TITLE: In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True?
QUESTION [75 upvotes]: Provided we have this truth table where "$p\implies q$" means "if $p$ then $q$":
$$\begin{array}{|c|c|c|}
\hline
p&q&p\implies q\\ \hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\\hline
\end{array}$$
My understanding is that "$p\implies q$" means "when there is $p$, there is q". The second row in the truth table where $p$ is true and $q$ is false would then contradict "$p\implies q$" because there is no $q$ when $p$ is present.
Why then, does the third row of the truth table not contradict "$p\implies q$"? If $q$ is true when $p$ is false, then $p$ is not a condition of $q$.
I have not taken any logic class so please explain it in layman's terms.


Administrative note. You may experience being directed here even though your question was actually about line 4 of the truth table instead. In that case, see the companion question In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? And even if your original worry was about line 3, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.

REPLY [3 votes]: The fact that the conditional $$P\to Q$$ is true whenever its antecedent $P$ is false (principle of explosion; vacuous truth) is actually so by definition:
      $P\to Q\,$ is a truth function that is tautologically equivalent to $\,\lnot P\lor Q.$
So, $P\to Q\,$ is false precisely when $P$ is true but $Q$ false.

To be clear: whenever $P$ is false, the assertion $P\implies Q\,$ gives no information about whether $Q$ is true.

Summarising these two explanations of the motivation for the above definition:
if we insist, to the contrary, that  False$\to$True  be false, then, unfortunately, these violations of natural deduction arise: $$\text{$A$ is true and $B$ is false $\implies\Big[(A\land B)\to A\Big]\;$is false!}$$ and $$\Big[\forall n\in\mathbb Z \;\big(n \text{ is a multiple of }4\, \to \,n \text{ is even}\big)\Big]\;\text{is false}!$$

It is worth noting that in logic/mathematics, $P$ need not cause $Q$ for $P$ to imply $Q,$ that is, for the material conditional $\,P\to Q\,$ to actually be true.
After all, the logical connective $\,\to,$ being a truth-functional operator, cares about truth states without considering the flow of time.