How to prove that the following three tautologies of classical logic are disallowed in intuitionistic logic?

Asked Oct 17 '18 at 10:31

Active Oct 17 '18 at 13:19

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I need to prove that the following three tautologies are disallowed in intuitionistic logic.

The tautologies are:

1-double negation $\neg \neg P\equiv P$

2-Law of excluded middle $P \vee\neg P$

3-Contraposition $(P \supset Q)\equiv (\neg Q \supset \neg P) $

can somebody explain how to do this?

edited Oct 17 '18 at 13:19

Graham Kemp

asked Oct 17 '18 at 10:31

KatherinD

3

Use Kripke semantics to produce suitable counter-models. – Mauro ALLEGRANZA Oct 17 '18 at 10:43
Example. – Mauro ALLEGRANZA Oct 17 '18 at 10:54
3

See also Kripke semantics for intuitionistic logic. – Mauro ALLEGRANZA Oct 17 '18 at 11:15
1

For an answer to your questions 1 and 2, see here. – Taroccoesbrocco Oct 17 '18 at 13:02
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Note that $P \supset \lnot \lnot P$ and $(P \supset Q) \supset (\lnot Q \supset \lnot P)$ are provable in intuitionistic logic. Only their converses are not provable in intuiti0nistic logic. – Taroccoesbrocco Oct 17 '18 at 13:04
@Taroccoesbrocco if I use the Kripke model here: https://math.stackexchange.com/questions/2622456/prove-the-undecidability-of-a-formula/2622484#2622484 and then say $0 ⊮ (A \to B) \equiv (\neg A \to \neg B)$ will my proof be correct? – KatherinD Oct 17 '18 at 15:30
@KatherinD - Good idea but it is slightly more complicated: consider the Kripke model over the domain ${0,1}$, as the one defined here. You should define the binary relation $\Vdash$ from ${0,1}$ to the set of propositional variables ${A,B}$ so that, according to the definition of intuitionistic Kripke model for propositional logic, $0 \not\Vdash (\lnot B \to \lnot A) \to (A \to B)$. – Taroccoesbrocco Oct 17 '18 at 20:14

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