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I need to prove, using the Hilbert System: $(\neg A → \neg B) → ((\neg B → A) → B)$

The Axioms are:

  1. $A→(B→ A)$
  2. $(A→(B → C)) → ((A→ B)→ (A→ C))$

  3. $(\neg A → \neg B) → (B → A)$

    MP: $\dfrac{A, A→ B}{B}$

Any idea?

Graham Kemp
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Aron Wolf
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2 Answers2

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I'm not 100% certain, but I think that in a calculus with these axioms the fact that the substitution $(\lnot T \to \lnot F) \to ((\lnot F \to T) \to F)$ (where $T$ stands for True and $F$ for False) results in the formula being invalidated means that it cannot be deduced.

Isti115
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  • Indeed. $(\neg A\to\neg B)\to((\neg B\to A)\to B)$ is not a tautology. However, $(\neg A\to\neg B)\to((\neg B\to A)\to A)$ is, so that is possibly what the OP was trying to prove. . – Graham Kemp Apr 28 '20 at 23:03
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I'm stuck with the last step, I don't know if I can leave it as it:

4.$(\neg A\rightarrow \neg B)$hypothesis

5.$(B\rightarrow A)$ MP(3,4)

6.$(B\rightarrow A)\rightarrow (\neg B \rightarrow (B\rightarrow A) )$ applying Axiom 1

7.$(\neg B \rightarrow (B\rightarrow A) )$ MP(1,4)

8.$(\neg B \rightarrow (B\rightarrow A) )\rightarrow ((\neg B \rightarrow B) \rightarrow (\neg B \rightarrow A)) $ Applying axiom 2

9.$(\neg B \rightarrow B) \rightarrow (\neg B \rightarrow A) $ MP(7,8)

10.$(\neg B \rightarrow A) $ because it is a tautology

... Stuck

Revolucion for Monica
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