Prove that $A  B=A\cap B^c$
Answer
Let $x \in A \setminus B$. By definition $x \in A$ and $x \not \in B$, or equivalently $x \in B^c$; it follows that $x \in A \cap B^c$, so $A \setminus B \subseteq A \cap B^c $.
Let $x \in A \cap B^c$. By definition $x \in A$ and $x \in B^c $, or equivalently $x \not \in B$; it follows that $x \in A \setminus B$, so $A \cap B^c \subseteq A \setminus B$.
The double inclusion holds, so $A \setminus B = A \cap B ^c$.

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