Modus Ponendo Ponens/Variant 1/Proof by Truth Table
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Theorem
- $p \vdash \paren {p \implies q} \implies q$
Proof
We apply the Method of Truth Tables.
$\begin{array}{|c|ccccc|} \hline p & (p & \implies & q) & \implies & q\\ \hline \F & \F & \T & \F & \F & \F \\ \F & \F & \T & \T & \T & \T \\ \T & \T & \F & \F & \T & \F \\ \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
As can be seen by inspection, when $p$ is true, the value of the main connective is also true.
$\blacksquare$