Proving $(A \cup B) \setminus (A \cap B) = (A \setminus B) \cup (B \setminus A)$

Question

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

$x\in (A\cup B) \wedge x\notin(A\cap B)$ (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

$x\in (A\cup B) \wedge x\notin(A\cap B) \iff \neg (x \in A \wedge x\notin B) \vee \neg (x \notin (A \cap B))$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff \neg (x \in A \wedge x\notin B) \vee (x \in (A \cap B))$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff \neg (x \in A \wedge x\notin B) \vee \neg(x \in A \wedge x \in B)$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff \neg (x \in A \wedge x\notin B) \vee (x \notin A \vee x \notin B)$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff \neg (x \in A \wedge x\notin B) \vee (x \notin A \vee x \notin B)$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff \neg (x \in A \wedge x\notin B) \vee (x \notin A \vee x \notin B)$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff (x \notin A \vee x\in B) \vee (x \notin A \vee x \notin B)$ $x\in (A\cup B) \wedge x\notin(A\cap B) \iff x \notin A \vee x\in B \vee x \notin A \vee x \notin B$

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

Where exactly did I make a mistake?

Answer 1

This would be one way of proving it:

$$x\in (A\cup B) \wedge x\notin(A\cap B)$$

$$\iff (x\in A\lor x\in B)\land \neg(x\in A\land x\in B)$$

$$\iff(x\in A\lor x\in B)\land(x\notin A\lor x\notin B)$$

$$\iff[(x\in A\lor x\in B)\land(x\notin A)]\lor[(x\in A\lor x\in B)\land(x\notin B)]$$

$$\iff(x\in A\land x\notin A)\lor(x\in B\land x\notin A)\lor (x\in A\land x\notin B)\lor(x\in B\land x\notin B)$$

$$\iff (x\in B\land x\notin A)\lor (x\in A\land x\notin B)$$

And, to continue:

$$\iff (x\in B\setminus A)\lor (x\in A\setminus B)$$

$$\iff x\in (B\setminus A)\cup(A\setminus B)$$

Answer 2

The statement $x\in (A\cup B)\backslash (A\cap B)$ is equivalent to the statement $x\in (A\backslash B)\cup (B\backslash A)$ because both are equivalent to $(x\in A)\not\equiv(x\in B)$.

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 $A\backslash B$; with the other circle, we get $B\backslash A$.)

Answer 3

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 $x\in A \cap B$, 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..

Answer 4

Here is my attempt

Let's take L.H.S

suppose $x \in (A-B) \cup (B - A)$

$\implies x \in (A-B) \lor x \in (B-A)$

$\implies x \in (A \cap B^\prime) \lor x \in (B \cap A^\prime)$

If $x \in (A \cap B^\prime) \implies x \in A \land x \not \in B \implies x \not \in (A \cap B)$, and similarly we can deduce from the second or condition that $x \in B \land x \not \in A$

In either case we have a proof that $x \not \in (A\cap B)$

now in my first case above I got $x \in A \implies x \in (A \cup B)$, in second case we got $x \in B \implies x \in (A \cup B)$

hence $x \in (A \cup B) \land x \not \in (A \cap B) \implies x \in (A\cup B)-(A \cap B)$