Show that $A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )$

You can either write a table with all the possible locations of $x$ with respect to $A,B,C$ or you can use these identities (which you should prove, as they are most useful):

• $(A\cup B)^c = A^c\cap B^c$

• $A\setminus B = A\cap B^c$

• $A\cup (B\cap C) = (A\cup B)\cap(A\cup C)$

• $A\cap (B\cup C) = (A\cap B)\cup(A\cap C)$

Now we can proceed:

$$ A\setminus(B\setminus C) = A\setminus (B \cap C^c) = A\cap (B\cap C^c)^c = A\cap(B^c\cup C) = (A\cap B^c)\cup(A\cap C) $$

Where $\cdot ^c$ is the complement.

Remember $x\in B\setminus C \iff x\in B\land x\not\in C$. Thus $x\not\in B\setminus C\iff x\not\in B\lor x\in C$ by de Morgan's laws.

So for some element $x$, $$ \begin{align*} x\in A\setminus (B\setminus C) &\iff x\in A \land (x\not\in B\setminus C)\\ &\iff x\in A \land (x\not\in B\lor x\in C). \end{align*} $$ But $$ \begin{align*} x\in (A\setminus B)\cup (A\cap C) &\iff x\in A\setminus B\lor x\in A\cap C\\ &\iff (x\in A\land x\not\in B)\lor (x\in A\land x\in C)\\ &\iff x\in A\land (x\not\in B\lor x\in C). \end{align*} $$

So you get $A \setminus ( B \setminus C ) = ( A \setminus B) \cup ( A \cap C )$.