Formal proof to $ A \setminus {B}^\complement = A \cap B $

Question

I got the below equality from somes draws of Venn diagram, but I'm not able to build a formal proof to this. How to make it?

$ A \setminus {B}^\complement = A \cap B $

Answer 1

$A\setminus B^\complement=A\setminus (X\setminus B)=A\cap(X\setminus (X\setminus B))=A\cap B$.

Answer 2

In order to prove $$A \setminus {B}^\complement = A \cap B$$

You need to show two things.

Every element of $$A \setminus {B}^\complement$$ is an element of $$A \cap B$$

and vice versa.

Let $x$ be an element of $$A \setminus {B}^\complement$$

That means $x$ is in $A$ and $x$ is not in ${B}^\complement$

Therefore $x$ is in $A$ and $x$ is in $B$ that is $x\in A\cap B$

You can finish the rest easily.

Answer 3

Just do definitions: $A \cap B = \{x \in U| x \in A \land x \in B\}$.

$A \setminus B^c = \{x\in A| x \not \in B^c\} = \{x\in U|x \in A \land x \not \in B^c\} = \{x\in U|x \in A \land (\lnot (x\not \in B))\} = \{x \in U|x \in A \land x \in B\}$>

So $A\cap B$ and $A\cap B^c$ both describe the same set by definition.