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
(define (boolean-product length)
"Return a list of all Boolean sequences of LENGTH."
(let more-rows ((row (- (expt 2 length) 1)))
(if (< row 0) (list)
(cons (let more-columns ((column (- length 1)))
(if (< column 0) (list)
(cons (even? (quotient row (expt 2 column)))
(more-columns (- column 1)))))
(more-rows (- row 1))))))
(define (truth-table procedure arity)
"Return a truth table for PROCEDURE with ARITY."
(map (lambda (inputs)
(append inputs (list (apply procedure inputs))))
(boolean-product arity)))
(define (tautology? procedure arity)
(not (member #f (map (lambda (row)
(car (reverse row)))
(truth-table procedure arity)))))
(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
\(\blacksquare\)