Negation of (p⇒q)⇒(q⇒p) is

~((p→q)→(q→p))≡∼((~p∨q)→(~q∨p))≡∼((p∧~q)∨(~q∨p))≡(~p∨q)∧(~p∧q)p q ~p ~q~p∨q~p∧q(~p∨q)∧(~p∧q) TTFFTFFTFFTFFFFTTFTTTFFTTTFF(~p∧q)≡ (~p∨q)∧(~p∧q)=q∧~p

Negation of (p⇒q)⇒(q⇒p) is

~((p→q)→(q→p))≡∼((~p∨q)→(~q∨p))≡∼((p∧~q)∨(~q∨p))≡(~p∨q)∧(~p∧q)p q ~p ~q~p∨q~p∧q(~p∨q)∧(~p∧q) TTFFTFFTFFTFFFFTTFTTTFFTTTFF(~p∧q)≡ (~p∨q)∧(~p∧q)=q∧~p

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$Negation of (p \Rightarrow q) \Rightarrow (q \Rightarrow p) is$

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$(~ p) \lor q$

$(~ q) \land p$

$p \lor (~ q)$

$q \land (~ p)$

answer is D.

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Detailed Solution

$\begin{array}{l} ~ ((p \to q) \to (q \to p)) \\ \equiv\sim ((~ p \lor q) \to (~ q \lor p)) \\ \equiv\sim ((p \land ~ q) \lor (~ q \lor p)) \equiv (~ p \lor q) \land (~ p \land q) \end{array}$

$p$	$q$	$~ p$	$~ q$	$~ p \lor q$	$~ p \land q$	$(~ p \lor q) \land (~ p \land q)$
T	T	F	F	T	F	F
T	F	F	T	F	F	F
F	T	T	F	T	T	T
F	F	T	T	T	F	F

$\begin{array}{l} (~ p \land q) \equiv (~ p \lor q) \land (~ p \land q) = q \land ~ p \end{array}$

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