D,D)==0 means that it correctly detected that it must abort
> its simulation. H(D,D)==1 means that H(D,D) never returns and
> indeed this is a verified fact.
>

No idea where you got that from.

H(D,D) means that D(D) will halt.

H(D,D) returning 0, even for your "Needed to Abort" criteria means that
H actually needed to abort it simulation

The meaning of that was aggreed to in:

On 3/17/24 6:11 AM, olcott wrote:
> On 3/17/2024 12:22 AM, Richard Damon wrote:
>> To me, for H to NEED to abort its simulation, that means that when
>> giving the input to a correct simulator, that simulator will not halt.
>>
> Yes that is correct.
>

So, since D(D) will halt, and thus when we give this input to a correct
simulator it will halt, it means that H did no "CORRECTLY" determine
that it needed to abort, and thus, you H isn't allowed to return 0 and
be correct, but it does, so it is just wrong.

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
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Date: Sun, 17 Mar 2024 18:43:36 -0500
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In-Reply-To: <ut7hkh$272r7$4@i2pn2.org>
 by: olcott - Sun, 17 Mar 2024 23:43 UTC

On 3/17/2024 2:51 PM, Richard Damon wrote:
> On 3/17/24 9:38 AM, olcott wrote:
>> On 3/17/2024 11:02 AM, Mikko wrote:
>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>
>>>> On 3
>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>> that H correctly determined that it had to abort the simulation
>>>> of its input to prevent the infinite execution of this input.
>>>
>>> What would H return if it incorrecly detected that it had to
>>> abort the simulation of its input to prevent the infinite execution
>>> of this input?
>>>
>>
>> H(D,D)==0 means that it correctly detected that it must abort
>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>> indeed this is a verified fact.
>>
>
> No idea where you got that from.
>

*I did not explain that clearly enough*
The needing to abort criteria is the same as the halt criteria
when H(D,D) is reporting on the behavior that it actually sees.

> H(D,D) means that D(D) will halt.
>
> H(D,D) returning 0, even for your "Needed to Abort" criteria means that
> H actually needed to abort it simulation
>
Yes

> The meaning of that was aggreed to in:
>
> On 3/17/24 6:11 AM, olcott wrote:
> > On 3/17/2024 12:22 AM, Richard Damon wrote:
> >> To me, for H to NEED to abort its simulation, that means that when
> >> giving the input to a correct simulator, that simulator will not halt.
> >>
> > Yes that is correct.
> >
>
> So, since D(D) will halt,
That is not the behavior that H(D,D) actually sees.

For every possible way that H can be encoded and D(D)
calls H(D,D) either H(D,D) aborts its simulation or D(D)
never stops running.

Aborting its simulation means that H(D,D) sees non-halting
behavior that must be aborted.

> and thus when we give this input to a correct
> simulator it will halt, it means that H did no "CORRECTLY" determine
> that it needed to abort, and thus, you H isn't allowed to return 0 and
> be correct, but it does, so it is just wrong.

Requiring H(D,D) to be clairvoyant and report on behavior
that is different than the behavior that it actually sees
is the actual error.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: news@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
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Date: Mon, 18 Mar 2024 01:25:07 +0100
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 by: immibis - Mon, 18 Mar 2024 00:25 UTC

On 18/03/24 00:43, olcott wrote:
> On 3/17/2024 2:51 PM, Richard Damon wrote:
>> On 3/17/24 9:38 AM, olcott wrote:
>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>
>>>>> On 3
>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>> that H correctly determined that it had to abort the simulation
>>>>> of its input to prevent the infinite execution of this input.
>>>>
>>>> What would H return if it incorrecly detected that it had to
>>>> abort the simulation of its input to prevent the infinite execution
>>>> of this input?
>>>>
>>>
>>> H(D,D)==0 means that it correctly detected that it must abort
>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>> indeed this is a verified fact.
>>>
>>
>> No idea where you got that from.
>>
>
> *I did not explain that clearly enough*
> The needing to abort criteria is the same as the halt criteria
> when H(D,D) is reporting on the behavior that it actually sees.
>
>> H(D,D) means that D(D) will halt.
>>
>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>> that H actually needed to abort it simulation
>>
> Yes
>
>> The meaning of that was aggreed to in:
>>
>> On 3/17/24 6:11 AM, olcott wrote:
>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>  >> To me, for H to NEED to abort its simulation, that means that when
>>  >> giving the input to a correct simulator, that simulator will not
>> halt.
>>  >>
>>  > Yes that is correct.
>>  >
>>
>> So, since D(D) will halt,
> That is not the behavior that H(D,D) actually sees.

H(D,D) does not see anything of the future. It only sees the past. It
has to predict the future, and *it predicts wrong.*

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 18:06:51 -0700
Organization: i2pn2 (i2pn.org)
Message-ID: <ut843b$27bqa$2@i2pn2.org>
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 by: Richard Damon - Mon, 18 Mar 2024 01:06 UTC

On 3/17/24 4:43 PM, olcott wrote:
> On 3/17/2024 2:51 PM, Richard Damon wrote:
>> On 3/17/24 9:38 AM, olcott wrote:
>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>
>>>>> On 3
>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>> that H correctly determined that it had to abort the simulation
>>>>> of its input to prevent the infinite execution of this input.
>>>>
>>>> What would H return if it incorrecly detected that it had to
>>>> abort the simulation of its input to prevent the infinite execution
>>>> of this input?
>>>>
>>>
>>> H(D,D)==0 means that it correctly detected that it must abort
>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>> indeed this is a verified fact.
>>>
>>
>> No idea where you got that from.
>>
>
> *I did not explain that clearly enough*
> The needing to abort criteria is the same as the halt criteria
> when H(D,D) is reporting on the behavior that it actually sees.

So, why does H(D,D) == 1 mean that H(D,D) never returns?
Shouldn't that be that H(D,D) == 1 mean that the input does halt?

>
>> H(D,D) means that D(D) will halt.
>>
>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>> that H actually needed to abort it simulation
>>
> Yes
>
>> The meaning of that was aggreed to in:
>>
>> On 3/17/24 6:11 AM, olcott wrote:
>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>  >> To me, for H to NEED to abort its simulation, that means that when
>>  >> giving the input to a correct simulator, that simulator will not
>> halt.
>>  >>
>>  > Yes that is correct.
>>  >
>>
>> So, since D(D) will halt,
> That is not the behavior that H(D,D) actually sees.

The criteria you agreed to wasn't about what H sees, but what a corect
simulator sees.

So, you seem to be confused again.

>
> For every possible way that H can be encoded and D(D)
> calls H(D,D) either H(D,D) aborts its simulation or D(D)
> never stops running.

But that doesn't prove that WHEN H(D,D) aborted its simulation, that the
input is was will never stop running when actually correctly simulate.

You seem to like living in the lie.

>
> Aborting its simulation means that H(D,D) sees non-halting
> behavior that must be aborted.

No, H(D,D) aborting its simulation means that it THINKS it sees
non-halting behavior, but in this case, it, LIKE YOU, is WROG.

>
>> and thus when we give this input to a correct simulator it will halt,
>> it means that H did no "CORRECTLY" determine that it needed to abort,
>> and thus, you H isn't allowed to return 0 and be correct, but it does,
>> so it is just wrong.
>
> Requiring H(D,D) to be clairvoyant and report on behavior
> that is different than the behavior that it actually sees
> is the actual error.
>

Nope. thinking it is an error is your error.

What is the formal grounds you are calling it an ERROR?

Name the defined meaning that this is actually an ERROR?

Your are just living in a fantasy world with your own made up rules and
you don't know any actual basis for your claims, so you just make up rules.

Or, is asking you to define your words and your reasoning an ERROR
because it is just to hard for stupid little old you?

You are just proving you are nothing but a pathetic ignorant
hypocritical pathological lying idiot who has no idea what he is talking
about.

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 23:23:54 -0500
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 by: olcott - Mon, 18 Mar 2024 04:23 UTC

On 3/17/2024 7:25 PM, immibis wrote:
> On 18/03/24 00:43, olcott wrote:
>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>> On 3/17/24 9:38 AM, olcott wrote:
>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>
>>>>>> On 3
>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>> that H correctly determined that it had to abort the simulation
>>>>>> of its input to prevent the infinite execution of this input.
>>>>>
>>>>> What would H return if it incorrecly detected that it had to
>>>>> abort the simulation of its input to prevent the infinite execution
>>>>> of this input?
>>>>>
>>>>
>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>> indeed this is a verified fact.
>>>>
>>>
>>> No idea where you got that from.
>>>
>>
>> *I did not explain that clearly enough*
>> The needing to abort criteria is the same as the halt criteria
>> when H(D,D) is reporting on the behavior that it actually sees.
>>
>>> H(D,D) means that D(D) will halt.
>>>
>>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>>> that H actually needed to abort it simulation
>>>
>> Yes
>>
>>> The meaning of that was aggreed to in:
>>>
>>> On 3/17/24 6:11 AM, olcott wrote:
>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>  >> To me, for H to NEED to abort its simulation, that means that when
>>>  >> giving the input to a correct simulator, that simulator will not
>>> halt.
>>>  >>
>>>  > Yes that is correct.
>>>  >
>>>
>>> So, since D(D) will halt,
>> That is not the behavior that H(D,D) actually sees.
>
> H(D,D) does not see anything of the future. It only sees the past. It
> has to predict the future, and *it predicts wrong.*

H(D,D) has the inductive basis to see that it must abort its simulation.
This means that the behavior that H(D,D) sees <is> non-halting.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: news@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 05:45:07 +0100
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 by: immibis - Mon, 18 Mar 2024 04:45 UTC

On 18/03/24 05:23, olcott wrote:
> On 3/17/2024 7:25 PM, immibis wrote:
>> On 18/03/24 00:43, olcott wrote:
>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>
>>>>>>> On 3
>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>
>>>>>> What would H return if it incorrecly detected that it had to
>>>>>> abort the simulation of its input to prevent the infinite execution
>>>>>> of this input?
>>>>>>
>>>>>
>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>> indeed this is a verified fact.
>>>>>
>>>>
>>>> No idea where you got that from.
>>>>
>>>
>>> *I did not explain that clearly enough*
>>> The needing to abort criteria is the same as the halt criteria
>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>
>>>> H(D,D) means that D(D) will halt.
>>>>
>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>>>> that H actually needed to abort it simulation
>>>>
>>> Yes
>>>
>>>> The meaning of that was aggreed to in:
>>>>
>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>  >> To me, for H to NEED to abort its simulation, that means that when
>>>>  >> giving the input to a correct simulator, that simulator will not
>>>> halt.
>>>>  >>
>>>>  > Yes that is correct.
>>>>  >
>>>>
>>>> So, since D(D) will halt,
>>> That is not the behavior that H(D,D) actually sees.
>>
>> H(D,D) does not see anything of the future. It only sees the past. It
>> has to predict the future, and *it predicts wrong.*
>
> H(D,D) has the inductive basis to see that it must abort its simulation.
> This means that the behavior that H(D,D) sees <is> non-halting.
>
Does it have the inductive basis to prove that the direct execution of
D(D) does not halt?

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 00:11:47 -0500
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In-Reply-To: <ut8gtt$m2e$1@dont-email.me>
 by: olcott - Mon, 18 Mar 2024 05:11 UTC

On 3/17/2024 11:45 PM, immibis wrote:
> On 18/03/24 05:23, olcott wrote:
>> On 3/17/2024 7:25 PM, immibis wrote:
>>> On 18/03/24 00:43, olcott wrote:
>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>
>>>>>>>> On 3
>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>
>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>> abort the simulation of its input to prevent the infinite execution
>>>>>>> of this input?
>>>>>>>
>>>>>>
>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>> indeed this is a verified fact.
>>>>>>
>>>>>
>>>>> No idea where you got that from.
>>>>>
>>>>
>>>> *I did not explain that clearly enough*
>>>> The needing to abort criteria is the same as the halt criteria
>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>
>>>>> H(D,D) means that D(D) will halt.
>>>>>
>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>>>>> that H actually needed to abort it simulation
>>>>>
>>>> Yes
>>>>
>>>>> The meaning of that was aggreed to in:
>>>>>
>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>  >> To me, for H to NEED to abort its simulation, that means that when
>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>> not halt.
>>>>>  >>
>>>>>  > Yes that is correct.
>>>>>  >
>>>>>
>>>>> So, since D(D) will halt,
>>>> That is not the behavior that H(D,D) actually sees.
>>>
>>> H(D,D) does not see anything of the future. It only sees the past. It
>>> has to predict the future, and *it predicts wrong.*
>>
>> H(D,D) has the inductive basis to see that it must abort its simulation.
>> This means that the behavior that H(D,D) sees <is> non-halting.
>>
> Does it have the inductive basis to prove that the direct execution of
> D(D) does not halt?

That is hidden from it.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 22:23:34 -0700
Organization: i2pn2 (i2pn.org)
Message-ID: <ut8j4m$27bqb$11@i2pn2.org>
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 by: Richard Damon - Mon, 18 Mar 2024 05:23 UTC

On 3/17/24 10:11 PM, olcott wrote:
> On 3/17/2024 11:45 PM, immibis wrote:
>> On 18/03/24 05:23, olcott wrote:
>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>> On 18/03/24 00:43, olcott wrote:
>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3
>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>
>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>> abort the simulation of its input to prevent the infinite execution
>>>>>>>> of this input?
>>>>>>>>
>>>>>>>
>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>> indeed this is a verified fact.
>>>>>>>
>>>>>>
>>>>>> No idea where you got that from.
>>>>>>
>>>>>
>>>>> *I did not explain that clearly enough*
>>>>> The needing to abort criteria is the same as the halt criteria
>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>
>>>>>> H(D,D) means that D(D) will halt.
>>>>>>
>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>>>>>> that H actually needed to abort it simulation
>>>>>>
>>>>> Yes
>>>>>
>>>>>> The meaning of that was aggreed to in:
>>>>>>
>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>  >> To me, for H to NEED to abort its simulation, that means that
>>>>>> when
>>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>>> not halt.
>>>>>>  >>
>>>>>>  > Yes that is correct.
>>>>>>  >
>>>>>>
>>>>>> So, since D(D) will halt,
>>>>> That is not the behavior that H(D,D) actually sees.
>>>>
>>>> H(D,D) does not see anything of the future. It only sees the past.
>>>> It has to predict the future, and *it predicts wrong.*
>>>
>>> H(D,D) has the inductive basis to see that it must abort its simulation.
>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>
>> Does it have the inductive basis to prove that the direct execution of
>> D(D) does not halt?
>
> That is hidden from it.

So, you are admitting NO.

H doesn't have an correct induction basis to make its decision.

Remember, HALTING is a property of direct execution.

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 00:37:01 -0500
Organization: A noiseless patient Spider
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 by: olcott - Mon, 18 Mar 2024 05:37 UTC

On 3/18/2024 12:23 AM, Richard Damon wrote:
> On 3/17/24 10:11 PM, olcott wrote:
>> On 3/17/2024 11:45 PM, immibis wrote:
>>> On 18/03/24 05:23, olcott wrote:
>>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>>> On 18/03/24 00:43, olcott wrote:
>>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 3
>>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>>
>>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>>> abort the simulation of its input to prevent the infinite
>>>>>>>>> execution
>>>>>>>>> of this input?
>>>>>>>>>
>>>>>>>>
>>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>>> indeed this is a verified fact.
>>>>>>>>
>>>>>>>
>>>>>>> No idea where you got that from.
>>>>>>>
>>>>>>
>>>>>> *I did not explain that clearly enough*
>>>>>> The needing to abort criteria is the same as the halt criteria
>>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>>
>>>>>>> H(D,D) means that D(D) will halt.
>>>>>>>
>>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria
>>>>>>> means that H actually needed to abort it simulation
>>>>>>>
>>>>>> Yes
>>>>>>
>>>>>>> The meaning of that was aggreed to in:
>>>>>>>
>>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>>  >> To me, for H to NEED to abort its simulation, that means that
>>>>>>> when
>>>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>>>> not halt.
>>>>>>>  >>
>>>>>>>  > Yes that is correct.
>>>>>>>  >
>>>>>>>
>>>>>>> So, since D(D) will halt,
>>>>>> That is not the behavior that H(D,D) actually sees.
>>>>>
>>>>> H(D,D) does not see anything of the future. It only sees the past.
>>>>> It has to predict the future, and *it predicts wrong.*
>>>>
>>>> H(D,D) has the inductive basis to see that it must abort its
>>>> simulation.
>>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>>
>>> Does it have the inductive basis to prove that the direct execution
>>> of D(D) does not halt?
>>
>> That is hidden from it.
>
> So, you are admitting NO.
>
> H doesn't have an correct induction basis to make its decision.
>
> Remember, HALTING is a property of direct execution.

Can D correctly simulated by H terminate normally?
01 int D(ptr x) // ptr is pointer to int function
02 {
03 int Halt_Status = H(x, x);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 void main()
10 {
11 H(D,D);
12 }

If H(D,D) does return 1 for halts then D gets stuck at line 05.
If H1(D,D) returns 1 for halts then this is the correct answer.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Sun, 17 Mar 2024 23:20:20 -0700
Organization: i2pn2 (i2pn.org)
Message-ID: <ut8mf4$27bqb$14@i2pn2.org>
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 by: Richard Damon - Mon, 18 Mar 2024 06:20 UTC

On 3/17/24 10:37 PM, olcott wrote:
> On 3/18/2024 12:23 AM, Richard Damon wrote:

>> So, you are admitting NO.
>>
>> H doesn't have an correct induction basis to make its decision.
>>
>> Remember, HALTING is a property of direct execution.
>
> Can D correctly simulated by H terminate normally?
> 01 int D(ptr x)  // ptr is pointer to int function
> 02 {
> 03   int Halt_Status = H(x, x);
> 04   if (Halt_Status)
> 05     HERE: goto HERE;
> 06   return Halt_Status;
> 07 }
> 08
> 09 void main()
> 10 {
> 11   H(D,D);
> 12 }
>
> If H(D,D) does return 1 for halts then D gets stuck at line 05.
> If H1(D,D) returns 1 for halts then this is the correct answer.
>

So, have you stopped beating your wife?

You can't asks your quesiton until you DEFINE D fully, which means you
have DEFINED H fully.

So, What DOES H do.

2 cases: (assuming H does return)

Case 1:
H(D,D) returns 1, saying it thinks D will Halt
THen D will loop, making wrong.
H1 can probably detect that, and returns 0, and is right.
H could also have returns 0 to be right, except that it doesn't

There is one correct answer for everyone, that is the value 0

Case 2
H(D,D) returns 0, saying it thinks D will not halt.
Then D will halt, making H wrong.
H1 will detect this and return 1 and be right.
H would also have been right to return 1, but it doesn;t, so it was wrong.

Again, there is one correct answer for everyone , and this time it is 1.

Remember, a machine only does what it does, so an H that returns 0 for
an input can't be considered to return 1 except in a total fantasy (or
an admttted alternate reality).

For H to return 1 when it returns 0 is just using illogical thinking.

WHich seems par for your course.

Thus, you are show to again just be LYING about your argument.

It isn't just an honest mistake, because you have been told many times,
so it based from just a simple error to a blantant disregard for the
truth, and thus even if beleived by you, is still a lie.

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: mikko.levanto@iki.fi (Mikko)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it sees
Date: Mon, 18 Mar 2024 13:54:28 +0200
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 by: Mikko - Mon, 18 Mar 2024 11:54 UTC

On 2024-03-17 16:38:30 +0000, olcott said:

> On 3/17/2024 11:02 AM, Mikko wrote:
>> On 2024-03-16 14:46:42 +0000, olcott said:
>>
>>> On 3/16/2024 6:51 AM, Mikko wrote:
>>>> On 2024-03-15 16:31:15 +0000, olcott said:
>>>>
>>>>> On 3/15/2024 5:31 AM, Mikko wrote:
>>>>>> On 2024-03-14 20:32:37 +0000, olcott said:
>>>>>>
>>>>>>> On 3/14/2024 7:19 AM, Mikko wrote:
>>>>>>>> On 2024-03-13 17:19:06 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3/13/2024 12:10 PM, Mikko wrote:
>>>>>>>>>> On 2024-03-12 15:06:49 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3/12/2024 4:32 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-03-11 15:06:10 +0000, olcott said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 3/11/2024 5:56 AM, Mikko wrote:
>>>>>>>>>>>>>> On 2024-03-11 01:24:55 +0000, olcott said:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On 3/10/2024 7:40 PM, immibis wrote:
>>>>>>>>>>>>>>>> On 10/03/24 20:16, olcott wrote:
>>>>>>>>>>>>>>>>> On 3/10/2024 2:09 PM, immibis wrote:
>>>>>>>>>>>>>>>>>> On 10/03/24 19:32, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 3/10/2024 1:08 PM, immibis wrote:
>>>>>>>>>>>>>>>>>>>> On 10/03/24 18:17, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not sound?
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Both H ⟨Ĥ⟩ ⟨Ĥ⟩ and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ correctly decide that:
>>>>>>>>>>>>>>>>>>>>>>> (a) Their input halts H.qy
>>>>>>>>>>>>>>>>>>>>>>> (b) Their input fails to halt or has a pathological
>>>>>>>>>>>>>>>>>>>>>>> relationship to itself H.qn.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> But the "Pathological Relationship" is ALLOWED.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> ZFC simply tossed out the Russell's Paradox question as unsound
>>>>>>>>>>>>>>>>>>>>> expressly disallowing the "Pathological Relationship".
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> So you are saying that some Turing machines are not real Turing machines?
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> I am only claiming that both H and Ĥ.H correctly say YES
>>>>>>>>>>>>>>>>>>>>> when their input halts and correctly say NOT YES otherwise.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> well the halting problem requires them to correctly say NO, so you
>>>>>>>>>>>>>>>>>>>> haven't solved it
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> All decision problem instances of program/input such that both
>>>>>>>>>>>>>>>>>>> yes and no are the wrong answer toss out the input as invalid.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> all decision problems are defined so that all instances are valid or
>>>>>>>>>>>>>>>>>> else they are not defined properly
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Not in the case of Russell's Paradox.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> And now we are back to: Every Turing machine and input pair defines an
>>>>>>>>>>>>>>>> execution sequence. Every sequence is either finite or infinite.
>>>>>>>>>>>>>>>> Therefore it is well-defined and there is no paradox.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Can you show me a Turing machine that specifies a sequence of
>>>>>>>>>>>>>>>> configurations that is not finite or infinite?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> When we construe every yes/no question that cannot possibly
>>>>>>>>>>>>>>> have a correct yes/no answer as an incorrect question
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> then we must correspondingly construe every decider/input
>>>>>>>>>>>>>>> pair that has no correct yes/no answer as invalid input.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Apparently the answer is "no" as no such Turing machine is shown
>>>>>>>>>>>>>> or mentioned above.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> *Every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer*
>>>>>>>>>>>>> This proves that there is something wrong with the question.
>>>>>>>>>>>>
>>>>>>>>>>>> No, it doesn't. It proves that there is something wrong with
>>>>>>>>>>>> every implementation of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩. Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets
>>>>>>>>>>>> the wrong answer the other answer is right.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> *Your reasoning is just like this reasoning*
>>>>>>>>>>> "This sentence is not true." is not true and that makes it true.
>>>>>>>>>>> You are stopping right there and not seeing the infinite cycle.
>>>>>>>>>>>
>>>>>>>>>>> *Please see my new post for a complete elaboration of this*
>>>>>>>>>>> [Proving my 2004 claim that some decider/input pairs are incorrect questions]
>>>>>>>>>>
>>>>>>>>>> Whenever Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ gets the wrong answer the other answer is right.
>>>>>>>>>
>>>>>>>>> *You have already affirmed otherwise*
>>>>>>>>>
>>>>>>>>> On 3/13/2024 11:45 AM, Mikko wrote:
>>>>>>>>>  > On 2024-03-12 15:31:58 +0000, olcott said:
>>>>>>>>>  >> *Formalized*
>>>>>>>>>  >> ∀ H ∈ Turing_Machine_Deciders
>>>>>>>>>  >> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>>>  >> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>>  >
>>>>>>>>>  > Yes, this is one way to state Linz' conclusion.
>>>>>>>>
>>>>>>>> That says nothing about Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ nor about correctness of
>>>>>>>> the other answer. Therefore "otherwise" is not affirmed there.
>>>>>>>>
>>>>>>>
>>>>>>> *The conventional spec is proved to be wrong for these instances*
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn     // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>>>> ∀Ĥ.H (Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ != Halts(⟨Ĥ⟩, ⟨Ĥ⟩))
>>>>>>
>>>>>> Nice to see that you don't disagree with my observation about your mistake.
>>>>>>
>>>>>
>>>>> Best selling author of Theory of Computation textbooks:
>>>>> *Introduction To The Theory Of Computation 3RD, by sipser*
>>>>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>>>>
>>>>> Date 10/13/2022 11:29:23 AM
>>>>> *MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
>>>>> (He has neither reviewed nor agreed to anything else in this paper)
>>>>> (a) If simulating halt decider H correctly simulates its input D until
>>>>> H correctly determines that its simulated D would never stop running
>>>>> unless aborted then
>>>>> (b) H can abort its simulation of D and correctly report that D
>>>>> specifies a non-halting sequence of configurations.
>>>>>
>>>>> *When we apply the abort criteria* (elaborated above)
>>>>> Will you halt if you never abort your simulation?
>>>>> *Then H(D,D) is proven to meet this criteria*
>>>>>
>>>>> This same thing applies to: H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn is correct according to the above criteria.
>>>>>
>>>>> *Lets stick with H(D,D)==0 is correct according to the above criteria*
>>>>> until we have mutual agreement on that because:
>>>>>
>>>>> I cannot afford to tolerate the [change the subject]
>>>>> form of rebuttal that wasted 15 years with Ben Bacarisse.
>>>>
>>>> Anyway it is nice to see that you agree even if that may be
>>>> a change of subject.
>>>>
>>>
>>> I have no idea what you are saying.
>>>
>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>> that H correctly determined that it had to abort the simulation
>>> of its input to prevent the infinite execution of this input.
>>
>> What would H return if it incorrecly detected that it had to
>> abort the simulation of its input to prevent the infinite execution
>> of this input?
>>
>
> H(D,D)==0 means that it correctly detected that it must abort
> its simulation. H(D,D)==1 means that H(D,D) never returns and
> indeed this is a verified fact.


Click here to read the complete article
Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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https://news.novabbs.org/devel/article-flat.php?id=56372&group=comp.theory#56372

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From: news@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 17:17:34 +0100
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 by: immibis - Mon, 18 Mar 2024 16:17 UTC

On 18/03/24 06:11, olcott wrote:
> On 3/17/2024 11:45 PM, immibis wrote:
>> On 18/03/24 05:23, olcott wrote:
>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>> On 18/03/24 00:43, olcott wrote:
>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 3
>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>
>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>> abort the simulation of its input to prevent the infinite execution
>>>>>>>> of this input?
>>>>>>>>
>>>>>>>
>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>> indeed this is a verified fact.
>>>>>>>
>>>>>>
>>>>>> No idea where you got that from.
>>>>>>
>>>>>
>>>>> *I did not explain that clearly enough*
>>>>> The needing to abort criteria is the same as the halt criteria
>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>
>>>>>> H(D,D) means that D(D) will halt.
>>>>>>
>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria means
>>>>>> that H actually needed to abort it simulation
>>>>>>
>>>>> Yes
>>>>>
>>>>>> The meaning of that was aggreed to in:
>>>>>>
>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>  >> To me, for H to NEED to abort its simulation, that means that
>>>>>> when
>>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>>> not halt.
>>>>>>  >>
>>>>>>  > Yes that is correct.
>>>>>>  >
>>>>>>
>>>>>> So, since D(D) will halt,
>>>>> That is not the behavior that H(D,D) actually sees.
>>>>
>>>> H(D,D) does not see anything of the future. It only sees the past.
>>>> It has to predict the future, and *it predicts wrong.*
>>>
>>> H(D,D) has the inductive basis to see that it must abort its simulation.
>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>
>> Does it have the inductive basis to prove that the direct execution of
>> D(D) does not halt?
>
> That is hidden from it.

So how does it prove that the direct execution of D(D) does not halt?

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 18:15:22 -0500
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 by: olcott - Mon, 18 Mar 2024 23:15 UTC

On 3/18/2024 11:17 AM, immibis wrote:
> On 18/03/24 06:11, olcott wrote:
>> On 3/17/2024 11:45 PM, immibis wrote:
>>> On 18/03/24 05:23, olcott wrote:
>>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>>> On 18/03/24 00:43, olcott wrote:
>>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 3
>>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>>
>>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>>> abort the simulation of its input to prevent the infinite
>>>>>>>>> execution
>>>>>>>>> of this input?
>>>>>>>>>
>>>>>>>>
>>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>>> indeed this is a verified fact.
>>>>>>>>
>>>>>>>
>>>>>>> No idea where you got that from.
>>>>>>>
>>>>>>
>>>>>> *I did not explain that clearly enough*
>>>>>> The needing to abort criteria is the same as the halt criteria
>>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>>
>>>>>>> H(D,D) means that D(D) will halt.
>>>>>>>
>>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria
>>>>>>> means that H actually needed to abort it simulation
>>>>>>>
>>>>>> Yes
>>>>>>
>>>>>>> The meaning of that was aggreed to in:
>>>>>>>
>>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>>  >> To me, for H to NEED to abort its simulation, that means that
>>>>>>> when
>>>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>>>> not halt.
>>>>>>>  >>
>>>>>>>  > Yes that is correct.
>>>>>>>  >
>>>>>>>
>>>>>>> So, since D(D) will halt,
>>>>>> That is not the behavior that H(D,D) actually sees.
>>>>>
>>>>> H(D,D) does not see anything of the future. It only sees the past.
>>>>> It has to predict the future, and *it predicts wrong.*
>>>>
>>>> H(D,D) has the inductive basis to see that it must abort its
>>>> simulation.
>>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>>
>>> Does it have the inductive basis to prove that the direct execution
>>> of D(D) does not halt?
>>
>> That is hidden from it.
>
> So how does it prove that the direct execution of D(D) does not halt?

It is not allowed to do that because that is false from
its point of view.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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From: news@immibis.com (immibis)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
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 by: immibis - Mon, 18 Mar 2024 23:34 UTC

On 19/03/24 00:15, olcott wrote:
> On 3/18/2024 11:17 AM, immibis wrote:
>> On 18/03/24 06:11, olcott wrote:
>>> On 3/17/2024 11:45 PM, immibis wrote:
>>>> On 18/03/24 05:23, olcott wrote:
>>>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>>>> On 18/03/24 00:43, olcott wrote:
>>>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3
>>>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>>>
>>>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>>>> abort the simulation of its input to prevent the infinite
>>>>>>>>>> execution
>>>>>>>>>> of this input?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>>>> indeed this is a verified fact.
>>>>>>>>>
>>>>>>>>
>>>>>>>> No idea where you got that from.
>>>>>>>>
>>>>>>>
>>>>>>> *I did not explain that clearly enough*
>>>>>>> The needing to abort criteria is the same as the halt criteria
>>>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>>>
>>>>>>>> H(D,D) means that D(D) will halt.
>>>>>>>>
>>>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria
>>>>>>>> means that H actually needed to abort it simulation
>>>>>>>>
>>>>>>> Yes
>>>>>>>
>>>>>>>> The meaning of that was aggreed to in:
>>>>>>>>
>>>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>>>  >> To me, for H to NEED to abort its simulation, that means
>>>>>>>> that when
>>>>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>>>>> not halt.
>>>>>>>>  >>
>>>>>>>>  > Yes that is correct.
>>>>>>>>  >
>>>>>>>>
>>>>>>>> So, since D(D) will halt,
>>>>>>> That is not the behavior that H(D,D) actually sees.
>>>>>>
>>>>>> H(D,D) does not see anything of the future. It only sees the past.
>>>>>> It has to predict the future, and *it predicts wrong.*
>>>>>
>>>>> H(D,D) has the inductive basis to see that it must abort its
>>>>> simulation.
>>>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>>>
>>>> Does it have the inductive basis to prove that the direct execution
>>>> of D(D) does not halt?
>>>
>>> That is hidden from it.
>>
>> So how does it prove that the direct execution of D(D) does not halt?
>
> It is not allowed to do that because that is false from
> its point of view.

If the direct execution of D(D) halts from its point of view, then why
does it report that it does not halt?

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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https://news.novabbs.org/devel/article-flat.php?id=56430&group=comp.theory#56430

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From: polcott2@gmail.com (olcott)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 18:45:09 -0500
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 by: olcott - Mon, 18 Mar 2024 23:45 UTC

On 3/18/2024 6:34 PM, immibis wrote:
> On 19/03/24 00:15, olcott wrote:
>> On 3/18/2024 11:17 AM, immibis wrote:
>>> On 18/03/24 06:11, olcott wrote:
>>>> On 3/17/2024 11:45 PM, immibis wrote:
>>>>> On 18/03/24 05:23, olcott wrote:
>>>>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>>>>> On 18/03/24 00:43, olcott wrote:
>>>>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>>>>
>>>>>>>>>>>> On 3
>>>>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>>>>
>>>>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>>>>> abort the simulation of its input to prevent the infinite
>>>>>>>>>>> execution
>>>>>>>>>>> of this input?
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>>>>> indeed this is a verified fact.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> No idea where you got that from.
>>>>>>>>>
>>>>>>>>
>>>>>>>> *I did not explain that clearly enough*
>>>>>>>> The needing to abort criteria is the same as the halt criteria
>>>>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>>>>
>>>>>>>>> H(D,D) means that D(D) will halt.
>>>>>>>>>
>>>>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria
>>>>>>>>> means that H actually needed to abort it simulation
>>>>>>>>>
>>>>>>>> Yes
>>>>>>>>
>>>>>>>>> The meaning of that was aggreed to in:
>>>>>>>>>
>>>>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>>>>  >> To me, for H to NEED to abort its simulation, that means
>>>>>>>>> that when
>>>>>>>>>  >> giving the input to a correct simulator, that simulator
>>>>>>>>> will not halt.
>>>>>>>>>  >>
>>>>>>>>>  > Yes that is correct.
>>>>>>>>>  >
>>>>>>>>>
>>>>>>>>> So, since D(D) will halt,
>>>>>>>> That is not the behavior that H(D,D) actually sees.
>>>>>>>
>>>>>>> H(D,D) does not see anything of the future. It only sees the
>>>>>>> past. It has to predict the future, and *it predicts wrong.*
>>>>>>
>>>>>> H(D,D) has the inductive basis to see that it must abort its
>>>>>> simulation.
>>>>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>>>>
>>>>> Does it have the inductive basis to prove that the direct execution
>>>>> of D(D) does not halt?
>>>>
>>>> That is hidden from it.
>>>
>>> So how does it prove that the direct execution of D(D) does not halt?
>>
>> It is not allowed to do that because that is false from
>> its point of view.
>
> If the direct execution of D(D) halts from its point of view, then why
> does it report that it does not halt?
>

D(D) only ever halts because H(D,D) correctly sees that
it must intervene and force its own D(D) to stop running.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: H(D,D)==0 is correct when reports on the actual behavior that it sees

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https://news.novabbs.org/devel/article-flat.php?id=56432&group=comp.theory#56432

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From: richard@damon-family.org (Richard Damon)
Newsgroups: comp.theory
Subject: Re: H(D,D)==0 is correct when reports on the actual behavior that it
sees
Date: Mon, 18 Mar 2024 17:21:28 -0700
Organization: i2pn2 (i2pn.org)
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 by: Richard Damon - Tue, 19 Mar 2024 00:21 UTC

On 3/18/24 4:15 PM, olcott wrote:
> On 3/18/2024 11:17 AM, immibis wrote:
>> On 18/03/24 06:11, olcott wrote:
>>> On 3/17/2024 11:45 PM, immibis wrote:
>>>> On 18/03/24 05:23, olcott wrote:
>>>>> On 3/17/2024 7:25 PM, immibis wrote:
>>>>>> On 18/03/24 00:43, olcott wrote:
>>>>>>> On 3/17/2024 2:51 PM, Richard Damon wrote:
>>>>>>>> On 3/17/24 9:38 AM, olcott wrote:
>>>>>>>>> On 3/17/2024 11:02 AM, Mikko wrote:
>>>>>>>>>> On 2024-03-16 14:46:42 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 3
>>>>>>>>>>> I am saying that H(D,D)==0 is correct in that H(D,D)==0 means
>>>>>>>>>>> that H correctly determined that it had to abort the simulation
>>>>>>>>>>> of its input to prevent the infinite execution of this input.
>>>>>>>>>>
>>>>>>>>>> What would H return if it incorrecly detected that it had to
>>>>>>>>>> abort the simulation of its input to prevent the infinite
>>>>>>>>>> execution
>>>>>>>>>> of this input?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> H(D,D)==0 means that it correctly detected that it must abort
>>>>>>>>> its simulation. H(D,D)==1 means that H(D,D) never returns and
>>>>>>>>> indeed this is a verified fact.
>>>>>>>>>
>>>>>>>>
>>>>>>>> No idea where you got that from.
>>>>>>>>
>>>>>>>
>>>>>>> *I did not explain that clearly enough*
>>>>>>> The needing to abort criteria is the same as the halt criteria
>>>>>>> when H(D,D) is reporting on the behavior that it actually sees.
>>>>>>>
>>>>>>>> H(D,D) means that D(D) will halt.
>>>>>>>>
>>>>>>>> H(D,D) returning 0, even for your "Needed to Abort" criteria
>>>>>>>> means that H actually needed to abort it simulation
>>>>>>>>
>>>>>>> Yes
>>>>>>>
>>>>>>>> The meaning of that was aggreed to in:
>>>>>>>>
>>>>>>>> On 3/17/24 6:11 AM, olcott wrote:
>>>>>>>>  > On 3/17/2024 12:22 AM, Richard Damon wrote:
>>>>>>>>  >> To me, for H to NEED to abort its simulation, that means
>>>>>>>> that when
>>>>>>>>  >> giving the input to a correct simulator, that simulator will
>>>>>>>> not halt.
>>>>>>>>  >>
>>>>>>>>  > Yes that is correct.
>>>>>>>>  >
>>>>>>>>
>>>>>>>> So, since D(D) will halt,
>>>>>>> That is not the behavior that H(D,D) actually sees.
>>>>>>
>>>>>> H(D,D) does not see anything of the future. It only sees the past.
>>>>>> It has to predict the future, and *it predicts wrong.*
>>>>>
>>>>> H(D,D) has the inductive basis to see that it must abort its
>>>>> simulation.
>>>>> This means that the behavior that H(D,D) sees <is> non-halting.
>>>>>
>>>> Does it have the inductive basis to prove that the direct execution
>>>> of D(D) does not halt?
>>>
>>> That is hidden from it.
>>
>> So how does it prove that the direct execution of D(D) does not halt?
>
> It is not allowed to do that because that is false from
> its point of view.
>

So, you admit it just lies to itself.

That seems to be the nature of most of your logic.

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