Question

Prove that A is symetric defference with B equals to A union B less A intersection B

Answer 1

We have to prove that,

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

Let $x\in(A\Delta B),$ i.e.,

$x\in((A-B)\cup(B-A))\\\\x\in(A-(A\cap B))\cup(B-(A\cap B))\\\\(x\in A\land x\notin (A\cap B))\lor(x\in B\land x\notin (A\cap B))\\\\(x\in A\lor x\in B)\land(x\notin(A\cap B))\\\\(x\in(A\cup B))\land(x\notin(A\cap B))\\\\x\in((A\cup B)-(A\cap B))$

This implies,

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

Hence the Proof!