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

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

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

Also see

 * Affirming the Consequent