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Math: sets

syndra Mon 22 Jun 19:32:29 2020 No.11825802 View Reply Original Report

Quoted By: >>11825847 >>11825869 >>11826037 >>11828412

Is it possible to prove this set equation,or is the equation false

A\(B\C)=(A\B)U(AUC)

Anonymous

Anonymous Mon 22 Jun 2020 19:45:41 No.11825842 Report

Quoted By: >>11825849 >>11825889

x in LHS <=> (x in A) and (x not in B\C) <=> (x in A) and not (x in B and x not in C) <=> (x in A) and (x not in B or x in C) <=> (x in A and x not in B) or (x in A and x in C) <= > (x in A\B) or (x in A?B) <=> x in (A\B)U(A?B)
it's true if you wrote it wrong and LHS is (A\B)U(A?B)

Anonymous

Anonymous Mon 22 Jun 2020 19:46:16 No.11825847 Report

Quoted By:

>>11825802
Yes, just use logic (not, or, and)

Anonymous

Anonymous Mon 22 Jun 2020 19:46:54 No.11825849 Report

Quoted By:

>>11825842
>A?B
my bad, it should be A?C instead

Anonymous

Anonymous Mon 22 Jun 2020 19:52:39 No.11825869 Report

Quoted By:

>>11825802
take B = A and C = empty,
then A\(B\C) = A\A = empty
but (A\B) U (A U C) = A

syndra

syndra Mon 22 Jun 2020 19:58:28 No.11825889 Report

Quoted By: >>11825902

>>11825842
Thanks for showing your work, tho I didn't write it wrong it is indeed (A\B)U(AUC)

Anonymous

Anonymous Mon 22 Jun 2020 20:02:06 No.11825902 Report

Quoted By:

>>11825889
then it's not true, counter example already in the thread

syndra

syndra Mon 22 Jun 2020 20:09:24 No.11825930 Report

Quoted By:

Thanks for the help ^^

Anonymous

Anonymous Mon 22 Jun 2020 20:35:04 No.11826037 Report

Quoted By:

>>11825802
This is false. Consider an example in which C is disjoint from A. The set on the left will contain no elements from C. The set on the right will contain every element in C

Anonymous

Anonymous Tue 23 Jun 2020 14:17:41 No.11828412 Report

Quoted By:

>>11825802
counter example:
A = {}, B = {}, C = {1}
Then
A\(B\C) = {}
But
(A\B)U(AUC) = {1}

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