Expanding $A \setminus B = A \setminus C$ using the definition $x \in A \setminus B \;\equiv\; x \in A \land \lnot(x \in B)$ and then simplifying results in $ \begin{align} & A \setminus B = A \setminus C \\ \equiv & \;\;\;\;\;\text{"extensionality"} \\ & \langle \forall x :: x \in A \setminus B \equiv x \in A \setminus C \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\setminus$, twice"} \\ & \langle \forall x :: x \in A \land \lnot(x \in B) \equiv x \in A \land \lnot(x \in C) \rangle \\ \equiv & \;\;\;\;\;\text{"logic: move common conjunct out of $\equiv$"} \\ & \langle \forall x :: x \in A \Rightarrow (\lnot(x \in B) \equiv \lnot(x \in C)) \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify by negating both sides of $\equiv$"} \\ & \langle \forall x :: x \in A \Rightarrow (x \in B \equiv x \in C) \rangle \\ \end{align} $
Now, in the same way expand $A \cap B = A \cap C$ using the definition $x \in A \cap B \;\equiv\; x \in A \land x \in B$ and compare the results.