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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?

ryang
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rb612
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  • It's vacuously true, because if it were false there should exist a counter-example. Such a counter-example cannot obviously exist if $p$ is false. – Bernard Jun 24 '18 at 19:39
  • You answered your own question in your second paragraph: if it's not false, it is true. – BDN Jun 24 '18 at 19:40
  • To assert $p \to q$ (i.e. "if $p$, then $q$") is not the same as "from $p$, we have proved $q$". When we try to prove something (e.g. $q$) we start from axioms or already proved theorems (e.g. $p$): in that case, we already know (or assume) that $p$ is true. – Mauro ALLEGRANZA Jun 24 '18 at 19:42
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    You can see the similar post : Still struggling to understand vacuous truths. – Mauro ALLEGRANZA Jun 24 '18 at 19:46
  • And also : Why is $p \rightarrow q$ true if $p$ is false and $q$ is true ? – Mauro ALLEGRANZA Jun 24 '18 at 19:50
  • I think you're getting confused because of your example :

    "If it rains, I go to the gym."

    This statement is true when it doesn't rain regardless of whether you go to gym or not.

     – AgentS Jun 24 '18 at 19:55
  • You probably want to look at the field of relevance logic. – Dave L. Renfro Jun 24 '18 at 20:05
  • @rsadhvika it's true in the fact that it's not inconsistent, but what happens when it doesn't rain doesn't prove what happens when it rains. That's why I say it's like inconclusive. For example, Goldbach's conjecture currently doesn't have a counterexample known. But that doesn't mean it's true. We haven't proved it yet. Hence it's in a "middle" state between true and false. So why isn't this the case here too? That it can exist in some state that isn't provably true or false. – rb612 Jun 24 '18 at 20:08
  • I sorta get what you're saying. I think in logic we focus on "structure" of statements, not semantics. – AgentS Jun 24 '18 at 20:12
  • Often when we are only interested in $q$ when $p$ is true, we replace $q$ with a predicate that is only defined when $p$ is true. E.g., rather than consider $x \geq 0 \rightarrow (\sqrt x)^2 = x$, we may write simply $(\sqrt x)^2 = x$. And one can make a case that ideally $(\sqrt x)^2 = x$ should be indeterminate or whatever when $x <0$. Also, it seems hard to fantasize about $p \rightarrow q$. But fantasizing about $p \wedge q$ is no substitute for fantasizing about $p \rightarrow q$ when $q \rightarrow p$ is evil yet $p \rightarrow q$ may be an important moral truth worth fantasizing about. – Stephen A. Meigs Jun 24 '18 at 23:03

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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!

ryang
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