Rule of Transposition/Variant 2/Formulation 1/Proof by Truth Table

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.

$\begin{array}{|cccc||cccc|} \hline \neg & p & \implies & q & \neg & q & \implies & p \\ \hline T & F & T & F & T & F & T & F \\ T & F & T & T & F & T & T & F \\ F & T & F & F & T & F & F & T \\ F & T & T & T & F & T & T & T \\ \hline \end{array}$