If A, B,and C are three sets,then A - (B cap | vedclass

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(A) By using De Morgan's Law for sets,we know that the complement of an intersection is the union of the complements: $(B \cap C)^c = B^c \cup C^c$. G

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$(A - B) \cup (A - C)$

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(A) By using De Morgan's Law for sets,we know that the complement of an intersection is the union of the complements: $(B \cap C)^c = B^c \cup C^c$. Given the expression $A - (B \cap C)$,we can rewrite this as $A \cap (B \cap C)^c$. Substituting the De Morgan's Law result,we get $A \cap (B^c \cup C^c)$. Using the distributive property of sets,this becomes $(A \cap B^c) \cup (A \cap C^c)$. Since $A \cap B^c = A - B$ and $A \cap C^c = A - C$,the expression simplifies to $(A - B) \cup (A - C)$.

If $A, B$,and $C$ are three sets,then $A - (B \cap C)$ is equal to

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