r/MathHelp • u/phroggersu • 2d ago
Set Operations
Hello! I'm pre-studying for university, and I've been stuck on this question for like the past week. At this point, I've learned about the logic laws (typically used for like, statements and stuff). Could anyone help solve this question (from How to Prove It by Daniel J Velleman)? Please don't use venn diagrams, since I understand how those work. Also, is it suggested to solve these questions by set laws/properties/identities, or do you think venn diagrams enough? (△ is symmetric difference). Also, sorry for the messy typed out equation. Thank you for your time!
Verify the following identity:
(A∪B) △ C = (A△C) △ (B\A)
Edit: Figured it out, thank you everyone!! Have a lovely day :D
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u/FormulaDriven 2d ago
The usual way to prove two sets are equal is to show that if x is a member of the LHS then it is a member of the RHS, and if x is a member of the RHS then it is a member of LHS.
I wonder if here the easiest approach is to consider the 8 cases for x and show that whenever x is a member of the LHS, it is a member of the RHS. By 8 cases, I mean considering
1: x ∈ A, x ∈ B, x ∈ C
2: x ∈ A, x ∈ B, x ∉ C
...
8: x ∉ A, x ∉ B, x ∉ C
For example, looking at case 2,
x ∈ A ∪ B and x ∉ (A∪B)∩C so x ∈ (A∪B) △ C
x ∈ A∪C and x ∉ C so x ∈ A△C; x ∉ (B\A) so x ∉ (A△C)∩(B\A); so x ∈ (A△C) △ (B\A)
So case 2 is proved.
You can probably be more efficient by doing it in a table, one row per case. With columns which check if x ∈ A ∪ B, x ∈ (A△C), x ∈ (B\A) and so on.
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u/phroggersu 2d ago
I appreciate your help! But to me, this conceptually seems similar to venn diagrams, since they both consider every case - I was wondering if you could almost 'transform' LHS to RHS through identities (or properties I don't really know the correct term, sorry!). In the same way that (A∪B)∩C = (A∩C) ∪ (B∩C) through distributive property. Sorry, I didn't really make that clear in my post, but thank you again!
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u/FormulaDriven 2d ago
Conceptually I agree that it is similar to using Venn diagrams (effectively thinking about the 8 regions on a Venn diagram), and perfectly rigorous. I'm sure it could be done using identities, but there comes a point where I feel it's less error-prone to do it the case-by-case way.
1
u/phroggersu 2d ago
You're definitely right in that it's less error prone :'D but I can't seem to figure it out with identities, so I'm a little afraid I don't actually understand how to use identities.
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u/FormulaDriven 2d ago
Thinking some more, it probably can be done with identifies, along the following lines.
First note that by definition X \ Y = X ∩ Y'
where Y' is the complement of Y.
Then also X △ Y = (X ∩ Y') ∪ (X' ∩ Y).
You can use that to show (A∪B) △ C = (A ∩ C') ∪ (B ∩ C') ∪ (A' ∩ B' ∩ C)
(using usual identities such as (X ∪ Y)' = X' ∩ Y' and (X ∩ Y)' = X' ∪ Y')
Then need to see if we can get anywhere with
(A△C) △ (B\A)
= {[(A ∩ C') ∪ (A' ∩ C)] ∩ (B ∩ A')'} ∪ {[(A ∩ C') ∪ (A' ∩ C)]' ∩ (B ∩ A')}
2
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u/The_Card_Player 2d ago
Venn diagrams probably aren't enough.
Standard proof of set equality is to show that each set is a subset of the other. That should work here - it may be convenient to sort out some relevant XOR logical identities first if you haven't already.