The negation of the statement (( A wedge ( B | vedclass

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Question

(A) Let $p = (( A \wedge ( B \vee C ))$ $\Rightarrow ( A \vee B ))$ $\Rightarrow A$. Using the implication rule $X \Rightarrow Y \equiv \sim X \vee Y$

Vedclass · Accepted Answer

equivalent to $\sim A$

Vedclass · Answer

(A) Let $p = (( A \wedge ( B \vee C ))$ $\Rightarrow ( A \vee B ))$ $\Rightarrow A$. Using the implication rule $X \Rightarrow Y \equiv \sim X \vee Y$,we have: $p \equiv \sim (( A \wedge ( B \vee C )) \Rightarrow ( A \vee B )) \vee A$. Applying the negation rule $\sim (X \Rightarrow Y) \equiv X \wedge \sim Y$: $p \equiv (( A \wedge ( B \vee C )) \wedge \sim ( A \vee B )) \vee A$. Using De Morgan's Law $\sim ( A \vee B ) \equiv \sim A \wedge \sim B$: $p \equiv (( A \wedge ( B \vee C )) \wedge ( \sim A \wedge \sim B )) \vee A$. Since $( A \wedge \sim A ) \equiv F$ (False),the expression simplifies to: $p \equiv ( F \wedge ( B \vee C ) \wedge \sim B ) \vee A \equiv F \vee A \equiv A$. Therefore,the negation of the statement is $\sim p \equiv \sim A$.

The negation of the statement $(( A \wedge ( B \vee C ))$ $\Rightarrow ( A \vee B ))$ $\Rightarrow A$ is

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$S_1$: If $-7$ is an integer,then $\sqrt{-7}$ is a complex number.
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