Understanding Vacuously True (Truth Table) I don't know very much about formal logic, and I'm trying to understand the concept of vacuously true statements. Consider the truth table below:
$$\begin{array} {|c|}
\hline
P & Q & P\implies Q & Q\implies P & P\iff Q  \\ \hline
T & T & T & T & T  \\ \hline
T & F & F & \color{blue} T & F\\ \hline
F & T & \color{blue}T & F & F\\ \hline
F & F & \color{blue} T & \color{blue}T & T
\end{array}$$.
The blue letters are definitions. To see why these definitions are the correct choices (as opposed to $F$s), suppose we changed the lower left $\color{blue} T$ to an $F$ (this would then force the lower right $\color{blue}T$ to an $F$), so $P\iff Q$ would be $F$ for $P$ and $Q$ both $F$, which isn't want we want. So I see why this makes sense to choose the lower  entries as $\color{blue}T$.
However, it isn't clear to me why $P\implies Q$ true for $P$ false and $Q$ true is the sensible choice (same for $Q\implies P$ true for $Q$ false and $P$ true). For if the $\color{blue}T$s in the middle rows were switched to $F$s, then $P\iff Q$ would still be $F$. So I don't see what the problem would be. Can someone please explain?

$P \Rightarrow Q$ is logically equivalent to $\sim P \vee Q$ (where $\sim$ is '"not" and $\vee$ is "or"), and $\sim P \vee Q$ is true whenever $P$ is false. Another thing to think about is providing counterexamples. $P\Rightarrow Q$ can only be proved false if you give an example where $P$ is true and $Q$ is not true. Any example where $P$ is false is not a counterexample, so $P\Rightarrow Q$ is vacuously true in that case.
