Fdtxjr

In this exercise sheet (German) there is the following problem: Prove that $(A cup B) setminus (A cap B) = (A setminus B) cup (B setminus A)$. There is a solution shown below (und means and, oder means or).

I don't understand how the transition from

$xin (Acup B) wedge xnotin(Acap B)$ (item 1 above)

to

$(x in A wedge x notin B) vee (x in B wedge x notin A)$ (item 2)

is made.

The only thing that comes to mind is De Morgan's law. Then

$xin (Acup B) wedge xnotin(Acap B) iff neg (x in A wedge xnotin B) vee neg (x notin (A cap B))$
$xin (Acup B) wedge xnotin(Acap B) iff neg (x in A wedge xnotin B) vee (x in (A cap B))$
$xin (Acup B) wedge xnotin(Acap B) iff neg (x in A wedge xnotin B) vee neg(x in A wedge x in B)$
$xin (Acup B) wedge xnotin(Acap B) iff neg (x in A wedge xnotin B) vee (x notin A vee x notin B)$
$xin (Acup B) wedge xnotin(Acap B) iff neg (x in A wedge xnotin B) vee (x notin A vee x notin B)$
$xin (Acup B) wedge xnotin(Acap B) iff neg (x in A wedge xnotin B) vee (x notin A vee x notin B)$
$xin (Acup B) wedge xnotin(Acap B) iff (x notin A vee xin B) vee (x notin A vee x notin B)$
$xin (Acup B) wedge xnotin(Acap B) iff x notin A vee xin B vee x notin A vee x notin B$

The problem is that I have two $xnotin A$, whereas in the solution from the exercise there is one $xin A$ and one $xnotin A$.

Where exactly did I make a mistake?

This would be one way of proving it:

$$xin (Acup B) wedge xnotin(Acap B)$$

$$iff (xin Alor xin B)land neg(xin Aland xin B)$$

$$iff(xin Alor xin B)land(xnotin Alor xnotin B)$$

$$iff[(xin Alor xin B)land(xnotin A)]lor[(xin Alor xin B)land(xnotin B)]$$

$$iff(xin Aland xnotin A)lor(xin Bland xnotin A)lor (xin Aland xnotin B)lor(xin Bland xnotin B)$$

$$iff (xin Bland xnotin A)lor (xin Aland xnotin B)$$

And, to continue:

$$iff (xin Bsetminus A)lor (xin Asetminus B)$$

$$iff xin (Bsetminus A)cup(Asetminus B)$$

The statement $xin (Acup B)backslash (Acap B)$ is equivalent to the statement $xin (Abackslash B)cup (Bbackslash A)$ because both are equivalent to $(xin A)notequiv(xin B)$.

Or if you prefer a proof by diagrams, both statements imply $x$ is in one of two intersecting circles that denote $A,,B$, but not in their intersection. (The part of one circle that doesn't intersect the other denotes $Abackslash B$; with the other circle, we get $Bbackslash A$.)

Once you know $x in A cup B$ and $x notin A cap B$, you have two cases: $x in A$ and $x in B$, from the union. If $x in A$ we know $x notin B$ (or else $xin A cap B$, which is not the case) and if $x in B$ in the same way : $x notin A$. Hence the step from (1) to (2) in your proof. No need for heavy formula manipulation, just simple reasoning..