Difference between when $p \implies q$ can be proved true/false and not?

Question

I completely understand why $p {\implies} q$ is not false when $p$ is false. If we take the statement, "If it rains, I don't go to the gym", and it's not raining and I go or don't go to the gym, the original statement is clearly not false. So, this, I understand.

However, when we are proving the validity of $p {\implies} q,$ there's some difference between the first two lines in the truth table and the last two lines.

I mean, if we assume $p$ and end up getting $q,$ we can prove $p {\implies} q$ true. This seems to me to correspond to the first line in the truth table, where $p$ and $q$ are true means $p {\implies} q$ true. Similarly, if we assume $p$ and end up getting $q$ false, we can prove $p {\implies} q$ false. This to me corresponds to the second line in the truth table, where $p$ is true and $q$ false means $p {\implies} q$ false.

However, if we assume that $p$ is false, obtaining $q$ true or $q$ false does not prove $$p {\implies} q$$ true or false. Here, we say that $p {\implies} q$ is vacuously true, but to me it would make more sense if it was "inconclusive"; couldn't $p {\implies} q,$ by the same logic, be "vacuously false" or just be inconclusive, since, whether it holds, using it in a proof would not result in a conclusion? Is there a reason why we say "vacuously true" specifically?

Answer 1

if we assume that $p$ is false, obtaining $q$ true or $q$ false does not prove $$p {\implies} q$$ true or false. Here, we say that $(p {\implies} q)$ is vacuously true, but to me it would make more sense if it was "inconclusive"

You've got it backwards: the sentence $(\text{false}{\implies} q)$ is vacuously true not because its truth is cannot be concluded (nor because $q$'s truth cannot be concluded), but by mere definition (those last two lines of its truth table), involving no proof.

In fact, it's the other way round: it is because of $(\text{false}{\implies} q)$'s vacuous truth that $q$'s truth cannot be concluded.

couldn't $(p {\implies} q),$ by the same logic, be "vacuously false" or just be inconclusive, since, whether it holds, using it in a proof would not result in a conclusion?

Why would we invoke $(\text{false}{\implies} q)$ when it is false or of indeterminate truth? Naturally, we invoke it in a proof precisely because it holds.

Besides, observe that invoking $(\text{false}{\implies} q)$ results in no conclusion about $q$'s truth because $(\text{false}{\implies} q)$ is compatible with both $q$ true and $q$ false, i.e., because $(\text{false}{\implies} q)$ is true regardless of $q$'s truth.

Finally, here's a concrete reason why the sentence $(\text{false}{\implies} q)$ is defined to be True: otherwise, the statement $$\forall z{\in}\mathbb C\,\Big(z\in\mathbb R\implies \Re(z^2)\ge0\Big)$$ would not hold!