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

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(C) Given expression: $(\sim(p \wedge q)) \vee q$Using De Morgan's Law: $(\sim p \vee \sim q) \vee q$Using Associative Law: $\sim p \vee (\sim q \vee

Vedclass · Accepted Answer

$p \rightarrow (p \vee q)$

Vedclass · Answer

(C) Given expression: $(\sim(p \wedge q)) \vee q$Using De Morgan's Law: $(\sim p \vee \sim q) \vee q$Using Associative Law: $\sim p \vee (\sim q \vee q)$Since $(\sim q \vee q) = t$ (tautology),the expression becomes $\sim p \vee t = t$.Thus,the expression is a tautology.Now,check the options:$A. q$ $\rightarrow (p \wedge q) \equiv \sim q \vee (p \wedge q) \equiv (\sim q \vee p) \wedge (\sim q \vee q) \equiv \sim q \vee p$ (Not a tautology)$B. p \rightarrow q \equiv \sim p \vee q$ (Not a tautology)$C. p \rightarrow (p \vee q) \equiv \sim p \vee (p \vee q) \equiv (\sim p \vee p) \vee q \equiv t \vee q \equiv t$ (Tautology)$D. p$ $\rightarrow (p$ $\rightarrow q) \equiv \sim p \vee (\sim p \vee q) \equiv \sim p \vee q$ (Not a tautology)Therefore,the expression is equivalent to $p \rightarrow (p \vee q)$.

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

Similar Questions

Consider the statement patterns:
$A. (q \to p) \lor (p \to q)$
$B. (\sim p \lor \sim q) \leftrightarrow \sim (p \land q)$
$C. [(p \lor q) \land \sim p] \land \sim q$
$D. (p \land q) \land (\sim p \lor \sim q)$
Which of the following is true regarding these statement patterns?

The negation of $(p \wedge q) \rightarrow (\sim p \vee r)$ is

Which one of the following statements is not a tautology?

The negation of $(p$ $\Rightarrow q)$ $\Rightarrow (q$ $\Rightarrow p)$ is

If $p: \forall n \in N, n^2+n$ is an even number and $q: \forall n \in N, n^2-n$ is an odd number,then the truth values of $p \wedge q, p \vee q$ and $p \rightarrow q$ are respectively:

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