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I am having a bit of trouble understanding the pasted excerpt. I think I might be missing something basic. As I understand it, the contrapositive of a conditional statement is where we take a conditional statement and both 1) flip the hypothesis and conclusion and 2) negate the q and p so we have ¬q -> ¬p

Looking at the truth table of the original p -> q I can convert each possibility to the contrapositive ¬q -> ¬p . So, for example, when p is True and q is False, the p -> q is false. I can now turn this case into the contrapositive by taking the q and negating it which is True and then take the p and negating it which is False.

What does this mean though that the contrapositive has the same truth value as p -> q? Like, the truth of table of p -> q was just a fact that was given to me. How do I know what the truth value for each possibility of ¬q -> ¬p is though? Is it simply that we can always convert the contrapositive back to the p -> q statement by "un-negating" the q and p in ¬q -> ¬p and then know the truth value based on original truth table for p -> q where we know it's only False when p is True and q is False?

Just generally confused. I hope my rambling question makes some sense.

Thanks in advance.

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    $\begingroup$ You can also ask yourself what would happen if not-q implied p instead, for then p implies q, which would lead to a contradiction. $\endgroup$ Commented May 31, 2021 at 19:48
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    $\begingroup$ Set up a truth table, with the 4 possibilities of ($p$ true or false) and ($q$ true or false). In the table, for each of the 4 possibilities, evaluate the two statements that are allegedly equivalent. The statements will be equivalent if and only if they evaluate the same in each of the 4 rows of the truth table. $\endgroup$ Commented May 31, 2021 at 19:49

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Well, you can easily verify that the two statements are equivalent using a truth table, as you have done.

You might also use the definition $$p\rightarrow q \equiv \sim p \cup q$$ to show that both are equivalent.

If you want to gain some intuitive sense, $p\rightarrow q$ means "if $p$, then $q$". This means that the truth of $p$ indicates the truth of $q$. Whereas the second statement means "if not $q$, then not $p$". The two are clearly equivalent. To see this take, as an example, $p\equiv$ "it is cloudy" and $q\equiv$ "it is raining".

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  • $\begingroup$ Thank you, that example does help! $\endgroup$
    – LuxuryMode
    Commented Jun 1, 2021 at 0:29
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I think something fundamental may be slipping past you. As I think you understand, the implication $p \Rightarrow q$ means that either the premise (which we're calling $p$) is false or the consequence (what we're calling $q$) is true (or possibly both).

But with this understanding, the truth values of the contrapositive are not arbitrary. Let's look at the contrapositive $\lnot q \Rightarrow \lnot p$. By our definition of an implication, this means the premise (whatever it may be) is false or the consequence (whatever it may be) is true. In this case, the premise is $\lnot q$, and that's false exactly when $q$ is true. Similarly, the consequence of the contrapositive is $\lnot p$, and that's true exactly when $p$ itself is false.

That means that $\lnot (\lnot q) \lor \lnot p$ (which is equivalent to the contrapositive) is true exactly when $\lnot p \lor q$ is true, so the contrapositive is true exactly when the original implication is true.

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    $\begingroup$ Thank you very much, I think this explained it. I think what I was missing was the actual "meaning" of an implication. I was looking at it as simply a table of rules where we know the only false case is when p is true and q is false. But there's a logic or sense to this in that when we have a premise and consequence, it can be true either because the premise is false or the conclusion is true. $\endgroup$
    – LuxuryMode
    Commented Jun 1, 2021 at 0:14
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    $\begingroup$ Hence the only case where the implication is false is when p is true and q is false. With this, we can then think of the contrapositive $\lnot q \Rightarrow \lnot p$ and apply the same idea. We have a premise, in this case $\lnot q$, and a consequence $\lnot p$. In this case, for the premise to be false, q must be true. For the consequence to be true, p must be false. $\endgroup$
    – LuxuryMode
    Commented Jun 1, 2021 at 0:14
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    $\begingroup$ This leaves us with the only case where the implication is false is when q is false (since in the case of the contraposition our premise is $\lnot q$ which evaluates to true) and when p is true which evaluates to false. $\endgroup$
    – LuxuryMode
    Commented Jun 1, 2021 at 0:15
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    $\begingroup$ So what we're doing in effect is saying if we took the truth table of if p then q and swapped the premise and consequence and also negated q and p, we will end up with the exact same truth values as the implication if p then q, namely that it is only false when q is false (making not q true) and when p is true (making not p false) which results in an implication where the hypothesis is true and the consequence is false which we know is false. Do I have this correctly understood now? $\endgroup$
    – LuxuryMode
    Commented Jun 1, 2021 at 0:15
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    $\begingroup$ That's exactly right. I'm glad I could help. $\endgroup$ Commented Jun 1, 2021 at 0:31

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