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Prove that the complement of an intersection is the union of the complements

Prove that A \smallsetminus (B \cap C) = (A \smallsetminus B) \cup (A \smallsetminus C).


Proof. First, let x be any element in A \smallsetminus (B \cap C). By definition of complement, this means that x \in A and x \notin (B \cap C). Since x \notin (B \cap C) we have either x \notin B or x \notin C which implies, coupled with the fact that x is in A, means x \in (A \smallsetminus B) or x \in (A \smallsetminus C), respectively. Since x is in at least one of these, x is in the union (A \smallsetminus B) \cup (A \smallsetminus C). Therefore, A \smallsetminus (B \cap C) \subseteq (A \smallsetminus B) \cup (A \smallsetminus C).
For the reverse inclusion, let x be an arbitrary element of (A \smallsetminus B) \cup (A \smallsetminus C). Then, either x \in (A \smallsetminus B) or x \in (A \smallsetminus C).
If x \in (A \smallsetminus B), then x \in A and x \notin B; hence, x \notin (B \cap C). Therefore, x is in A \smallsetminus (B \cap C). On the other hand, if x \in (A \smallsetminus C), then x \in A and X \notin C; hence, x \notin (B \cap C). This again implies x is in A \smallsetminus (B \cap C). Therefore, (A \smallsetminus B) \cup (A \smallsetminus C) \subseteq A \smallsetminus (B \cap C).
Hence, A \smallsetminus (B \cap C) = (A \smallsetminus B) \cup (A \smallsetminus C).∎

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