Answer
Both sides contain all elements that belong to an **odd number** of the sets \(A, B, C\)
So:
\[
\boxed{(A \triangle B) \triangle C = A \triangle (B \triangle C)}
\]
✔️ **Proven** — symmetric difference is **associative**.
Work Step by Step
We are asked to prove the identity:
\[
(A \triangle B) \triangle C = A \triangle (B \triangle C)
\]
This is the **associative law** for symmetric difference.
---
### ✅ Step-by-Step Proof:
Recall the definition:
\[
A \triangle B = (A - B) \cup (B - A)
\]
We will prove associativity by using **element argument** — pick any \(x\) and show it belongs to both sides.
---
### 🔹 Let \(x \in (A \triangle B) \triangle C\)
We interpret symmetric difference as:
- \(x\) is in **exactly one** of the sets
So:
- \(x \in (A \triangle B) \triangle C\) means:
- \(x\) is in **exactly one** of \((A \triangle B)\) and \(C\)
But \(x \in A \triangle B\) means:
- \(x\) is in \(A\) or \(B\), but not both
So:
- \(x\) is in exactly one of \(A\), \(B\), and \(C\)
- That is, \(x\) belongs to an **odd number** of the sets \(A, B, C\)
---
### 🔹 Now let’s analyze \(x \in A \triangle (B \triangle C)\)
By same logic:
- \(x\) is in exactly one of \(A\) and \((B \triangle C)\)
- And \(x \in B \triangle C\) means: in **exactly one** of \(B\), \(C\)
So again:
- \(x\) is in an **odd number** of \(A, B, C\)
