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AU(B-C) = {x/ x∈A v x∈(B-C)}
= {x/ x∈A v (x∈B ∧ ¬x∈C)}
= {x/ (x∈A v x∈B) ∧ (x∈A ∨ ¬x∈C)}
= {x/ (x∈A v x∈B) ∧ ¬(¬x∈A ∧ x∈C)}
= {x/ (x∈A v x∈B) ∧ ¬(x∈C ∧ ¬x∈A)}
={x/ x∈(AUB) ∧ ¬x∈(C-A)}
=(AUB) - (C-A)
2da forma
Se sabe que B-C = B ∩ C⁽
AU(B-C) = AU(B ∩ C⁽)
= (AUB)∩(AUC⁽)
=(AUB)∩(A⁽ ∩ C)⁽
=(AUB) ∩ (C-A)⁽
=(AUB)-(C-A)
DEMOSTRADO