# 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

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