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, $A \smallsetminus (B \cap C) = (A \smallsetminus B) \cup (A \smallsetminus C)$ .∎

2 comments

October 26, 2016 at 11:05 am

Anonymous says:

sorry, i cant understand Since x \notin (B \cap C) we have either x \notin B or x \notin C which
why this happens?

- May 15, 2021 at 12:39 pm
  
  Anonymous says:
  
  I have the same exact question.

Stumbling Robot

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

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