- #1

solakis1

- 422

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\(\displaystyle x\in A\cap B\leftrightarrow x\in A\wedge x\in B\)

\(\displaystyle x\in A\cup B\leftrightarrow x\in A\vee x\in B\)

\(\displaystyle x\in A-B\leftrightarrow x\in A\wedge x\notin B\)

\(\displaystyle A=B\leftrightarrow(\forall x(x\in A\leftrightarrow x\in B))\)

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

™

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