The

reversal

of the antecedent and the consequent of the conditional

\[\begin{equation*} P \implies Q \end{equation*} \]

in the form

\[\begin{align*} & P \impliedby Q \\ \qor & Q \implies P, \end{align*} \]

which, unlike the contrapositive, is

so

\[\begin{equation*} (Q \implies P) \> \not \equiv \> (P \implies Q), \end{equation*} \]

but is

equivalent to its inverse,

so

\[\begin{equation*} (Q \implies P) \> \equiv \> (\lnot P \implies \lnot Q). \end{equation*} \]

For example, the statement

“if you are dancing, you are sweating”

is *not* equivalent to

“if you sweating, you are dancing”,

as you might be

sweating for a different reason.

Prove the statement

\[\begin{equation*} (P \implies Q) \> \not \equiv \> (Q \implies P). \end{equation*} \]

By the definition of the conditional and the biconditional,

\[\begin{align*} (P \implies Q) & \equiv (Q \implies P) \\ (\lnot P \lor Q) & \equiv (\lnot Q \lor P) \\ (\lnot P \lor Q) \implies (\lnot Q \lor P) \big) & \land \big( (\lnot Q \lor P) \implies (\lnot P \lor Q) \big) \\ \big( \lnot (\lnot P \lor Q) \lor (\lnot Q \lor P) \big) & \land \big( \lnot (\lnot Q \lor P) \lor (\lnot P \lor Q) \big), \end{align*} \]

which is *not* a tautology, as per

<<scheme/tautology?>> (not (tautology? (lambda (p q) (and (or (not (or (not p) q)) (or (not q) p)) (or (not (or (not q) p)) (or (not p) q)))) 2))

#t

*\(\blacksquare\)*

© 2024 Rudolf Adamkovič under GNU General Public License version 3.

Made with Emacs and secret alien technologies of yesteryear.