(p wedge sim q wedge sim r) vee (sim p wedge | vedclass

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(D) Let the given expression be $S = (p \wedge \sim q \wedge \sim r) \vee (\sim p \wedge q \wedge \sim r) \vee (\sim p \wedge \sim q \wedge r)$.This e

Vedclass · Accepted Answer

$(\sim ((p \wedge q) \vee (q \wedge r) \vee (r \wedge p)) \wedge (p \vee q \vee r))$

Vedclass · Answer

(D) Let the given expression be $S = (p \wedge \sim q \wedge \sim r) \vee (\sim p \wedge q \wedge \sim r) \vee (\sim p \wedge \sim q \wedge r)$.This expression represents the condition where exactly one of the three propositions $p, q, r$ is true.Consider the expression $(p \vee q \vee r) \wedge \sim ((p \wedge q) \vee (q \wedge r) \vee (r \wedge p))$.$(p \vee q \vee r)$ is true if at least one of $p, q, r$ is true.$((p \wedge q) \vee (q \wedge r) \vee (r \wedge p))$ is true if at least two of $p, q, r$ are true.Therefore,their conjunction is true if and only if exactly one of $p, q, r$ is true.This matches the given expression $S$.Thus,the correct option is $D$.

$(p \wedge \sim q \wedge \sim r) \vee (\sim p \wedge q \wedge \sim r) \vee (\sim p \wedge \sim q \wedge r)$ is equivalent to-

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