if $P \implies Q$ why does $\bar{Q} \implies \bar{P}$

Question

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...

Would an example help? "If I am human then I eat" is equivalent to "If I don't eat then i'm not human". — MathMajor, Dec 25 '16 at 04:15
@user401334 the conditional statement $\overline{Q}\Rightarrow \overline{P}$ is called the contrapostive statement of the given statement $P\implies Q$. — Juniven Acapulco, Dec 25 '16 at 04:15
If I am in Amsterdam, I am in the Netherlands. If I am not in the Netherlands, then I am also not in Amsterdam! — Job Bouwman, Dec 25 '16 at 06:16
I don't know the level of your knowledge in formal logic. I would suggest you read https://en.wikipedia.org/wiki/Contraposition to help you start to undersand why the 2 implications you discuss are equivalent. — MasB, Dec 25 '16 at 21:43

Alex · Answer 1 · 2016-12-25T04:44:24.383

3

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.

edited Dec 25 '16 at 04:44

answered Dec 25 '16 at 04:15

Alex

4,873

A related important point is: "a falsehood implies anything"; this is discussed e.g. in another question: link – Alex Dec 25 '16 at 20:34

score 2 · Answer 2 · answered Dec 25 '16 at 04:19

$$A\implies B\quad\equiv\quad\lnot A\lor B$$ 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 $P\implies Q$.

Jon · Answer 3 · 2016-12-25T21:28:33.337

0

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.

edited Dec 25 '16 at 21:28

answered Dec 25 '16 at 21:03

Jon

1,215

score 0 · Answer 4 · answered Dec 25 '16 at 21:17

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.

if $P \implies Q$ why does $\bar{Q} \implies \bar{P}$

4 Answers4

Linked