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I am having a hard time proving the following $$ (a \to b) \vdash (\lnot a \to \lnot b) $$

I followed the book advice and first proved that $ (a \to b) \vdash (\lnot \lnot a \to \lnot \lnot b) $ using the Deduction Theorem but I am stuck afterwards. I have been trying to use every Axioms/Modus/Rules I know but always end up with unsatisfactory result.

A hint would be appreciated,

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    $\begingroup$ That's because it's false. It is true that one can infer $\neg b \to \neg a$ from $a \to b$ though (in classical logic anyway). $\endgroup$ – Ian Aug 26 '16 at 1:21
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    $\begingroup$ For an example of why it is false: If $n$ is a multiple of $10$ then $n$ is also a multiple of $2$. This is a true statement ($n=10k\Rightarrow n=2\cdot (5k)$). However, the statement "if $n$ is not a multiple of $10$ then it is not a multiple of $2$" is false. Take $6$ for a counterexample. $6$ happens to not be a multiple of $10$ however it is indeed a multiple of $2$. $\endgroup$ – JMoravitz Aug 26 '16 at 1:26
  • $\begingroup$ See my answer to a similar question at math.stackexchange.com/questions/1551320/… $\endgroup$ – Dan Christensen Aug 26 '16 at 4:34
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If it is raining, then there are clouds out. ($a\implies b$)

If it is not raining, then there are no clouds out ($\neg a \implies \neg b$)

The first is true and the second isn't. There is something close to what you want, called contrapositive.

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