Principle of Commutation/Formulation 1/Proof 2

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Theorem

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


Proof

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.


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

$\blacksquare$