Using truth table, prove that \(\sim p \wedge q \equiv (p \vee q) \wedge \sim p\)
Q.3 Using truth table, prove that \(\sim p \wedge q \equiv (p \vee q) \wedge \sim p\)
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Answer: Columns 4 and 6 of the truth table are identical, so the two statements are logically equivalent.
| \(p\) | \(q\) | \(\sim p\) | \(\sim p \wedge q\) | \(p \vee q\) | \((p \vee q) \wedge \sim p\) |
|---|---|---|---|---|---|
| T | T | F | F | T | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | F | F | F |
All entries in the columns of \(\sim p \wedge q\) and \((p \vee q) \wedge \sim p\) are the same. Hence \(\sim p \wedge q \equiv (p \vee q) \wedge \sim p\).
Doubtly AI
· 28 Oct 2026
AI-verified · Oct 2026
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