Conditional and Inverse are not Equivalent

Theorem

A conditional statement:

$p \implies q$

is not logically equivalent to its inverse:

$\lnot p \implies \lnot q$


Proof

We apply the Method of Truth Tables to the proposition:

$\paren {p \implies q} \iff \paren {\lnot p \implies \lnot q}$

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

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

$\blacksquare$


Also see


Sources

  • 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
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