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

Given expression is p ⇔ ( q ⇒ p ) ∼ ( 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

Mathematics - Mathematical Reasoning Question with Solution | TestHub

MathematicsMathematical ReasoningGeneralHard2 minPYQ_2023

MathematicsHardsingle choice

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

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Answer:

Solution:

Given expression is $p \Leftrightarrow (q \Rightarrow p)$
$\sim (p \leftrightarrow (q \to p))$
$\sim (p \leftrightarrow q) = (p \land \sim q) \lor (q \land \sim p)$
$\sim (p \leftrightarrow (q \to p))$ $= (p \land \sim (q \to p)) \lor ((q \to p) \land \sim p)$
$(p \land \sim (q \to p)) = p \land (q \land \sim p)$ $= (p \land \sim p) \land q = c$
$(q \to p) \land \sim p = (\sim q \lor p) \land \sim p$ $=\sim p \land (\sim q \lor p)$
$= (\sim p \land \sim q) \lor (\sim p \land p) =\sim p \land \sim q$
$\sim (p \leftrightarrow (\overset{q}{˙} \to p)) = c \lor (\sim p \land \sim q) =\sim p \land \sim q$

Stream:BITSATSubject:MathematicsTopic:Mathematical ReasoningSubtopic:General

⏱ 2mℹ️ Source: PYQ_2023

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