Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"?

Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"?

I am trying to understand why this does always apply, in terms of pure logic. Can you please explain it to me?

• – user122283
Commented Jun 14, 2014 at 16:00
• Have you compared their truth-tables? Commented Jun 14, 2014 at 16:00
• You could also search this site for the term contrapositive. Commented Jun 14, 2014 at 16:04
• @amWhy great idea, thank you! Commented Jun 14, 2014 at 16:05
• @HaraldHanche-Olsen Never heard of it before, thank you! Commented Jun 14, 2014 at 16:05

Suppose it's true that:

1. If you hit my glass table with a hammer, it will break.
2. My glass coffee table is not broken.

If you accept premisses 1 and 2, wouldn't you conclude from this that

3. My glass coffee table has not been hit with a hammer?

If we let $P$ represent “my glass table was hit with a hammer” and $Q$ be “my glass table is broken”, then (1) is $P\implies Q$, and (2) is $\lnot Q$.

From (1) and (2) we can conclude that the table has not been hit with a hammer; that's $\lnot P$. One way to write this is $$((P\implies Q) \land \lnot Q) \implies \lnot P$$

but is $$(P\implies Q)\implies (\lnot Q \implies \lnot P)$$ which says that if we know $P\implies Q$ (that's (1)) then if we know $\lnot Q$ (that's (2)) then we can conclude $\lnot P$ (that's (3)).

On the other hand, it is not the case that

$$(P\implies Q)\implies (\lnot P \implies \lnot Q)$$

This says that if (1) is true, and you know my table was not hit with a hammer, then you know it is not broken. But no, that's wrong, because actually it is broken because my brother-in-law got drunk and fell on it; no hammer was involved.

It is given that $P\to Q$. By material implication, $\neg P\vee Q$. By commutativity, $Q\vee\neg P$. Again, by material implication, $\neg Q\to \neg P$.

• By "material implication" he means: the definition of $P\to Q$ is $\neg P\vee Q$. Since you said "pure logic" this is the thing to use. Commented Jun 14, 2014 at 16:06
• But this relationship holds in Lukasiewicz three-valued logic where (P→Q) is not logically equivalent to (¬P∨Q). Commented Jun 15, 2014 at 13:02

$$\begin{array} {|c|c|c|c|} \hline P & Q & P \Rightarrow Q & \lnot Q \Rightarrow \lnot P \\ \hline \top & \top & \top & \top \\ \top & \bot & \bot & \bot \\ \bot & \top & \top & \top \\ \bot & \bot & \top & \top \\ \hline \end{array}$$

Equal.

You can check it simply by writing up every situation in a truth-table and see that they produce the same results for any choices of $P$ and $Q$.

$P \Rightarrow Q$ is only false when $P$ is true and $Q$ is false.

Likewise $\neg Q \Rightarrow \neg P$ is only false when $\neg Q$ is true and $\neg P$ is false, thus when $Q$ is false and $P$ is true.