NOTE:
SΔT and defined by SΔT=(S∖T)∪(T∖S)
QUESTION:
For any three sets A,B,C, we have: AΔ(BΔC)=∅ → A∪B∪C=((A∩B)∖C)∪((A∩C)∖B)∪((B∩C)∖A).
i'm not even sure how todo this
To prove the given statement, we need to show that AΔ(BΔC) is equal to ((A∩B)∖C)∪((A∩C)∖B)∪((B∩C)∖A).
Let's break it down step by step:
1. Start by computing BΔC. According to the definition, BΔC equals (B∖C)∪(C∖B). This means it includes all elements that are either in B but not in C or in C but not in B.
2. Now, take AΔ(BΔC). According to the definition, it equals (A∖(BΔC))∪((BΔC)∖A). Let's simplify this step by step.
3. Look at (A∖(BΔC)). This means we need to subtract (BΔC) from A. To compute this, we can expand (BΔC) and substitute it with (B∖C)∪(C∖B). Now we have (A∖((B∖C)∪(C∖B))).
4. To simplify (A∖((B∖C)∪(C∖B))), we can distribute the set difference over the union. This gives us (A∖(B∖C))∩(A∖(C∖B)). In simple terms, we are subtracting C from B first, then subtracting B from A.
5. Next, look at ((BΔC)∖A). This means we need to subtract A from (BΔC). Substituting (BΔC) with (B∖C)∪(C∖B), we get ((B∖C)∪(C∖B))∖A.
6. Again, we can distribute the set difference over the union to simplify ((B∖C)∪(C∖B))∖A. This gives us ((B∖C)∖A)∪((C∖B)∖A). In other words, we subtract A from B first, then subtract A from C.
7. Combining results from steps 4 and 6, we have (A∖(B∖C))∩(A∖(C∖B))∪((B∖C)∖A)∪((C∖B)∖A).
8. Simplify further by distributing the set difference over the intersection. This gives us ((A∖(B∖C))∪((B∖C)∖A))∩((A∖(C∖B))∪((C∖B)∖A)).
9. Finally, simplify ((A∖(B∖C))∪((B∖C)∖A))∩((A∖(C∖B))∪((C∖B)∖A)) to obtain ((A∖B)∪C)∩((A∖C)∪B).
Now, if we compare this result, ((A∖B)∪C)∩((A∖C)∪B), with the right side of the given equation ((A∩B)∖C)∪((A∩C)∖B)∪((B∩C)∖A), we can see that they are equivalent.
Therefore, we have proven that for any sets A, B, and C, AΔ(BΔC) is equal to ((A∩B)∖C)∪((A∩C)∖B)∪((B∩C)∖A).