(A-B)∪(C-A)∪(C-B)=(A∪C)-(A∩B) | Mathematics | Page 1

slate epoch Dec 23, 2023, 9:07 AM

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i'm stuck showing this equality between sets.
(A-B)∪(C-A)∪(C-B) = (A∩B')∪(C∩A')∪(C∩B') = (A∩B')∪(C∩(A'∪B')) =
((A∩B') ∪ C)∩((A∩B')∪(A'∪B')) = ((A∩B') ∪ C)∩(B'∪A')
how should continue from here

jaunty trenchBOT Dec 23, 2023, 9:07 AM

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slate epoch Dec 23, 2023, 10:44 AM

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I know how to show left to right i'm just stuck showing from right side to the left side

vagrant nova Dec 23, 2023, 10:54 AM

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there is something inside you it's hard to explain

slate epoch Dec 23, 2023, 10:55 AM

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tell me more

vagrant nova Dec 23, 2023, 10:56 AM

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slate epoch Dec 23, 2023, 10:57 AM

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where did you get this clip of me

drowsy spoke Dec 23, 2023, 10:59 AM

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Lmao

vagrant nova Dec 23, 2023, 11:04 AM

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slate epoch I know how to show left to right i'm just stuck showing from right side to the l...

If you showed left to right, then you can basically copy your work and start backwards to show right to left

drowsy spoke Dec 23, 2023, 11:08 AM

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Yea he is right, why don’t you do that?

slate epoch Dec 23, 2023, 11:11 AM

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I feel so stupid rn.

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it didn't cross my mind to use the union of the empty set

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(A-B)∪(C-A)∪(C-B) = (A∩B')∪(C∩A')∪(C∩B') = (A∩B')∪(C∩(A'∪B')) = (∅U(A∩B'))∪(C∩(A'∪B')) = (A∩A')U(A∩B'))∪(C∩(A'∪B')) = (A∩(A'∪B'))∪(C∩(A'∪B')) = (A'∪B')∩(A∪C) = (A∪C)∩(A∩B)' = (A∪C) - (A∩B).

drowsy spoke Dec 23, 2023, 11:14 AM

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Do you still need help?

slate epoch Dec 23, 2023, 11:15 AM

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No thanks a lot.

vagrant nova Dec 23, 2023, 11:15 AM

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slate epoch Dec 23, 2023, 11:16 AM

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.close

#(A-B)∪(C-A)∪(C-B)=(A∪C)-(A∩B)