# Deductive Logic : If A -> B it can be deduced neg(A) -> neg(B)

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,

• 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). – Ian Aug 26 '16 at 1:21
• 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 $2$. – JMoravitz Aug 26 '16 at 1:26
• See my answer to a similar question at math.stackexchange.com/questions/1551320/… – Dan Christensen Aug 26 '16 at 4:34

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