Showing $a \setminus (a \setminus b) = a \cap b$ using only set notation.

Question

I need to prove that $a \setminus (a \setminus b) = a \cap b$ only through set notations.

I have reached the fact that
$a \setminus (a \setminus b)$ = {x | x $\in$ A $\land$ x $\notin$ (A $\setminus$ B}

Which I then simplify using the De Morgan' formula to {x | x $\in$ A $\land$ (x $\notin$ A $\lor$ x $\in$ B)}.

It is evident that this is the same as A $\cap$ B (by constructing diagrams and truth tables). But how do I get this expression to {x | x $\in$ A $\land$ x $\in$ B}? Please help!

Answer 1

Equivalent are:

• $x\in A\wedge(x\notin A\vee x\in B)$
• $(x\in A\wedge x\notin A)\vee(x\in A\wedge x\in B)$
• $x\in A\wedge x\in B$

Answer 2

a - (a - b) = a - (a $\cap$ b$^c$) =
a - (a$^c$ $\cup$ b)$^c$ = a $\cap$ (a$^c$ $\cup$ b)
= (a $\cap$ a$^c$) $\cup$ (a $\cap$ b) = a $\cap$ b