Conditional and Converse are not Equivalent

Theorem
A conditional statement:
 * $p \implies q$

is not logically equivalent to its converse:
 * $q \implies p$

Proof
We apply the Method of Truth Tables to the proposition:
 * $\paren {p \implies q} \iff \paren {q \implies p}$

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

As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.

Also see

 * Affirming the Consequent