Let $A$, $B$ and $C$ be sets. Prove that $(A - B)\cap C = (A\cap C) - B = (A\cap C) - (B\cap C)$ I was trying to prove the proposed relations as exhibited next:
$(A-B)∩ C \ =\
\sim((A-B) \implies \sim C) \ =\
\sim(\sim(A-B)∪ C)$
But I failed so far.

Let us prove the first claim first:
\begin{align*}
(A - B)\cap C & = A\cap B^{c}\cap C\\\\
& = (A\cap C)\cap B^{c}\\\\
& = (A\cap C) - B
\end{align*}
Now let us prove the second claim (in the reverse order):
\begin{align*}
(A\cap C) - (B\cap C) & = (A\cap C)\cap(B\cap C)^{c}\\\\
& = (A\cap C)\cap(B^{c}\cup C^{c})\\\\
& = (A\cap B^{c}\cap C)\cup(A\cap C\cap C^{c})\\\\
& = (A\cap B^{c}\cap C)\cup\varnothing\\\\
& = (A\cap C)\cap B^{c}\\\\
& = (A\cap C) - B
\end{align*}
