The Boolean expression (p wedge q) Rightarrow | vedclass

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(A) Given expression: $(p \wedge q) \Rightarrow ((r \wedge q) \wedge p)$Using the implication law $A \Rightarrow B \equiv \sim A \vee B$:$\sim(p \wedg

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$(p \wedge q) \Rightarrow (r \wedge p)$

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(A) Given expression: $(p \wedge q) \Rightarrow ((r \wedge q) \wedge p)$Using the implication law $A \Rightarrow B \equiv \sim A \vee B$:$\sim(p \wedge q) \vee ((r \wedge q) \wedge p)$Using the associative and commutative properties:$\sim(p \wedge q) \vee ((r \wedge p) \wedge (p \wedge q))$Using the distributive law $X \vee (Y \wedge Z) \equiv (X \vee Y) \wedge (X \vee Z)$:$(\sim(p \wedge q) \vee (r \wedge p)) \wedge (\sim(p \wedge q) \vee (p \wedge q))$Since $\sim A \vee A \equiv t$ (tautology):$(\sim(p \wedge q) \vee (r \wedge p)) \wedge t$$\equiv \sim(p \wedge q) \vee (r \wedge p)$Converting back to implication form:$(p \wedge q) \Rightarrow (r \wedge p)$

The Boolean expression $(p \wedge q) \Rightarrow ((r \wedge q) \wedge p)$ is equivalent to:

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