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Set Equality

Date: 10/12/98 at 07:28:32
From: lior
Subject: Groups algebra (discrete math)

I have to prove the following equation, using groups algebra only. By 
groups algebra, I mean: DeMorgan's rules, absorption rules, and so 
on). The equation is:

   (A-B)-C = (A-C)-(B-C)

Thank you very much for your help.

Date: 10/12/98 at 10:12:21
From: Doctor Anke
Subject: Re: Groups algebra (discrete math)

Hello Lior,

I assume that by A, B, and C, you meant sets (it is not necessary for 
them to have any special property). So we will do this proof in
two parts: first we will show that (A-B)-C is a subset of (A-C)-(B-C),
and then we will show that (A-C)-(B-C) is a subset of (A-B)-C. Showing 
these two things is equivalent to showing equality between the two 

1. (A-B)-C subset of (A-C)-(B-C)
To show this, we choose an arbitrary x in (A-B)-C. Now we have to show 
that x is in (A-C)-(B-C). Since x is in (A-B)-C, it holds that x is in 
A, x is not in B, and x is not in C. Therefore, it is also true that 
x is in A-C and x is not in B-C, so x is in (A-C)-(B-C), as well.

2. (A-C)-(B-C) subset of (A-B)-C
We proceed as in the above case. Let x be an arbitrary element in 
(A-C)-(B-C). From this we can conclude that x is in A-C and x is not in 
B-C. Since x is not in C and x is not in B-C, x is also not in B. So we 
know: x is in A, x is not in B, and x is not in C. Putting this 
together gives us that x is in (A-B)-C.

I hope this made things clearer. I didn't state each time which law I 
used, but it shouldn't be too difficult to find this out and write it 
down. If you have further questions, please feel free to write again.

- Doctor Anke, The Math Forum   
Associated Topics:
High School Discrete Mathematics
High School Sets

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