Prove that $A \smallsetminus (A \smallsetminus B) = A \cap B$

Question

$A$ and $B$ are any sets, prove that $A \smallsetminus (A \smallsetminus B) = A \cap B.$ This formula makes sense when represented on a Venn diagram, but I am having trouble with proving it mathematically.

I have tried letting $x$ be an element of $A$ and continue from there, but it doesn't seem to work out as a valid proof anyways.

Could anyone please point me in the right direction?

Many thanks.

Answer 1

Let $x \in A \setminus (A \setminus B).$

Then $x$ is an element such that $x \in A$ and $x \notin A \setminus B$. But if $x \notin A\setminus B$, with some additional work, we realize this implies that $x \in A$ and $ x \in B$. So $x \in A \cap B$.

Vice-versa: let $x \in A \cap B$ so $x \in A$ and $x \in B$. This implies that $x \notin A \setminus B$. But given that $x \in A$ and $x \notin A \setminus B$ this implies that $x \in A \setminus (A \setminus B)$.

Answer 2

Notice that \begin{eqnarray*} x\in A\setminus(A\setminus B) &\Leftrightarrow& (x\in A)\wedge \neg(x\in A\setminus B)\\ &\Leftrightarrow & (x\in A)\wedge \neg((x\in A)\wedge \neg (x\in B))\\ &\Leftrightarrow & (x\in A)\wedge ((x\notin A)\lor(x\in B))\\ &\Leftrightarrow & ((x\in A)\wedge (x\notin A))\lor ((x\in A)\wedge(x\in B))\\ &\Leftrightarrow & (x\in A)\wedge (x\in B)\\ &\Leftrightarrow & x\in A\cap B. \end{eqnarray*}

Answer 3

Did you know that

$$A \smallsetminus B = A \cap \overline{B} \;\;?$$

$\begin{align} A \smallsetminus (A \smallsetminus B) &= A \smallsetminus (A \cap \overline{B}) \\ \\ &= A \cap \overline{ (A \cap \overline{B})}\\ \\ &= A \cap (\overline{A} \cup B)\\ \\ &= (A \cap \overline{A} )\cup (A \cap B) \\ \\ &= A \cap B \end{align}$

Answer 4

Let $x\in A\cap B$.
Then $x\in A$ and $x\in B$.
Then $x\not \in A\setminus B$,
so $x\in A\setminus (A\setminus B)$.

Conversely, if $x\in A\setminus (A\setminus B)$,
then $x\in A$ and $x\not \in (A\setminus B)$.
So $x\in A$ and $x\in B$.
That is $x\in A\cap B$...