[Discrete Mathematics] Prove (AΔC)Δ(B\A) = (A∪B)ΔC.

TheAgileWarrior249

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This is from "How to Prove It", by Velleman.
The problem is asking me to prove that (AΔC)Δ(B\A) = (A∪B)ΔC
I expanded (AΔC)Δ(B\A) = [((AvC)^¬(A^C))v(B\A)]^¬[((AvC)^¬(A^C))^(B\A)].
After simplifying this answer, I should simply get (A∪B)ΔC = ((AvB)vC)^¬((AvB)^C). But I've attempted this many times, and did not reach this conclusion. Any help would be appreciated.. thanks.
*I know the proper form is x ∈ A∪B = (x ∈ A) v (x ∈ B), but I'm just writing A v B for short. Also, this is what I tried (sorry for the bad handwriting):
 
This is from "How to Prove It", by Velleman.
The problem is asking me to prove that (AΔC)Δ(B\A) = (A∪B)ΔC
Sorry to say that I cannot follow the notation you posted. I can give some notional help.
\(\displaystyle A\Delta C=(A\cap C^c)\cup(C\cap A^c)\)______, ____\(\displaystyle B\setminus A=B\cap A^c\)

\(\displaystyle (A\Delta C)\Delta (B\setminus A)=\left[(A\Delta C)\cap(B\setminus A)^c\right]\cup\left[(A\Delta C)^c\cap(B\setminus A)\right] \)

\(\displaystyle (B\setminus A)^c=B^c\cup A\)____,_____\(\displaystyle (A\Delta C)^c=(A^c\cup C)\cap(A\cup C^c)\)

Now see if you can expand and factor.
 
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