# How to prove that the set-theoretic difference operation $\setminus$ cannot be defined through $\cap$ and $\cup$

How does one go about proving that the set-theoretic difference operation $$\setminus$$ cannot be defined through the operations $$\cap$$ and $$\cup$$?

My thoughts: I first assumed $$A$$ and $$B$$ are two non-disjoint non-empty sets since if they are disjoint and non-empty, then we have that $$A\setminus B= A =A\cap A=(A\cap B) \cup A$$. Therefore we have defined, in this case, $$\setminus$$ in terms of $$\cap$$ and $$\cup$$.

Next, I drew three Venn diagrams for $$A\cap B$$, $$A\cup B$$ and $$A\setminus B$$ and made the observation that $$A \setminus B$$ involves an exclusion of a part of $$A$$. When I looked at the definitions, I could see this clearer:

$$A\cap B = \{x\mid (x\in A) \land (x\in B)\}$$ $$A\cup B = \{x\mid (x\in A) \lor (x\in B)\}$$ $$A\setminus B = \{x\mid (x\in A) \land \mathbf{(x\notin B)}\}$$

From this, I decided to conclude that since the intersection and union operations have the condition that $$(x\in B)$$ whereas the set difference operation requires $$(x\notin B)$$, we cannot define set difference through the operations union and intersection only.

This is as far as I could go. How can I prove this formally?

Let $$A = B = \{ 0 \}$$. Then all of the sets $$A,~ B,~ A \cup A,~ A \cap A,~ A \cup B,~ A \cap B,~ B \cup A,~ B \cap A,~ B \cup B,~ B \cup B$$ are equal to $$\{ 0 \}$$, and so any set built out of $$A$$, $$B$$ and the operations $$\cup$$ and $$\cap$$ is equal to $$\{ 0 \}$$.
But $$B \setminus A = \varnothing \ne \{ 0 \}$$, and so the set difference operator $$\setminus$$ can't be expressed in terms of $$\cup$$ and $$\cap$$ alone.
The union and intersection operators are monotone. If $$A\subseteq A'$$ and $$B\subseteq B'$$ then $$A\cup B\subseteq A'\cup B'$$ and $$A\cap B\subseteq A'\cap B'$$. If $$f(A,B)$$ is a formal combination of $$A$$s and $$B$$s with unions and intersections, then it is monotone: $$f(A,B)\subseteq f(A',B')$$ under the above assumptions. But $$A\setminus B$$ is not a monotone operation.