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

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Sum

Using the algebra of statement, prove that

[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p

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Solution

L.H.S.

= [p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p]

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

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

≡ [p ∧ (q ∨ r)] ∨ [~ (q ∨ r) ∧ p]         ....[De Morgan’s Law]

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

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

≡ p ∧ t          ......[Complement Law]

≡ p                .....[Identity 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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