Proof. We will prove that if \(A\), \(B\), and \(C\) are sets, then \((A \cap B) \times C = (A \times C) \cap (B \times C)\) by using element chasing to show that each set is a subset of the other.

First, let \((x,y)\) be an element of \((A \cap B) \times C)\). So \(x \in A \cap B\) and \(y \in C\), i.e., \(x \in A\) and \(x \in B\). Since \(x \in A\) and \(y \in C\), \((x,y) \in A \times C\). Furthermore, since \(x \in B\) and \(y \in C\), \((x,y)\) is also in \(B \times C\). Since \((x,y) \in A \times C\) and \((x,y) \in B \times C\), we conclude that \((x,y) \in (A \times C) \cap (B \times C)\).

Now, suppose \((x,y) \in (A \times C) \cap (B \times C)\). Then \((x,y) \in A \times C\) and \((x,y) \in B \times C\). For this to be true, \(x\) must be a member of \(A\) and \(B\), and \(y\) must be in \(C\). Since \(x \in A\) and \(x \in B\), \(x \in A \cap B\). Finally, since \(x \in A \cap B\) and \(y \in C\), the pair \((x,y) \in (A \cap B) \times C\).

Having shown that each product is a subset of the other, we have shown that if \(A\), \(B\), and \(C\) are sets, then \((A \cap B) \times C = (A \times C) \cap (B \times C)\). □