The Boolean expression (p wedge sim q) vee q | vedclass

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Question

(A) Let the expression be $E = (p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.Using the distributive law,$(p \wedge \sim q) \vee q \equiv (p \vee q)

Vedclass · Accepted Answer

$p \vee q$

Vedclass · Answer

(A) Let the expression be $E = (p \wedge \sim q) \vee q \vee (\sim p \wedge q)$.Using the distributive law,$(p \wedge \sim q) \vee q \equiv (p \vee q) \wedge (\sim q \vee q)$.Since $(\sim q \vee q) \equiv T$ (Tautology),we have $(p \vee q) \wedge T \equiv p \vee q$.Now,substitute this back into the expression:$E \equiv (p \vee q) \vee (\sim p \wedge q)$.Using the distributive law again,$(p \vee q) \vee (\sim p \wedge q) \equiv (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$.Since $(p \vee \sim p) \equiv T$,we have $(T \vee q) \wedge (p \vee q)$.Since $(T \vee q) \equiv T$,we have $T \wedge (p \vee q) \equiv p \vee q$.Thus,the expression is equivalent to $p \vee q$.

The Boolean expression $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$ is equivalent to:

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