Symmetric difference is associative

We want to show that : A \triangle (B \triangle C) = (A \triangle B) \triangle C

Part I: TPT: LHS  \subset RHS

Proof of Part I: let x \in LHS

Then, x \in A \triangle (B \triangle C). By definition of symmetric difference,

x \in \{ A - (B \triangle C)\} \bigcup \{(B \triangle C) -A \}

Hence, x \in A, x \notin (B \triangle C), OR x \in B \triangle C, x \in A

That is, x \notin A \bigcap (B \triangle C).

Hence, x \notin A, and x \notin B \triangle C

Hence, x \notin A and x \in B \bigcap C.

Hence, x \in (B \bigcap C)-A.

Hence, x \in B, x \in C, but x \notin A.

Therefore, x \in B, but x \notin A \bigcap C. —- Call this I.

Consider y \in (A \triangle B) \triangle C.

Therefore, y \notin (A \triangle B) \bigcap C.

Therefore, y \notin C, and y \notin A \triangle B

Therefore, y \notin C, and y \in A \bigcap B.

Therefore, y \in (A \bigcap B)-C.

Hence, y \in A, y \in B, but y \notin C.

That is, y \in B, y \notin A \bigcap C. Call this II.

From I and II, LHS \subset RHS.

Part II: TPT: RHS \subset LHS.

Quite simply, reversing the above steps we prove part II.

QED.

Cheers,

Nalin Pithwa