Question:
The compound statement \([\sim (p\rightarrow q)\vee \sim (p\wedge r)]\wedge p\) is logically equivalent to:
The compound statement \([\sim (p\rightarrow q)\vee \sim (p\wedge r)]\wedge p\) is logically equivalent to:
Show Hint
Negate the implication and the conjunction, then distribute the conjunction with p over the disjunction.
Updated On: Nov 12, 2026
- \(p\wedge q\wedge r\)
- \(\sim p\vee \sim q\vee \sim r\)
- \(p\wedge (\sim q\vee \sim r)\)
- \(p\wedge \sim (q\wedge \sim r)\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
We use \(\sim(p\rightarrow q)\equiv p\wedge\sim q\) and De Morgan's law \(\sim(p\wedge r)\equiv\sim p\vee\sim r\).
Step 2: Rewrite the bracket
\(\sim(p\rightarrow q)\vee\sim(p\wedge r)\equiv(p\wedge\sim q)\vee(\sim p\vee\sim r)\).
Step 3: Take conjunction with p
\([(p\wedge\sim q)\vee\sim p\vee\sim r]\wedge p\).
Distribute: \((p\wedge\sim q)\vee(\sim p\wedge p)\vee(\sim r\wedge p)\).
Step 4: Simplify
Since \(\sim p\wedge p\) is a contradiction, the result is \((p\wedge\sim q)\vee(p\wedge\sim r)\equiv p\wedge(\sim q\vee\sim r)\).
Final Answer:
The statement is equivalent to p and (not q or not r). \[ \boxed{\text{(C) }p\wedge(\sim q\vee\sim r)} \]
We use \(\sim(p\rightarrow q)\equiv p\wedge\sim q\) and De Morgan's law \(\sim(p\wedge r)\equiv\sim p\vee\sim r\).
Step 2: Rewrite the bracket
\(\sim(p\rightarrow q)\vee\sim(p\wedge r)\equiv(p\wedge\sim q)\vee(\sim p\vee\sim r)\).
Step 3: Take conjunction with p
\([(p\wedge\sim q)\vee\sim p\vee\sim r]\wedge p\).
Distribute: \((p\wedge\sim q)\vee(\sim p\wedge p)\vee(\sim r\wedge p)\).
Step 4: Simplify
Since \(\sim p\wedge p\) is a contradiction, the result is \((p\wedge\sim q)\vee(p\wedge\sim r)\equiv p\wedge(\sim q\vee\sim r)\).
Final Answer:
The statement is equivalent to p and (not q or not r). \[ \boxed{\text{(C) }p\wedge(\sim q\vee\sim r)} \]
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