Converse

Define

The

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

so

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

(Levin, 2021, sec. 0.2)

Example

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

Explore

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

\(\blacksquare\)