Rule of Transposition/Variant 1/Formulation 1/Proof 2

Theorem

 * $p \implies \neg q \dashv \vdash q \implies \neg p$

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.

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