I am asked to prove that, if $A, B$ and $C$ are sets, then $$A-(B\cap C)=(A-B)\cap(A-C).$$ However, I think that either I have made an error in my working, or the wording of the problem contains a typographical error.
My working so far is as follows: $$\begin{align}x\in(A-(B\cap C)) &\iff (x\in A)\wedge(x\not\in(B\cap C))\\ &\iff(x\in A)\wedge((x\not\in B)\vee(x\not\in C))\\ &\iff((x\in A)\wedge(x\not\in B))\vee((x\in A)\wedge(x\not\in C))\\ &\iff(x\in(A-B))\vee(x\in(A-C))\\ &\iff x\in((A-B)\cup(A-C)) \end{align}$$
First line to second line: by De Morgan's Law.
Second line to third line: by distributivity of set intersection over set union.
Third line to fourth line: by definitions of intersection and set difference.
If my working is correct, then I have shown that $$A-(B\cap C)=(A-B)\cup(A-C).$$ Am I correct?
$\endgroup$ 72 Answers
$\begingroup$You are right. As an other proof:
$$A-(B\cap C)=A\cap (B\cap C)^c= A\cap (B^c\cup C^c)=$$
$$(A\cap B^c)\cup (A\cap C^c)=(A-B)\cup(A-C).$$
$$x \in (A-B) \cap (A-C) \Rightarrow (x \in A \wedge x \notin B) \wedge (x \in A \wedge x \notin C) \\ \Rightarrow x \in A \wedge x \notin B \wedge x \notin C \Rightarrow x \in A-(B \cup C) \Rightarrow x \in A-(B \cap C)$$
$$x \in A-(B \cap C) \Rightarrow x \in A \wedge x \notin B \cap C \Rightarrow x \in A-(B \cap C)$$
$\endgroup$ 5
