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

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(A) Given expression: $(p \wedge (\sim q)) \vee ((\sim p) \wedge q) \vee ((\sim p) \wedge (\sim q))$Group the terms containing $(\sim p)$:$(p \wedge (

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$(\sim p) \vee (\sim q)$

Vedclass · Answer

(A) Given expression: $(p \wedge (\sim q)) \vee ((\sim p) \wedge q) \vee ((\sim p) \wedge (\sim q))$Group the terms containing $(\sim p)$:$(p \wedge (\sim q)) \vee ((\sim p) \wedge (q \vee (\sim q)))$Since $(q \vee (\sim q)) = t$ (tautology):$(p \wedge (\sim q)) \vee ((\sim p) \wedge t)$$(p \wedge (\sim q)) \vee (\sim p)$Using the distributive law $(A \vee (B \wedge C) = (A \vee B) \wedge (A \vee C))$:$((\sim p) \vee p) \wedge ((\sim p) \vee (\sim q))$Since $((\sim p) \vee p) = t$:$t \wedge ((\sim p) \vee (\sim q))$$= (\sim p) \vee (\sim q)$

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

Similar Questions

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