# Set theory

#### solakis

##### Active member
Given:
$$\displaystyle x\in A\cap B\leftrightarrow x\in A\wedge x\in B$$
$$\displaystyle x\in A\cup B\leftrightarrow x\in A\vee x\in B$$
$$\displaystyle x\in A-B\leftrightarrow x\in A\wedge x\notin B$$
$$\displaystyle A=B\leftrightarrow(\forall x(x\in A\leftrightarrow x\in B))$$

Then prove using only the above and the laws of logic that:

$$\displaystyle (A\cup B)-(A\cap B)=(A-B)\cup(B-A)$$