I need to show
$(A \setminus B) \cup (B \setminus A) \cup (A \cap B) = A \cup B $, so pretty clear that this holds
I would like to show it with an algebraic proof
I started :
$x∈(A \setminus B) \cup (B \setminus A)\cup(A \cap B) $
$\rightarrow x \in (A \setminus B) \vee x \in (B \setminus A) \vee (x \in A \cap B$)
$ \rightarrow (x \in A \wedge x \notin B) \vee (x \in B \wedge x \notin A) \vee (x \in A \wedge x \in B)$
However I am stuck here, I do not know how to apply the distributive laws here.
Any help with the distributive laws would be highly appreciated.
Best Answer
Just for ease of writing I assume the following:
$$p : x \in A$$ and, $$q: y \in B$$
Your last statement would be equivalent to the following:
$$(p \wedge \neg q) \vee (q \wedge \neg p) \vee (p \wedge q)$$
$$\equiv (p \wedge \neg q) \vee (q \wedge \neg p) \vee (p \wedge q) \vee (p \wedge q)$$
$$\equiv ((p \wedge \neg q) \vee (p \wedge q)) \vee ((q \wedge \neg p) \vee (q \wedge p)) $$
$$\equiv (p \wedge (q \vee \neg q)) \vee (q \wedge (p \vee \neg p))$$
$$ \equiv p \vee q $$
$$= (x \in A) \vee (x \in B) \implies A \cup B$$