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How do I show the following by producing formal or informal proofs?

  1. ⊢ (¬A → ¬B) → (B → A )

  2. ⊢ ¬B → (B → A)

I can use the Modus ponus (MP) rule, and deduction theorem (DT). And I have these 3 axioms:

  1. α → (β → α) --- (A1)

  2. (α → (β → γ)) → ((α → β) → (α → γ)) --- (A2)

  3. (¬β → ¬α) → ((¬β → α) → β) --- (A3)

Thank you!

Djy
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2 Answers2

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For the first one:

$1\ \neg A \to \neg B \ Premise$

$2 \ B \ Premise$

$3 \ B \to (\neg A \to B) \ Axiom \ 1$

$4 \ \neg A \to B \ MP \ 2,3$

$5 \ (\neg A \to \neg B) \to ((\neg A \to B) \to A \ Axiom \ 3$

$6 \ (\neg A \to B) \to A \ MP \ 1,5$

$7 \ A \ MP \ 4,6$

This shows that $\{ \neg A \to \neg B, B \vdash A\}$

So, with two applications of DT, you're there.

Bram28
  • 103,721
  • thank you very much! Guess i need to practice more questions to get the hang of it – Djy Sep 24 '19 at 11:20
  • Why was this answer downvoted?... That fact on its own inspired me to read the answer. That reading on its own gave me a reason to upvote. – drhab Oct 16 '19 at 07:58
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Hint:

1) With DT we firstly conclude that it is enough to prove $\left\{ \neg A\to\neg B\right\} \vdash B\to A$ and based on that secondly that it is enough to prove $\left\{ \neg A\to\neg B,B\right\} \vdash A$.

2) With DT we firstly conclude that it is enough to prove $\left\{ \neg B\right\} \vdash B\to A$ and based on that secondly that it is enough to prove $\left\{ \neg B,B\right\} \vdash A$.

Now give that a try and tell us where you get stuck in the sequel.

drhab
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