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# Prove distributive laws for unions and intersections of sets

Prove that and .

Proof (). Let be an arbitrary element of . This means that and . Further, since we know or . But then, this implies or depending on whether or , respectively. Then, since or , we have . Thus, .
For the opposite inclusion, we let be an arbitrary element of . This means that or . If then and , while if we have and . Thus, we have no matter what, and either or . Since or , we know . Since we already had that no matter what, we now have . Therefore, .
Hence, .∎

Proof (). Let be an arbitrary element of . This means or . If then and . Hence, . Otherwise, if then and . Hence, and ; therefore, . Therefore, .
For the reverse inclusion, let be an arbitrary element of . This means and . This implies that or and (since if then the fact that and means must be in both and ). If , then . On the other hand, if and , then . Hence, . Therefore, .
Therefore, .∎

### One comment

1. David He says:

Hey Rori,

Noticed that on the second proof in the line “then x \in B and c \in C.” should be x \in C

Cheers!