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Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).

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प्रश्न

Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).

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उत्तर

L.H.S. = (p ∨ q) ∧ (~ p ∨ ~ q)

≡ [(p ∨ q)] ∧ [(p ∨ q) ∧ ~ q]      ...[Distributive law]

≡ [(p ∧ ~ p) ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ (q ∧ ~ q)]        ...[Distributive law]

≡ [F ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ F]    ...[Complement law]

≡ (q ∧ ~ p) ∨ (p ∧ ~ q)   ...[Identity law]

≡ (p ∧ ~ q) ∨ (~ p ∧ q)    ...[Commutative law]

≡ R.H.S.

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Notes

The question is modified.

  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Mathematical Logic - Exercise 1.9 [पृष्ठ २२]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 1 Mathematical Logic
Exercise 1.9 | Q 2.3 | पृष्ठ २२

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