Justification of truth table of conditional statement if $p$ then $q$ 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?
 A: There are 16 possible truth tables for two binary variables $p$ and $q$ that take the values $T$ or $F$.
Assuming that we want:
I. $p = T$, $q = T$ and $p \implies q = T$
II. $p = T$, $q = F$ and $p \implies q = F$
we have that from the 16 possible truth tables only these 4 comply with conditions I and II:
\begin{array}{|c|c|c|c|c|c|}
p & q & 1 & 2 & 3 & 4\\
\hline
T & T & T & T & T & T\\
T & F & F & F & F & F\\
F & T & F & T & F & T\\
F & F & F & F & T & T\\
\end{array}
The truth table $1$ corresponds to the truth table of $p \land q$ so we discard that this table is suitable for $p \implies q$.
The truth table $2$ corresponds to $q$ so we discard that this table is suitable for $p \implies q$.
The truth table $3$ corresponds to the truth table of $p \Longleftrightarrow q$ so we discard that this table is suitable for $p \implies q$.   
Therefore table $4$ is the only truth table suitable for $p \implies q$ if we think that conditions I and II must be fulfill and $p$ and $q$ are two binary variables that take the values $T$ or $F$.
That is the justification for the truth table of $p \implies q$.
A: 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.
A: You can memorise easily it:
$$a \rightarrow b = \overline a \lor b$$
A: $$ \left( p \implies q \right) \equiv \mbox{“p implies q”}  \equiv \mbox{“if p then q”} $$
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. 
