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

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Question

(C) We are given the logical equivalence: $(p \wedge r) \Leftrightarrow (p \wedge (\sim q)) \equiv (\sim p)$.Let us test the options by substituting $

Vedclass · Accepted Answer

$q$

Vedclass · Answer

(C) We are given the logical equivalence: $(p \wedge r) \Leftrightarrow (p \wedge (\sim q)) \equiv (\sim p)$.Let us test the options by substituting $r = \sim q$:$(p \wedge (\sim q)) \Leftrightarrow (p \wedge (\sim q)) \equiv T$ (Tautology),which is not $(\sim p)$.Let us test $r = q$:$(p \wedge q) \Leftrightarrow (p \wedge (\sim q))$.If $p = T$,then $(T \wedge q) \Leftrightarrow (T \wedge (\sim q)) \equiv q \Leftrightarrow (\sim q)$,which is $F$.If $p = F$,then $(F \wedge q) \Leftrightarrow (F \wedge (\sim q)) \equiv F \Leftrightarrow F$,which is $T$.This matches the truth table for $(\sim p)$.Thus,$r = q$ is the correct choice.

$(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ is equivalent to $(\sim p)$ when $r$ is.

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