The proposition ∼(p Leftrightarrow q) is equivalent to

Question

The proposition ∼(p Leftrightarrow q) is equivalent to (A) (p ∨ ∼ q) ∧(q ∧ ∼ p) (B) (p ∧ ∼ q) ∨(q ∧ ∼ p) (C) (p ∧ ∼ q) ∧(q ∧ ∼ p) (D) None of the above

Tardigrade Admin · Accepted Answer

∼(p Leftrightarrow q) ≡ ∼[(p ⇒ q) ∧(q ⇒ p)] ≡ ∼(p ⇒ q) ∨(∼(q ⇒ p)) (∵ by De-Morgan's law ) ≡(p ∧ ∼ q) ∨(q ∧ ∼ p)

Q. The proposition $\sim (p \Leftrightarrow q)$ is equivalent to

$(p \lor \sim q) \land (q \land \sim p)$

$(p \land \sim q) \lor (q \land \sim p)$

$(p \land \sim q) \land (q \land \sim p)$

None of the above

$\sim (p \Leftrightarrow q) \equiv\sim [(p \Rightarrow q) \land (q \Rightarrow p)]$
$\equiv\sim (p \Rightarrow q) \lor (\sim (q \Rightarrow p))$
( $∵$ by De-Morgan's law )
$\equiv (p \land \sim q) \lor (q \land \sim p)$

Q. The proposition ∼(p⇔q) is equivalent to

(p∨∼q)∧(q∧∼p)

(p∧∼q)∨(q∧∼p)

(p∧∼q)∧(q∧∼p)