# Modus Ponendo Ponens/Sequent Form/Proof 2

## Theorem

$p \implies q, p \vdash q$

## Proof

We apply the Method of Truth Tables.

$\begin{array}{|c|ccc||c|} \hline p & p & \implies & q & q\\ \hline F & F & T & F & F \\ F & F & T & T & T \\ T & T & F & F & F \\ T & T & T & T & T \\ \hline \end{array}$

As can be seen, when $p$ is true, and so is $p \implies q$, then $q$ is also true.

$\blacksquare$