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Exercise 1.4 · Q113

Q.Without using truth table, prove that: (p∨q)∧(p∨∼q)≡p(p \vee q) \wedge (p \vee \sim q) \equiv p

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(p∨q)∧(p∨∼q)≡p∨(q∧∼q)(p\vee q)\wedge(p\vee\sim q) \equiv p\vee(q\wedge\sim q) [Distributive law]

≡p∨c\equiv p\vee c [Complement law: q∧∼q≡cq\wedge\sim q\equiv c] …

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