I am studying entailment in classical first-order logic.

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

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

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

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

**Answer**

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 P→Q of the form T→T is also true.

When x=π we have P,Q both false, and P→Q of the form F→F is true.

When x=√2 we have P false and Q true, so P→Q of the form F→T is again true.

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

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