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?
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http://en.wikipedia.org/wiki/Transposition_(logic). – Jun 14 '14 at 16:00
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Have you compared their truth-tables? – amWhy Jun 14 '14 at 16:00
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You could also search this site for the term contrapositive. – Harald Hanche-Olsen Jun 14 '14 at 16:04
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@amWhy great idea, thank you! – muffel Jun 14 '14 at 16:05
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@HaraldHanche-Olsen Never heard of it before, thank you! – muffel Jun 14 '14 at 16:05
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For which logic? – Doug Spoonwood Jun 15 '14 at 13:03
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@Doug He probably means the ordinary first-order logic. – Jun 15 '14 at 13:13
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Suppose it's true that:
- If you hit my glass table with a hammer, it will break.
- 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.
MJD
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It is given that $P\to Q$. By material implication, $\neg P\vee Q$. By commutativity, $Q\vee\neg P$. Again, by material implication, $\neg Q\to \neg P$.
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2By "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. – GEdgar Jun 14 '14 at 16:06
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But this relationship holds in Lukasiewicz three-valued logic where (P→Q) is not logically equivalent to (¬P∨Q). – Doug Spoonwood Jun 15 '14 at 13:02
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$$\begin{array} {|c|c|c|c|} \hline P & Q & P \Rightarrow Q & \lnot Q \Rightarrow \lnot P \\ \hline \top & \top & \top & \top \\ \top & \bot & \bot & \bot \\ \bot & \top & \top & \top \\ \bot & \bot & \top & \top \\ \hline \end{array}$$
Equal.
DanielV
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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.
zo0x
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