# 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.

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

$\blacksquare$