Prove the subsequent Boolean expression:

(x∨y) ∧ (x∨~y) ∧ (~x∨z) = x∧z

Ans: In the following expression, LHS is equal to:  

(x∨y)∧(x∨ ~y)∧(~x ∨ z) = [x∧(x∨ ~y)] ∨ [y∧(x∨ ~y)] ∧(~x ∨ z)

= [x∧(x∨ ~y)] ∨ [y∧(x∨ ~y)] ∧(~x ∨ z)

= [(x∧x) ∨ (x∧~y)] ∨ [(y∧x)∨ (y∧~y)] ∧(~x ∨ z)

= [x ∨ (x∧~y)] ∨ [(y∧x)∨ 0] ∧(~x ∨ z)

= [x ∨ (y∧x)] ∧(~x ∨ z)  [x ∨ (x∧~y) =x]

= x ∧(~x ∨ z)   [x ∨ (x∧y) =x]

= [x ∧~x)] ∨ (x ∧ z)  [x ∨ (x∧~y) =x]

= 0 ∨ (x ∧ z) = (x ∧ z) = RHS