Justification of truth table of conditional statement if $p$ then $q$

Question

I have gone through various sites but i can't understand the justification for the truth table of "If $p$ then $q$". Is it accepted by the mathematicians without any proof or justification? \begin{array}{c|c|c} p & q & p\rightarrow q \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array} Last two values of truth table seems a bit confusing how is true?

Answer 1

For the third when $p$ is false prove $p\to q$ by contradiction. So accept $p$ and assume $\neg q$. As $\neg p$ is true, $p$ and $\neg p$ are a contradiction.
Thus the assumption $\neg q$ is false. This proves $p \to q$ when p is false.

The fourth is simular.

Answer 2

You can memorise easily it: $$a \rightarrow b = \overline a \lor b$$

Answer 3

$$ \left( p \implies q \right) \equiv \mbox{“p implies q”} \equiv \mbox{“if p then q”} $$

So if the statement "$p \implies q$" holds, we have:

• If $p$ is true then the statement tells us $q$ must be true.

• If $p$ is false then the statement does not tell us anything about $q$.

The only way to prove the statement "$p \implies q$" is wrong is to show that $p$ holds but $q$ does not. That is why the truth table only indicates the statement is "false/wrong" in that one row.