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

[TheAgileWarrior249]

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):

 

[pka]

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.