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