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).

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