if $P \implies Q$ why does $\bar{Q} \implies \bar{P}$ I recently started reading the book: A concise introduction to Pure Mathematics. I'm enjoying it so far, but unfortunately I've run into something I don't quite understand. 
Let P,Q be mathematical statements such that 
$$ P \implies Q $$ it is then stated that if this is the case, then $$ \bar{Q} \implies \bar{P} $$
From the way they describe it in the book, they make it seem like the latter implication is quite obvious. However, I don't understand why this is...
 A: These two compound statements $P\implies Q$ and $\bar{Q}\implies \bar{P}$ are in fact equivalent. One easy way to see this is to notice that they have identical truth tables. Namely:


*

*$P=\mbox{true}$, $Q=\mbox{true}$: both compound statements are true.

*$P=\mbox{true}$, $Q=\mbox{false}$: both compound statements are false.

*$P=\mbox{false}$, $Q=\mbox{true}$: both compound statements are true.

*$P=\mbox{false}$, $Q=\mbox{false}$: both compound statements are true.

A: $$A\implies B\quad\equiv\quad\lnot A\lor B$$
By letting $A=\lnot Q$ and $B=\lnot P$, by the above definition we have
$$\lnot Q\implies\lnot P\quad\equiv\quad\lnot(\lnot Q)\lor\lnot P\quad\equiv\quad\lnot P\lor Q$$ which satisfies then that $P\implies Q$.
A: In another way, we can assume that: $\lnot(\bar{Q}\implies \bar{P})$  
Therefore we can deduce that: $\;\bar{Q} \land \lnot \bar{P}$ which is equivalent to $\bar{Q} \land P\;$ which implies $Q$ (because $P\implies Q$). And we got that $Q\land\bar Q$ . Contradiction.
A: I like to think of statements like little light bulbs. Let's say you have 2 bulbs in front of you. P is the one on the left, and Q is the one on the right. P $\implies$ Q means that whenever the light on the left is on, the one on the right on too. The one on the right can be on when the one on the left is off, but when the left is on then the right is on. So now lets see the contrapositive. If the right is off, then the left is off too; because had the left been on, the right would be on too.
