On 4/22/2023 1:01 PM, Richard Damon wrote:
> On 4/22/23 1:13 PM, olcott wrote:
>> On 4/22/2023 11:56 AM, Richard Damon wrote:
>>> On 4/22/23 12:45 PM, olcott wrote:
>>>> On 4/22/2023 11:36 AM, Richard Damon wrote:
>>>>> On 4/22/23 12:27 PM, olcott wrote:
>>>>>> On 4/22/2023 11:12 AM, Richard Damon wrote:
>>>>>>> On 4/22/23 11:39 AM, olcott wrote:
>>>>>>>> On 4/22/2023 9:57 AM, Richard Damon wrote:
>>>>>>>>> On 4/22/23 10:48 AM, olcott wrote:
>>>>>>>>>> On 4/22/2023 9:38 AM, Richard Damon wrote:
>>>>>>>>>>> On 4/22/23 10:28 AM, olcott wrote:
>>>>>>>>>>>> On 4/22/2023 6:17 AM, Richard Damon wrote:
>>>>>>>>>>>>> On 4/21/23 11:40 PM, olcott wrote:
>>>>>>>>>>>>>> On 4/21/2023 9:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 4/21/23 9:41 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 4/21/2023 7:49 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 4/21/23 8:33 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> There exists a G such that G is logically equivalent
>>>>>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> *If we assume that there is such a G in F that means
>>>>>>>>>>>>>>>>>> that*
>>>>>>>>>>>>>>>>>> G is true means there is no sequence of inference
>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>> G is false means there is a sequence of inference
>>>>>>>>>>>>>>>>>> steps that satisfies G in F.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> *Thus the above G simply does not exist in F*
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I finally learned enough model theory to correctly link
>>>>>>>>>>>>>>>> provability to
>>>>>>>>>>>>>>>> truth in the conventional model theory way.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Doesn't seem so, you don't seem to understand the
>>>>>>>>>>>>>>> difference. You seem to confuse Truth with Knowledge.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I finally approximated {G asserts its own unprovability
>>>>>>>>>>>>>>>> in F}
>>>>>>>>>>>>>>>> using conventional math symbols in their conventional way.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Except that isn't what G is, you only think that because
>>>>>>>>>>>>>>> you can't actually understand even the outline of Godel's
>>>>>>>>>>>>>>> proof, so you take pieces out of context.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> G never asserts its own unprovability.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The statement that we now have a statement that asserts
>>>>>>>>>>>>>>> its own unprovablity, as a simplification describing a
>>>>>>>>>>>>>>> statment DERIVED from G, and that derivation happens in
>>>>>>>>>>>>>>> Meta-F, and is about what can be proven in F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Since Godel's G isn't of that form, but only can be
>>>>>>>>>>>>>>>>> used to derive a statment IN META-F that says that G is
>>>>>>>>>>>>>>>>> not provable in F, your argument says nothing about
>>>>>>>>>>>>>>>>> Godel's G.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> F ⊢ GF ↔ ¬ProvF (┌GF┐).
>>>>>>>>>>>>>>>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom
>>>>>>>>>>>>>>>> I have finally created a G that is equivalent to
>>>>>>>>>>>>>>>> Panu Raatikainen's SEP article.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Did you read that article?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Also, you don't understand what those terms mean,
>>>>>>>>>>>>>>>>> because G being true doesn't mean there is no sequence
>>>>>>>>>>>>>>>>> of inference steps that satisfies G in F, but there is
>>>>>>>>>>>>>>>>> no FINITE sequence of inference steps that satisfies G
>>>>>>>>>>>>>>>>> in F.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Because we can see that every finite or infinite
>>>>>>>>>>>>>>>> sequence in F that
>>>>>>>>>>>>>>>> satisfies the RHS of ↔ contradicts the LHS a powerful F
>>>>>>>>>>>>>>>> can infer that G
>>>>>>>>>>>>>>>> is utterly unsatisfiable even for infinite sequences in
>>>>>>>>>>>>>>>> this more
>>>>>>>>>>>>>>>> powerful F.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Nope. Show the PROOF.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You don't know HOW to do a proof, you can only do arguement.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> A proof is any sequence of steps that shows that its
>>>>>>>>>>>>>> conclusion is a
>>>>>>>>>>>>>> necessary consequence of its premises.\
>>>>>>>>>>>>>
>>>>>>>>>>>>> Boy are you wrong.
>>>>>>>>>>>>>
>>>>>>>>>>>>> A proof is a FINITE sequence of steps that shows that a
>>>>>>>>>>>>> given statement is a necessary consequence of the defined
>>>>>>>>>>>>> system.
>>>>>>>>>>>>>
>>>>>>>>>>>>> "Proof" doesn't have a  "Premise", it has a system.
>>>>>>>>>>>>>
>>>>>>>>>>>>> The statement may have conditions in it restricting when
>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> ∃G ∈ F (G ↔ (G ⊬ F))
>>>>>>>>>>>>>> There exists a G in F such that G is logically equivalent
>>>>>>>>>>>>>> to its own unprovability in F
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is true then there is no sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F making G untrue.
>>>>>>>>>>>>>
>>>>>>>>>>>>> no FINITE sequence, making G UNPROVABLE, and there IS an
>>>>>>>>>>>>> INFINITE sequence making it TRUE.
>>>>>>>>>>>>>
>>>>>>>>>>>>> This is possible.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> If G is false then there is a sequence of inference steps
>>>>>>>>>>>>>> that satisfies G in F making G true.
>>>>>>>>>>>>>
>>>>>>>>>>>>> If G is false, then there is a finite sequence proving G,
>>>>>>>>>>>>> which forces G to be true, thus this is a contradiction.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Because the RHS of ↔ contradicts the LHS there is no such
>>>>>>>>>>>>>> G in F.
>>>>>>>>>>>>>> Thus the above G simply does not exist in F.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nope, because we can have an infinite sequence that isn't
>>>>>>>>>>>>> finite, G can be True but not Provable.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> If G is false and ↔ is true this makes the RHS false which
>>>>>>>>>>>> negates the RHS making it say (G ⊢ F) which makes G true in F.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Right, G can't be false, but it can be True.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Thus ↔ cannot be satisfied thus no such G exists in F.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Why do you say that?
>>>>>>>>>
>>>>>>>>> I don't think you know what you terms mean.
>>>>>>>>>
>>>>>>>>> There exists a G in F such that G is true if and only if G is
>>>>>>>>> Unprovable.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Logical equality
>>>>>>>> p q p ↔ q
>>>>>>>> T T   T // G is true if and only if G is Unprovable.
>>>>>>>> T F   F //
>>>>>>>> F T   F //
>>>>>>>> F F   T // G is false if and only if G is Provable.
>>>>>>>> https://en.wikipedia.org/wiki/Truth_table#Logical_equality
>>>>>>>>
>>>>>>>> Row(1) There exists a G in F such that G is true if and only if
>>>>>>>> G is
>>>>>>>> unprovable in F making G unsatisfied thus untrue in F.
>>>>>>>>
>>>>>>>> Row(4) There exists a G in F such that G is false if and only if
>>>>>>>> G is
>>>>>>>> provable in F making G satisfied thus true in F.
>>>>>>>>
>>>>>>>> If either Row(1) or Row(4) are unsatisfied then ↔ is false.
>>>>>>>
>>>>>>> But if neither row values can ACTUALLY EXIST, then the equality
>>>>>>> is true.
>>>>>>>
>>>>>> If either Row(1) or Row(4) cannot have the same value for p and q
>>>>>> (for whatever reason) then ↔ is unsatisfied and no such G exists
>>>>>> in F.
>>>>>>
>>>>> So, you don't understand how truth tables work.
>>>>>
>>>>> You don't need to have all the rows with true being possible, you
>>>>> need all the rows that are possible to be True.
>>>>>
>>>>
>>>> To the best of my knowledge
>>>> ↔ is also known as logical equivalence meaning that the LHS and the RHS
>>>> must always have the same truth value or ↔ is not true.
>>>>
>>>
>>> Right, and for that statement, the actual G found in F, the ONLY
>>> values that happen is G is ALWAYS true, an Unprovable is always true.
>>>
>>> Thus the equivalence is always true.
>> I don't think that is the way that it works.
>> We must assume that the RHS is true and see how that effects the LHS
>> We must assume that the RHS is false and see how that effects the LHS
>> ((True(RHS) → True(LHS)) ∧ (False(RHS) → False(LHS))) ≡ (RHS ↔ LHS)
>> False(RHS) → True(LHS) refutes (RHS ↔ LHS)
>>
>
> Nope, that isn't how it works.
>
> Can you show me something that says that is how it works?
>

I tried and in the first search all of the articles seemed to dodge
rather than address this point. I have always understood ↔ to mean that
the LHS and the RHS must always have the same Boolean value.

What sources do you have that directly contradict this?

p ↔ q would seem to mean ((p → q) ∧ (q → p))

Logical implication
p---q---p ⇒ q
T---T-----T // (p ⇒ q) ∧ (q ⇒ p) thus (p ↔ q)
T---F-----F // ¬(p ⇒ q) ∧ (q ⇒ p) thus ¬(p ↔ q)
F---T-----T // (p ⇒ q) ∧ ¬(q ⇒ p) thus ¬(p ↔ q)
F---F-----T // (p ⇒ q) ∧ (q ⇒ p) thus (p ↔ q)

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer