Without using truth table, show that 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)

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Solution

L.H.S.

≡ p ↔ q

≡ (p → q) ∧ (q → p)

≡ (~p ∨ q) ∧ (~q ∨ p)

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

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

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

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

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

≡ R.H.S.

  Is there an error in this question or solution?
Chapter 1: Mathematical Logic - Exercise 1.9 [Page 22]

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