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Using the algebra of statement, prove that (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q) - Mathematics and Statistics

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प्रश्न

Using the algebra of statement, prove that

(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)

योग

उत्तर

L.H.S.

= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)

≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)]    ....[Associative Law]

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

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

≡ (p ∧ q) ∨ (~q ∧ t)      .....[Complement Law]

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

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

≡ (p ∨ ~ q) ∧ t              ....[Complement Law]

≡ p ∨ ~ q                   .....[Identity Law]

= R.H.S.

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अध्याय 1: Mathematical Logic - Exercise 1.9 [पृष्ठ २२]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
अध्याय 1 Mathematical Logic
Exercise 1.9 | Q 2.2 | पृष्ठ २२

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