Negation of the Boolean expression p⇔(q⇒p) is

~(p↔(q→p))~(p↔q)=(p∧~q)∨(q∧~p)~(p↔(q→p))=(p∧~(q→p))∨((q→p)∧~p)(p∧~(q→p))=p∧(q∧~p)=(p∧~p)∧q=c(q→p)∧~p=(~q∨p)∧~p=∼p∧(~q∨p) =(~p∧~q)∨(~p∧p)=∼p∧~q~(p↔(q→p))=c∨(~p∧~q)=∼p∧~q

Negation of the Boolean expression p⇔(q⇒p) is

~(p↔(q→p))~(p↔q)=(p∧~q)∨(q∧~p)~(p↔(q→p))=(p∧~(q→p))∨((q→p)∧~p)(p∧~(q→p))=p∧(q∧~p)=(p∧~p)∧q=c(q→p)∧~p=(~q∨p)∧~p=∼p∧(~q∨p) =(~p∧~q)∨(~p∧p)=∼p∧~q~(p↔(q→p))=c∨(~p∧~q)=∼p∧~q

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Negation of the Boolean expression $p \Leftrightarrow (q \Rightarrow p)$ is

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$(~ p) \land q$

$p \land (~ q)$

$(~ p) \lor (~ q)$

$(~ p) \land (~ q)$

answer is D.

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Detailed Solution

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