Rule of Transposition/Variant 2/Formulation 1/Proof by Truth Table
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Theorem
- $\neg p \implies q \dashv \vdash \neg q \implies p$
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.
$\begin{array}{|cccc||cccc|} \hline
\neg & p & \implies & q & \neg & q & \implies & p \\
\hline
\T & \F & \T & \F & \T & \F & \T & \F \\
\T & \F & \T & \T & \F & \T & \T & \F \\
\F & \T & \F & \F & \T & \F & \F & \T \\
\F & \T & \T & \T & \F & \T & \T & \T \\
\hline
\end{array}$
$\blacksquare$