Without using truth table show that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ ( ~ p ∧ q) - Mathematics and Statistics

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Sum

Without using truth table show that

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

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Solution

L.H.S

= (p ∨ q) ∧ (~ p ∨ ~ q)

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

= R.H.S

  Is there an error in this question or solution?
Chapter 1.1: Mathematical Logic - Q.5

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