how to prove that (P⊃Q)≡(¬Q⊃¬P) ( P ⊃ Q ) ≡ ( ¬ Q ⊃ ¬ P ) is disallowed in intuitionistic logic?

Question

this is what I've tried;

define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by 0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then

1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )

can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?

Answer 1

I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.

Note that intuitionistically, as classically, $(A \supset B) \supset (\lnot B \supset \lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(\lnot B \supset \lnot A) \supset (A \supset B)$.

I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:

• Since $1\Vdash B$, we have $0\not\Vdash\lnot B$ and $1\not\Vdash\lnot B$, so $0\Vdash (\lnot B \supset \lnot A)$.
• On the other hand, $0\Vdash A$ but $0\not\Vdash B$, so $0\not\Vdash (A \supset B)$.

Therefore, $0\not\Vdash (\lnot B \supset \lnot A) \supset (A \supset B)$, and so $0\not\Vdash (A \supset B) \equiv (\lnot B \supset \lnot A)$.

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You also seem to consider $(B \supset A) \equiv (\lnot B \supset \lnot A)$, but this is even classically false!