Logic of Set Operations: Proof

[solakis1]

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:
™
\(\displaystyle (A\cup B)-(A\cap B)=(A-B)\cup(B-A)\)

 

[jvicens]

To prove this statement, we will use the laws of logic and the given definitions of set operations.

First, we will rewrite the left side of the equation using the given definitions:
(A\cup B)-(A\cap B)=(x\in A\vee x\in B)-(x\in A\wedge x\in B)

Next, we will use the distributive law of logic to rewrite the right side of the equation:
(x\in A\vee x\in B)-(x\in A\wedge x\in B)=((x\in A\vee x\in B)\wedge \neg(x\in A\wedge x\in B))

Now, we will use the definition of set difference to rewrite the right side of the equation:
((x\in A\vee x\in B)\wedge \neg(x\in A\wedge x\in B))=((x\in A\wedge \neg(x\in A\wedge x\in B))\vee (x\in B\wedge \neg(x\in A\wedge x\in B)))

Using De Morgan's laws, we can simplify the right side of the equation:
((x\in A\wedge \neg(x\in A\wedge x\in B))\vee (x\in B\wedge \neg(x\in A\wedge x\in B)))=((x\in A\wedge \neg x\in B)\vee (x\in B\wedge \neg x\in A))

Finally, we can use the definition of set union to rewrite the right side of the equation:
((x\in A\wedge \neg x\in B)\vee (x\in B\wedge \neg x\in A))=((x\in A-B)\vee (x\in B-A))

Therefore, we have shown that (A\cup B)-(A\cap B)=(A-B)\cup(B-A) using only the given definitions and laws of logic.