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- From: Benjamin Werner <benjamin.werner AT inria.fr>
- To: Stefano Berardi <stefano AT di.unito.it>
- Cc: Wan Hai <wan.whyhigh AT gmail.com>, Coq Club <coq-club AT pauillac.inria.fr>
- Subject: Re: [Coq-Club] how to prove ~(A/\B)->(~A\/~B)
- Date: Tue, 27 Nov 2007 17:52:23 +0100
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A more elementary proof is showing that ~(A/\B)->(~A\/~B)
can be derived from the scheme (A->B)\/(B->A) (for any A and B).
This scheme is well-known to be weaker than EM. For instance,
it is valid in any linear (totally ordered) Heyting Algebra.
However, I am indeed a little worried that more and more coq-club
messages look like students looking for solutions to their homeworks.
Should we have a policy of answering only to people who are either
known or have presented themselves ?
Best,
Benjamin
Le 27 nov. 07 à 17:31, Stefano Berardi a écrit :
Wan Hai wrote:
Dear All.
I have two questions:
1) Can "~(A/\B)->(~A\/~B)" be proved without using classical logic,
such as A\/~A ? This can be easily proved using "forall A:Prop, A\/ ~A"
2) Does "~(A/\B)->(~A\/~B)" have the same power of "A \/ ~A"? Or say,
(forall A B:Prop, ~(A/\B)->(~A\/~B)) -> (forall A:Prop, A\/~A) ?
For question 1, I think the answer is no. For question 2, I have
worked on it for several hours, but haven't worked it out.
Can someone give me a clue?
Thanks a lot!
related questions are adressed in a LICS2004 paper "An Arithmetical Hierarchy of the Law of Excluded Middle and Related Principles". The answer is that (forall A B:Prop, ~(A/\B)->(~A\/~B)) is an intermediate principle between intuitionistic and classical logic.
(1) Indeed, (forall A B:Prop, ~(A/\B)->(~A\/~B)) is false in the standard realization model of second order arithmetic, this is why (1) is false.
(2) (forall A B:Prop, ~(A/\B)->(~A\/~B)) is true in the realization model in which a map f can depend on A,B:Prop, but all maps of the model are bounded by some recursive map. In this model, however,
Exists x.A(x,y) v ~Exists x.A(x,y)
is false, because in general there is no bound recursive in y on some x such that A(x,y). Therefore
(forall A B:Prop, ~(A/\B)->(~A\/~B)) -> (forall A:Prop, A\/~A)
is false in this model, hence not provable intuitionistic logic.
Stefano
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- [Coq-Club] how to prove ~(A/\B)->(~A\/~B), Wan Hai
- Re: [Coq-Club] how to prove ~(A/\B)->(~A\/~B),
Stefano Berardi
- Re: [Coq-Club] how to prove ~(A/\B)->(~A\/~B), Benjamin Werner
- Re: [Coq-Club] how to prove ~(A/\B)->(~A\/~B),
Stefano Berardi
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