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devel / comp.theory / Re: Favorite computation formalism? (The "Best Test" for CS)

SubjectAuthor
* Favorite computation formalism? (The "Best Test" for CS)Rock Brentwood
`- Favorite computation formalism? (The "Best Test" for CS)Andy Walker

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Re: Favorite computation formalism? (The "Best Test" for CS)

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Subject: Re: Favorite computation formalism? (The "Best Test" for CS)
From: rockbrentwood@gmail.com (Rock Brentwood)
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 by: Rock Brentwood - Tue, 15 Aug 2023 22:57 UTC

On Wednesday, October 19, 2022 at 8:10:31 PM UTC-5, Ben Bacarisse wrote:
> Hey! A topical post!
> > Lambda Calculus with infinitary terms (and the conditional operator).
> > Notation: x = A, B means (lambda x B) A
> Curious notation. Is it your own? Presumably
>
> x=A, y=B, C
>
> means
>
> (lambda x ((lambda y C) B)) A

.... which is all it can mean, since x = A, y = B, C means x = A, (y = B, C), and x = a, f (which is often written as "let x = a in f") stands for (λxf)a ... which is standard notation, in case you forgot. It's just a more natural abbreviated form of it, without the "let" or "in" as you (presumably) already know.

So x = A, (y = B, C) mean x = A, ((λyC) B) means λx((λyC)B)A.
> rather than the more usual

No. There is no "more usual".
let x = A in let y = B in C means λx((λyC)B)A.

> This raises the question of how you write
>
> (lambda x (lambda y C)) A B

the same way as you always do:
(λx(λyC))BA
or equivalently as
(x = B, λyC)A
which (in the less abbreviated form) is
(let x = B in λyC)A

Perhaps you missed the part up at the top where it said:
> Lambda Calculus *with* infinitary terms
that means *extension of* not *replacement for*.

> I'm not seeing the advantages of this notation.

An upward extension of a given language is *always* more useful or "advantageous" than the language it extends - almost by definition of "useful" ... provided the language, itself, has the same meaning (and parsing) in the extension as it does in the original ... since it includes the original language, itself, and adds more. That's universally true, regardless of the context or situation and independently of how "useful" or "advantage" is defined.

Re: Favorite computation formalism? (The "Best Test" for CS)

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From: anw@cuboid.co.uk (Andy Walker)
Newsgroups: comp.theory
Subject: Re: Favorite computation formalism? (The "Best Test" for CS)
Date: Wed, 16 Aug 2023 12:03:06 +0100
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 by: Andy Walker - Wed, 16 Aug 2023 11:03 UTC

On 15/08/2023 23:57, Rock Brentwood wrote:
> An upward extension of a given language is *always* more useful or
> "advantageous" than the language it extends - almost by definition of
> "useful" ... provided the language, itself, has the same meaning (and
> parsing) in the extension as it does in the original ... since it
> includes the original language, itself, and adds more. That's
> universally true, regardless of the context or situation and
> independently of how "useful" or "advantage" is defined.

Not so. Certainly an upwards extension has extra uses and/or
advantages; but it comes with extra costs and disadvantages, so is
not necessarily overall more useful or advantageous. The extended
language is typically harder to learn, to write documentation for,
to compile, to debug, to maintain, .... All too often, what started
as a simple one-person project that was Really Useful turns into a
huge monolith that only a major company can handle.

The trouble is, perhaps, that "just one more" feature always
seems better, but over a period that too often grows into a hundred
new features, each of which someone somewhere finds useful but the
vast majority of which are useless to the vast majority of users.
There are examples of such bloatware all over computing, mathematics
and the Real World.

--
Andy Walker, Nottingham.
Andy's music pages: www.cuboid.me.uk/andy/Music
Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Sinding

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