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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
Provided a homogeneous system of equations (2), we will have one of the two probabilities for the number of solutions. 1. Accurately one solution, the trivial solution 2.
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