证明-基本概念

 1. 验证“分配律”

\[A\cap(B\cup C) = (A\cap B)\cup(A\cap C) \]

\[A\cup(B\cap C) = (A\cup B)\cap(A\cup C) \]

和DeMorgan律

\[A - (B\cup C) = (A - B)\cap (A - C) \]

\[A - (B\cap C) = (A - B)\cup (A - C) \]

 证明 

$$x\in A\cap(B\cup C)\iff x\in B\text{或}x\in C,\text{且}x\in A\iff x\in A\text{且}x\in B,\text{或}x\in A\text{且}x\in C\iff x\in (A\cap B)\cup(A\cap C)$$ $$x\in A\cup(B\cap C)\iff x\in A,\text{或}x\in B\text{且}x\in C\iff x\in A\text{或}x\in B,\text{且}x\in A\text{或}x\in C\iff x\in (A\cup B)\cap(A\cup C)$$ $$\begin{gather*} x\in A - (B\cup C)\iff x\in A,\text{且}x\notin B\cup C\iff x\in A,\text{且}x\notin B\text{且}x\notin C\iff\\ x\in A\text{且}x\notin B,\text{且}x\in A\text{且}x\notin C\iff x\in(A - B)\cap (A - C) \end{gather*}$$ $$\begin{gather*} x\in A - (B\cap C)\iff x\in A,\text{且}x\notin B\cap C\iff x\in A,\text{且}x\notin B\text{或}x\notin C\iff\\ x\in A\text{且}x\notin B,\text{或}x\in A\text{且}x\notin C\iff x\in(A - B)\cup (A - C) \end{gather*}$$

$\square$