Re: [isabelle] Infinitely recursive lambda expression or not?

This document seems a good introduction on LCF vs HOL and the idea of getting one with the other:
(the one I will read)

For HOL and LCF as an Isabelle theory, this one:

Quote of the abstract of the former:

The LCF theorem prover provides a logic of domain theory and is useful for reasoning about nontermination, general recursive definitions and infinite-valued datatypes like lazy lists. Because of the continued presence of bottom (undefined) elements, it is clumsy for reasoning about finite-valued datatypes and strict func- tions. By contrast, the HOL theorem prover provides a logic of set theory (without a notion of undefinedness) and supports reasoning about finite-valued datatypes
and primitive recursive functions well.

I was not suspecting this kind of difference between HOL and LCF provers and their logics.

On Fri, 27 Dec 2013 18:50:26 +0100, Lawrence Paulson <lp15 at> wrote:

You could try Isabelle/HOLCF, but I’m not sure how well documented and supported it is.

Larry Paulson

On 27 Dec 2013, at 17:47, Yannick Duchêne (Hibou57) <yannick_duchene at> wrote:

Le Fri, 27 Dec 2013 18:33:49 +0100, Lawrence Paulson <lp15 at> a écrit:

From a computational point of view, "f a = (a ∨ f a)” must be regarded as undefined, because the recursion is not well-founded. There are logics where you could then prove that f(True)=True and f(False)=undefined.

Larry Paulson

Interesting. Just out of curiosity and learn more (really not to use it), what are the names of these logics?

“Syntactic sugar causes cancer of the semi-colons.” [1]
“Structured Programming supports the law of the excluded muddle.” [1]
[1]: Epigrams on Programming — Alan J. — P. Yale University

Yannick Duchêne

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