Conditional and Converse are not Equivalent

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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.

$\blacksquare$


Also see


Sources