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

Respuesta :

Answer:

Step-by-step explanation:

Consider the sets A and B

(A βˆ’ (A ∩ B)) ∩ (B βˆ’ (A ∩ B))

= (A ∩ (A ∩ B)c) ∩ (B ∩ (A ∩ B)c) by the set difference law

= (A ∩ (Ac ∩ B)c) ∩ (B ∩ (Ac ∩ B)c) by De Morgan's law

= {(A ∩ Ac) βˆͺ (A ∩ Bc)} ∩ {(B ∩ Ac) βˆͺ (B ∩ Bc)} by the distributive law

= {βˆ… βˆͺ (A ∩ Bc)} ∩ {(B ∩ Ac) βˆͺ βˆ…} by complementation

= {A ∩ Bc} ∩ {B ∩ Ac} by identity law

= (A ∩ Ac) ∩ (B ∩ Ac) by the associative law

= βˆ… ∩ βˆ… by complementation

= βˆ… by the universal bound law

Therefore, Β (A βˆ’ (A ∩ B)) ∩ (B βˆ’ (A ∩ B)) = βˆ…

Answer:

Considere los conjuntos A y B

(A βˆ’ (A ∩ B)) ∩ (B βˆ’ (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) βˆͺ (A ∩ Bc)} ∩ {(B ∩ Ac) βˆͺ (B ∩ Bc)} por la ley distributiva

= {βˆ… βˆͺ (A ∩ Bc)} ∩ {(B ∩ Ac) βˆͺ βˆ…} complementando

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

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

= βˆ… ∩ βˆ… complementando

= βˆ… por la ley universal consolidada

Step-by-step explanation: