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    Post intersection and union proof

    Prove by contradiction or otherwise that (A intersect B)=(A intersect C) and (A union B)=(A union C) if and only if B=C.



    SO... I know we have to show this both ways since it's an iff.
    Going the first way- Let (A intersect B)=(A intersect C) and (A union B)=(A union C). then for any x in A intersect B, x is in A intersect C. and for any x in A union b, x is in A union C. For x in A intersect B, x is in A and x is in B. for X in A intersect C, x is in A and x is in C. Since x is in A, B, C, and A intersect B equals A intersect C, then sets B and C must be equal.

    For A union B and A union C... is where I get a little confused... and also going the other way, letting B=C and proving these two things are true.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Laurali224 View Post
    Prove by contradiction or otherwise that (A intersect B)=(A intersect C) and (A union B)=(A union C) if and only if B=C.



    SO... I know we have to show this both ways since it's an iff.
    Going the first way- Let (A intersect B)=(A intersect C) and (A union B)=(A union C). then for any x in A intersect B, x is in A intersect C. and for any x in A union b, x is in A union C. For x in A intersect B, x is in A and x is in B. for X in A intersect C, x is in A and x is in C. Since x is in A, B, C, and A intersect B equals A intersect C, then sets B and C must be equal.

    For A union B and A union C... is where I get a little confused... and also going the other way, letting B=C and proving these two things are true.
    So A\cap B=A\cap C and A\cup B=A\cup C.

    Let x\in B then x\in A\cup B\implies x\in A\cup C. Suppose then that x\notin C then x\notin A\cap C and thus x\notin A\cap B\implies x\notin B. Contradiction.

    Similarly, let x\in C then x\in A\cup C\implies x\in A\cup B. Suppose that x\notin B then x\notin A\cap B\implies x\notin A\cap C and so x\notin C. Contradiction
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