Prove that: $(A-B)\cap B=\phi$ Prove that: $(A-B)\cap B=\phi$.
My Attempt:
Let $x$ be an element of $(A-B)\cap B$. Then,
$$x\in (A-B)\cap B \iff x\in (A-B) \textrm {and} x\in B$$
$$\iff (x\in A \textrm {and} x\notin B) \textrm {and} x\in B$$.

Your approach is correct. No element can be in both $B$ and $B^c$. Formally we can write, by definition of set subtraction,
\begin{align*}
A - B = A \cap B^c.
\end{align*}
Since $(A \cap B^c) \cap B$ is a subset of $B^c \cap B = \emptyset$, we are done.
