Modus Ponendo Ponens/Variant 2/Proof 2

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Theorem

$\vdash p \implies \paren {\paren {p \implies q} \implies q}$


Proof

We apply the Method of Truth Tables.

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

As can be seen by inspection, the main connective is true throughout.

$\blacksquare$