I am studying entailment in classical first-order logic.
The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is:
$$\begin{array}{|c|c|c|}
\hline
p&q&p\Rightarrow q\\ \hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\\hline
\end{array}$$
I 'get' lines 1, 2, and 3, but I do not understand line 4.
Why is the statement $(p \Rightarrow q)$ True if both p and q are False?
We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).
Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.
Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True? And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
Best Answer
Here is an example. Mathematicians claim that this is true:
If $x$ is a rational number, then $x^2$ is a rational number
But let's consider some cases. Let $P$ be "$x$ is a rational number". Let $Q$ be "$x^2$ is a rational number".
When $x=3/2$ we have $P, Q$ both true, and $P \rightarrow Q$ of the form $T \rightarrow T$ is also true.
When $x=\pi$ we have $P,Q$ both false, and $P \rightarrow Q$ of the form $F \rightarrow F$ is true.
When $x=\sqrt{2}$ we have $P$ false and $Q$ true, so $P \rightarrow Q$ of the form $F \rightarrow T$ is again true.
But the assertion in bold I made above means that we never ever get the case $T \rightarrow F$, no matter what number we put in for $x$.