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Simplify the given expression. Cite a property from Theorem 6.2.2 for each step. (A − (A ∩ B)) ∩ (B − (A ∩ B)) Let A and B be any sets. Then (A − (A ∩ B)) ∩ (B − (A ∩ B)) = = (A ∩ (A ∩ B)c) ∩ (B ∩ (A ∩ B)c) = A ∩ ((A ∩ B)c ∩ (B ∩ (A ∩ B)c)) = A ∩ (((A ∩ B)c ∩ B) ∩ (A ∩ B)c) = A ∩ ((B ∩ (A ∩ B)c) ∩ (A ∩ B)c) = A ∩ (B ∩ ((A ∩ B)c ∩ (A ∩ B)c)) = A ∩ (B ∩ (A ∩ B)c) = (A ∩ B) ∩ (A ∩ B)c = ∅

User Be Kind
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Answer:

Considere los conjuntos A y B

(A − (A ∩ B)) ∩ (B − (A ∩ B))

= (A ∩ (A ∩ B)c) ∩ (B ∩ (A ∩ B)c) por la ley de diferencia establecida

= (A ∩ (Ac ∩ B)c) ∩ (B ∩ (Ac ∩ B)c) por la ley de De Morgan

= {(A ∩ Ac) ∪ (A ∩ Bc)} ∩ {(B ∩ Ac) ∪ (B ∩ Bc)} por la ley distributiva

= {∅ ∪ (A ∩ Bc)} ∩ {(B ∩ Ac) ∪ ∅} complementando

= {A ∩ Bc} ∩ {B ∩ Ac} por ley de identidad

= (A ∩ Ac) ∩ (B ∩ Ac) por la ley asociativa

= ∅ ∩ ∅ complementando

= ∅ por la ley universal consolidada

Explanation:

User Koppor
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