In classical logic, why is (p⇒q)(p\Rightarrow q) True if both pp and qq are False?

I am studying entailment in classical first-order logic.

The Truth Table we have been presented with for the statement (pq) (a.k.a. ‘p implies q‘) is:

I ‘get’ lines 1, 2, and 3, but I do not understand line 4.

Why is the statement (pq) True if both p and q are False?

We have also been told that (pq) is logically equivalent to ( p||q) (that is ¬pq).

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 (pq) 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.


Here is an example. Mathematicians claim that this is true:

If x is a rational number, then x2 is a rational number

But let’s consider some cases. Let P be “x is a rational number”. Let Q be “x2 is a rational number”.
When x=3/2 we have P,Q both true, and PQ of the form TT is also true.
When x=π we have P,Q both false, and PQ of the form FF is true.
When x=2 we have P false and Q true, so PQ of the form FT is again true.

But the assertion in bold I made above means that we never ever get the case TF, no matter what number we put in for x.

Source : Link , Question Author : Ethan , Answer Author : GEdgar

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