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?

All you need is a counterexample.
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.
