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

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Question

(C) We simplify the expression using logical laws: $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$ Using the distributive law on the first two terms

Vedclass · Accepted Answer

$p \vee q$

Vedclass · Answer

(C) We simplify the expression using logical laws: $(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$ Using the distributive law on the first two terms: $\equiv ((p \vee q) \wedge (\sim q \vee q)) \vee (\sim p \wedge q)$ Since $(\sim q \vee q) \equiv T$ (Complement law): $\equiv ((p \vee q) \wedge T) \vee (\sim p \wedge q)$ $\equiv (p \vee q) \vee (\sim p \wedge q)$ Applying the distributive law again: $\equiv (p \vee q \vee \sim p) \wedge (p \vee q \vee q)$ Since $(p \vee \sim p) \equiv T$ and $(q \vee q) \equiv q$ (Idempotent law): $\equiv (T \vee q) \wedge (p \vee q)$ Since $(T \vee q) \equiv T$: $\equiv T \wedge (p \vee q)$ $\equiv p \vee q$ Therefore,the expression is equivalent to $p \vee q$.

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

Similar Questions

If $p, q$ and $r$ are three propositions,then which of the following combination of truth values of $p, q$ and $r$ makes the logical expression $\{(p \vee q) \wedge ((\sim p) \vee r)\} \rightarrow ((\sim q) \vee r)$ false?

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