Question

For any two sets A and B prove the followin:(A-B)intersection (B-A)=EMPTY SET​

Answer 1

We have to prove,

$(A-B)\cap(B-A)=\O$

Let there exists an element x in the set mentioned in the LHS.

$\textsf{Let}\ \ x\in(A-B)\cap(B-A).$

We know,  $P\cap Q=\{x:x\in P\ \land\ x\in Q\}.$

[$\land$ just means 'and'.]

So we get,

$x\in(A-B)\ \land\ x\in(B-A)$

And we know that  $P-Q=\{x:x\in P\ \land\ x\notin Q\}$

So,

$x\in A\ \land x\notin B\ \land\ x\in B\ \land\ x\notin A\\ \\ \implies\ x\in A\ \land x\notin A\ \land\ x\in B\ \land\ x\notin B$

This brings a contradiction, and hence, a confusion too.

What does $x\in A\ \land\ x\notin A$  mean?!

Similarly, what is  $x\in B\ \land\ x\notin B$

To avoid this contradiction, the solution continues as,

$x\in A\ \land x\notin A\ \land\ x\in B\ \land\ x\notin B\\ \\ \implies\ x\in A\ \land\ x\in A'\ \land\ x\in B\land\ x\in B'\quad\quad[\because\ x\notin P\ \Rightarrow\ x\in P']\\ \\ \implies\ x\in(A\cap A')\ \land\ x\in(B\cap B')\\ \\ \implies\ x\in \O\ \land\ x\in\O\\ \\ \implies\ x\in\O$

Therefore,

$(A-B)\cap(B-A)=\O$

Hence Proved!

Similarly,

$\textsf{Let}\ \ x\in(A-B)\cap(B-A)\\ \\ \implies\ x\in(A-B)\ \land\ x\in(B-A)$

Since  $P-Q=P-(P\cap Q),$

$\implies\ x\in(A-(A\cap B))\ \land\ x\in(B-(B\cap A))\\ \\ \implies\ x\in A\ \land\ x\notin(A\cap B)\ \land\ x\in B\ \land\ x\notin(B\cap A)\\ \\ \implies\ x\in A\ \land\ x\in B\ \land\ x\notin(A\cap B)\ \ \ [\because\ A\cap B=B\cap A]\\ \\ \implies\ x\in(A\cap B)\ \land\ x\notin(A\cap B)\\ \\ \implies\ x\in(A\cap B)\ \land\ x\in(A\cap B)'\\ \\ \implies\ x\in((A\cap B)\cap(A\cap B)')\\ \\ \implies\ x\in \O$