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.

$\blacksquare$

Also see

• Affirming the Consequent

Sources

• 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (2nd ed.) ... (previous) ... (next): $\S \text{II}.14$: Application of laws of sentential calculus in inference
• 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.5$: Theorems and Proofs
• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(3)$