Is $A-\left(A-B\right)=B$ true?

I am trying to prove that $A-\left(A-B\right)=B$ but I've arrived to this instead:

$A-\left(A-B\right)=A\cap \overline{\left(A\cap \overline{B}\right)}=A\cap \left(\overline{A}\cup B\right)=\left(A \cap \overline{A} \right)\cup \left( A \cap B\right)=\emptyset \cup \left( A \cap B\right)=\left( A \cap B\right) \neq B$

Is this correct? The first step is to use a logical equivalence, then de Morgan's law, distributive laws and other logical equivalences, I think I am following the rules. If this happens to be correct should the equal sign have to change to a subset sign so the statement holds?

Kind Regards

Solution 1:

You correctly proved $A-(A-B)=A\cap B$. Consequently, $A-(A-B)=B$ is true if and only if $A\cap B=B$. You may know that the latter is true if and only if $B\subseteq A$. Hence in general, $A-(A-B)=B$ is not true.

Solution 2:

$A-(A-B) A,B \subset E.$

$A-B=\{x \in A , x\notin B\}$

$A-(A-B)= \{x \in A, x \notin (A-B)\}$

$x \notin (A-B)\implies x\notin A $ or ($x \in A$ and $x \in B)$

We have $x \in A$, so we also must have $x \in B$.

So $x\in A-(A-B) \implies x \in A \cap B$.

Equivalence doesn't necessarily appply.

$A \cap B=B$ iff $B \subset A $.

Solution 3:

If $A$ is empty and $B$ is not, the equality is obviously false