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.
$\blacksquare$