set theory proof of $A\cap B = A \setminus(A\setminus B)$ Please proof that $A\cap B=A\setminus(A\setminus B)$ this really gave me sleepless night. I tried using the set intersection properties but I still got confused.
 A: To show $A\cap B = A -(A-B)$, you show two subsets relations
$A\cap B \supseteq A -(A-B)$ and $A\cap B \subseteq A -(A-B)$.
To show $A\cap B \supseteq A -(A-B)$, let $x \in A-(A-B)$, then $x\in A$ and $x\notin A-B$. Here if $x\notin B$, then $x\in A-B$ and contradiction arises, so $x\in B$. so $x\in A$ and $x\in B$, i.e. $x\in A\cap B$. So we show $A\cap B \supseteq A -(A-B)$.
To show $A\cap B \subseteq A -(A-B)$, let $x\in A\cap B$, then $x\in A$. Here if $x\in A-B$, then $x\notin B$ and contradiction arises, so $x\notin A-B$. Together we have $x\in A$ and $x\notin A-B$, i.e. $x\in A-(A-B)$. So $A\cap B \subseteq A -(A-B)$. Hence this completes the proof.
A: Notice that
$$\begin{align}A \setminus (A \setminus B) & = A \cap (A \cap B^c)^c = A \cap (A^c \cup B) \\ & = (A \cap A^c) \cup (A \cap B) = \varnothing \cup (A \cap B) = A \cap B.\end{align}$$
I have used two properties here - the de Morgan's laws for sets and the distributivity of $\cup$ and $\cap$.
A: Any element of $A$ is either also an element of $B$, or it is not.
so, $A\cap B$ means $x$ is an element of BOTH $A$ and $B$.
$A\setminus B$ means that $x$ is an element of $A$ BUT NOT $B$
$A\setminus(A\setminus B)$ means that $x$ is an element of $A$ BUT NOT $A\setminus B$, and so  $x$ is an element of $A$ BUT NOT an element of $A$ BUT NOT $B$.
And the clever bit is that 'BUT NOT an element of $A$ BUT NOT $B$' means 'BOTH $A$ and $B$', which is $A\cap B$.
A: $$A, B \subseteq E$$
$$\begin{align} A \setminus (A \setminus B) = \{o\in E  \;|\; (o \in A)\;\land\;(o \in (A \setminus B)^c) \} = \{o\in E  \;|\; (o \in A)\;\land\;(\neg (o \in A \;\land\;o \in B^c )) \} =  \{o\in E  \;|\; (o \in A)\;\land\;(o \in A^c \;\lor\;o \in B ) \} = \{o\in E  \;|\; (o \in A \;\land\; o \in A^c )\lor(o \in A\;\land\;o \in B ) \} = \{o\in E  \;|\; \text{⊥ (false)}\lor(o \in A\;\land\;o \in B ) \} = \{o\in E  \;|\;(o \in A\;\land\;o \in B ) \} = A \cap B \end{align} $$
