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.