Rule of Transposition/Formulation 2/Proof by Truth Table

Theorem

 * $\vdash \left({p \implies q}\right) \iff \left({\neg q \implies \neg p}\right)$

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 models.

$\begin{array}{|ccc|c|ccccc|} \hline p & \implies & q) & \iff & (\neg & q & \implies & \neg & p) \\ \hline F & T & F & T & T & F & T & T & F \\ F & T & T & T & F & T & T & T & F \\ T & F & F & T & T & F & F & F & T \\ T & T & T & T & F & T & T & F & T \\ \hline \end{array}$