Let Delta in {wedge, vee, Rightarrow, Leftrig | vedclass

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(C) First,simplify the expression $(p \vee q) \Rightarrow q$: $(p \vee q) \Rightarrow q \equiv \sim(p \vee q) \vee q$ $\equiv (\sim p \wedge \sim q) \

Vedclass · Accepted Answer

$\Rightarrow$

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(C) First,simplify the expression $(p \vee q) \Rightarrow q$: $(p \vee q) \Rightarrow q \equiv \sim(p \vee q) \vee q$ $\equiv (\sim p \wedge \sim q) \vee q$ $\equiv (\sim p \vee q) \wedge (\sim q \vee q)$ $\equiv (\sim p \vee q) \wedge t \equiv \sim p \vee q$. Now,we test the options for $(p \wedge q) \Delta (\sim p \vee q)$. For option $C$ $(\Rightarrow)$: $(p \wedge q) \Rightarrow (\sim p \vee q) \equiv \sim(p \wedge q) \vee (\sim p \vee q)$ $\equiv (\sim p \vee \sim q) \vee (\sim p \vee q)$ $\equiv \sim p \vee (\sim q \vee q)$ $\equiv \sim p \vee t \equiv t$. Since the result is a tautology,$\Delta$ is $\Rightarrow$.

Let $\Delta \in \{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$ be such that $(p \wedge q) \Delta ((p \vee q) \Rightarrow q)$ is a tautology. Then $\Delta$ is equal to

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