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Re: [Coq-Club] how to prove ~(A/\B)->(~A\/~B)


chronological Thread 
  • 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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