Check the distributive laws for \(\cup\) and \(\cap\) and DeMorgan’s laws.
Solution
For any element \(x\), membership in a union corresponds to logical “or” and membership in an intersection corresponds to logical “and”. Hence \[ \begin{aligned} x\in A\cap(B\cup C) &\iff x\in A\ \text{and}\ (x\in B\ \text{or}\ x\in C)\\ &\iff (x\in A\cap B)\ \text{or}\ (x\in A\cap C)\\ &\iff x\in (A\cap B)\cup(A\cap C), \end{aligned} \] so \[ A\cap(B\cup C)=(A\cap B)\cup(A\cap C). \] Similarly, \[ A\cup(B\cap C)=(A\cup B)\cap(A\cup C). \] These are the two distributive laws.
For DeMorgan’s laws, \[ \begin{aligned} x\in A-(B\cup C) &\iff x\in A,\ x\notin B,\ x\notin C\\ &\iff x\in(A-B)\cap(A-C), \end{aligned} \] so \[ A-(B\cup C)=(A-B)\cap(A-C). \] Likewise, \[ A-(B\cap C)=(A-B)\cup(A-C). \] Equivalently, relative to a fixed universe \(U\), \[ (B\cup C)^c=B^c\cap C^c, \qquad (B\cap C)^c=B^c\cup C^c. \]