Proof using Element Chasing the associative law of symetric difference

Hi there,
I have to prove using Element Chasing that (A delta B) delta C = A delta ( B delta C)
I have the Venn drawn, and was told to start with:
let x be an element of (A delta B) delta C
if x is an element of (A delta B) union(and) C but not in x element (A delta B) intersection(or) C
(Thinking)

I am not just sure really where to go next in the proof. (Worried)
The help would be ever so appreciated. (Evilgrin)

suppose x is in (AΔB)ΔC.

this means x is in AΔB, or x is in C, but not BOTH (it's in exactly ONE of these sets).

case 1: x is in AΔB, x is not in C.

case 1a: x is in A, x is not in B, x is not in C.

since x is not in B, and not in C, x is not in BUC, and therefore not in BΔC, which is a subset of BUC.

thus in this case, x is in A, but not in BΔC, so x is in AΔ(BΔC).

case 1b: x is not in A, x is in B, x is not in C.

in this case, x is in B-C, which is a subset of BΔC (BΔC = (B-C).

so x is not in A, and is in BΔC, so x is in AΔ(BΔC).

case 2: x is not in AΔB, x is in C.

case 2a: x is is not in AUB, x is in C.

in this case, x is not in A, and not in B, but in C, so x is in C-B, and thus in BΔC, so x is not in A, but in BΔC, so is in AΔ(BΔC).

case 2b: x is in AUB, x is in C.

here, x is clearly in A∩B (since A∩B = (AUB) - (AΔB)). thus x is in A,B and C, which means x is in B∩C, which means x is not in BΔC.

thus, in this case, x is in A, but not in BΔC, so x is in AΔ(BΔC).

***************

that's HALF the proof: (AΔB)ΔC is a subset of AΔ(BΔC). you do the other half.