Clarification on contrapositive logic

Question

I have read a little around proof by contrapositive and follow the reasoning given by the recommended answer in the referenced question below but still have a slight nagging bit of confusion. I also get that when we say $P \Rightarrow Q$ we want it to always be the case with no counter-examples.

Now, if we don't know beforehand whether $P => Q$ it could turn out that $Q$ is sometimes implied by $P$, but sometimes isn't. So we have one of two cases in my understanding:

So I get that in case 1 if $P \Rightarrow Q$ then we can show $ \lnot Q \Rightarrow \lnot P$ but am a little unclear about how we would know to use contrapositive given we might not know whether we are in case 1 or case 2 ex-ante.

So for the example in the picture:

$P$ is "I am coughing"

$Q$ is "I am ill"

So we'd like to show that if I am coughing then I am ill. Of course, we know here that I might be ill without coughing but in general we might not know this sort of information when we set out to try to show $P \Rightarrow Q$ via contrapositive, so what would we do?

Apologies if this isn't terribly clear (or if I've not even asked a proper question), it perhaps reflects some of my own confusion on the matter and so anything that can clarify my scattered thoughts is appreciated.

Reference question: here.

Answer 1

Your question is a little unclear, but if I understand you correctly, you are making the following argument:

In case 2 we do not have that $P \rightarrow Q$, and yet when we take the contrapositive, it seems like we have $\neg Q \rightarrow \neg P$ ... and so we would have $P \rightarrow Q$ after all! Something seems to be wrong?!

Well, in the picture of the contrapositive of Case 2, focus on the dark blue region: that region is outside non-P, but inside non-Q. Hence, you do not have $\neg Q \rightarrow \neg P$, and therefore also not $P \rightarrow Q$

The fact that the region that you are pointing to is inside the non-P region means nothing: sure, in that region we have $\neg P$ and we have $Q$ ... so? That does not mean $P \rightarrow Q$ nor does it refute $P \rightarrow Q$

You also write:

So I get that in case 1 if $P \Rightarrow Q$ then we can show $ \lnot Q \Rightarrow \lnot P$ but am a little unclear about how we would know to use contrapositive given we might not know whether we are in case 1 or case 2 ex-ante.

To show $P \rightarrow Q$ you can try to show $\neg Q \rightarrow \neg P$. If you can, great: you have just shown $P \rightarrow Q$ by showing its contrapositive $\neg Q \rightarrow \neg P$. But if you can't show $\neg Q \rightarrow \neg P$ (as in case 2), then you can't show $P \rightarrow Q$ either ... well... that's good! Since in case 2 $P \rightarrow Q$ does not hold either! So, there is no problem here.

Finally, you write:

So we'd like to show that if I am coughing then I am ill. Of course, we know here that I might be ill without coughing but in general we might not know this sort of information when we set out to try to show $P \Rightarrow Q$ via contrapositive, so what would we do?

If you want to show that "if I am coughing then I am ill", then knowing that I might be ill without coughing means nothing. Again, that does not refute the claim that "if I am coughing then I am ill". Indeed, if you want to show the claim using the contrapositive, you need to show that if you are not ill, you are not coughing. This is the case in case 1, so if you are dealing with case 1, you should be able to show that $\neg Q \rightarrow \neg P$, and thereby $P \Rightarrow Q$. But it is not the case in case 2 (see the dark blue region), and hence in case 2 you cannot show that $\neg Q \rightarrow \neg P$, which again is a good thing, since in case 2 we do not have $P \Rightarrow Q$. So once again, the logic works as it should!

BTW: Not knowing whether $P \rightarrow Q$ is the case or not allows for the third case, namely where the whole $Q$ circle is outside the $P$ circle.

Answer 2

$p{\implies}q$ would be represented only by: $P\subseteq Q$

$\require{enclose}\enclose{box}{\enclose{circle}{\enclose{circle}{\raise{3ex}~\lower{1ex}~P~~}\raise{4ex}~\lower{3ex}~Q~~}}$

Only in such a setting does $x\in P$ guarantee that $x\in Q$.

Also, in such a setting $x\notin Q$ guarantees that $x\notin P$.

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  If set $P$ contains elements not in $Q$ this guarantee would clearly not be the case; it would not represent the implication.   In such a setting the contraposition would not hold either.

$\require{enclose}\enclose{box}{\enclose{circle}{\raise{3ex}~\lower{1ex}~P\quad}\hspace{-2ex}\enclose{circle}{\quad\raise{4ex}~\lower{3ex}~Q~~}}$