# How do we know that the contrapositive, ¬q → ¬p, of a conditional statement p → q always has the same truth value as p → q?

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.

• 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. Commented May 31, 2021 at 19:48
• 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. Commented May 31, 2021 at 19:49

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".

• Thank you, that example does help! Commented Jun 1, 2021 at 0:29

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.

• 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. Commented Jun 1, 2021 at 0:14
• 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. Commented Jun 1, 2021 at 0:14
• 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. Commented Jun 1, 2021 at 0:15
• 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? Commented Jun 1, 2021 at 0:15
• That's exactly right. I'm glad I could help. Commented Jun 1, 2021 at 0:31
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