Let p and q be two statements. Then sim(p wed | vedclass

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(C) We are given the expression $\sim(p \wedge (p \rightarrow \sim q))$.Using De Morgan's Law,$\sim(A \wedge B) \equiv \sim A \vee \sim B$:$\sim(p \we

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$(\sim p) \vee q$

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(C) We are given the expression $\sim(p \wedge (p \rightarrow \sim q))$.Using De Morgan's Law,$\sim(A \wedge B) \equiv \sim A \vee \sim B$:$\sim(p \wedge (p$ $\rightarrow \sim q)) \equiv \sim p \vee \sim(p$ $\rightarrow \sim q)$.Since $p \rightarrow q \equiv \sim p \vee q$,we have $p \rightarrow \sim q \equiv \sim p \vee \sim q$.Substituting this into the expression:$\equiv \sim p \vee \sim(\sim p \vee \sim q)$.Applying De Morgan's Law again:$\equiv \sim p \vee (p \wedge q)$.Using the Distributive Law,$A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$:$\equiv (\sim p \vee p) \wedge (\sim p \vee q)$.Since $\sim p \vee p \equiv t$ (tautology):$\equiv t \wedge (\sim p \vee q)$.$\equiv \sim p \vee q$.

Let $p$ and $q$ be two statements. Then $\sim(p \wedge (p \Rightarrow \sim q))$ is equivalent to:

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