> [!NOTE] Theorem > Let $A,B$ be [[Sets|sets]]. Let $A-B$ denote a [[Set Difference|difference]]. Let $A \cap B$ denote their [[Set Intersection|intersection]]. Then $A-B$ and $A\cap B$ form a [[Partition of a Set|partition]] of $A.$ **Proof**: We have $A-B = \{ a\in A: a\not \in B \}$ and $A \cap B= \{ a\in A: a\in B \}.$ Thus $(A\cap B)\cap (A-B)= \{ a\in A: a\in B \land a \not \in B \}=\emptyset.$Also $(A\cap B)\cup (A-B)=\{ a\in A: b\in B \lor b\not \in B \} = A.$