Prove by contradiction $[(A\cap B)−(B\cap C)]−(A\cap C)'=\phi$.

Question

Prove by contradiction that for all sets A, B, and C

$$[(A\cap B)−(B\cap C)]−(A\cap C)'=\phi$$

My teacher's hint is: If a set is assumed to be nonempty, then we can assert the existence of an element, x, in that set.

I started by using element identities on the left hand side but got stuck: $$[(A∩B)−(B∩C)]\cap(A∩C)$$ $$[(A∩B)\cap(B∩C)^{'}]\cap(A∩C)$$

Answer 1

I am assuming $(A\cap C)'$ is the complement of $(A\cap C)$.

Let $x\in [(A\cap B)\setminus (B\cap C)]\setminus (A\cap C)' $. Then $x\in [(A\cap B)\setminus (B\cap C)] $ and $x\notin (A\cap C)' $. Thus $x\in (A\cap B) $ , $x\notin (B\cap C)$ and $x\notin (A\cap C)' $.

Since $x\in (A\cap B) $, $x\in A$ and $x\in B$.

Since $x\notin (B\cap C)$ , $x\notin B$ or $x\notin C$.

Since $x\notin (A\cap C)'$, $x\in A\cap C$. That is $x\in A$ and $x\in C.$

Clearly the three statements above are contradictory.

Answer 2

Say you have an element $x$ in $(A \cap B) \setminus (B \cap C)$ intersected with $A \cap C$. Then $x$ is in $A \cap B$ and in $A \cap C$. So it is in $A \cap B \cap C$, and in particular in $ B \cap C$ : here's the contradiction.