Question
Using truth table prove that : $\sim p \wedge q=(p \vee q) \wedge \sim p$
| I | II | III | IV | V | VI |
| 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 |
From column (IV) and (VI), we get
∴ ∼p ˄ q ≡ (p ˅ q) ˄ ∼p
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