Show $A \cap (B-C) = (A \cap B) - (A \cap C)$ using an alternative proof

Question

I am trying to prove that $$A \cap (B-C) = (A \cap B) - (A \cap C)$$ holds for any sets $A,B,C$. This is equal to $$A \cap (B-C) = (A \cap B)\cap\color{red}{(A\cap C)'}.$$

I have seen a similar question: Proving : $A \cap (B-C) = (A \cap B) - (A \cap C)$ but I am trying to use another approach, if it is correct.

We pick $$x\in A\cap(B-C)\to x\in A\wedge(x\in B\wedge x\in C')\underbrace{\to}_{\text{Identity}}(x\in A\wedge x\in B)\wedge x\in C'\wedge(x\in A\vee x\in A').$$ Then the first part is the same as $x\in A\cap B$. So we focus on: $x\in C'\wedge(x\in A\vee x\in A')$.

A friend of mine distributed so I ask you if from $$(x\in C'\wedge x\in A)\vee(x\in C'\vee x\in A')\tag{1}$$ we can conclude $$\color{red}{x\in(A\cap C)'}.$$

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From $(1)$ we get that $x\in A-C\vee x\in(A\cap C)'$, but I can't proceed further.

I know the following laws:

• Addition: $p\to p\vee q$. EDIT. Since by this property we can finish quickly let me assume that we can't use Addition law.
• Simplification: $p\wedge q\to p$.

Answer 1

Notice that
1. $ x \in C' \rightarrow x \in (A \cap C)' $ and
2. $ x \in A' \rightarrow x \in (A \cap C)' $

What you need follows from these two (try to see how).

Answer 2

(A Inter B) - (A Inter C)

= (A Inter B) Inter (A Inter C)'

= (A Inter B) Inter ( A' Union C')

( by De Morgan)

= [(A Inter B ) Inter A'] Union [ ( A Inter B) Inter C']

( by distributivity)

= Empty Set Union [ A Inter ( B Inter C')]

( by associativity of intersection)

= A Inter ( B Inter C')

= A Inter ( B - C )