For any two sets A and B, prove that :
(i) `(AcupB)-A=B-A`
(ii) `(A-B)capA=AcapB'`
(iii) `A capB' = phi rArr A sube B`
(iv) `A' cupB=UrArrA subeB`.

Explanation

(i) `(A cupB)-A`
`=(AcupB)capA'" "(becauseX-Y=XcapY')`
`=(A capA')cup (B cap A')` (From distributive law)
`= phi cup(B capA')" "(because A capA'=phi)`
`= B cap A' = B-A`.
(ii) `(A-B)capA`
`=A-B" "(becauseA-B subeA)`
`=A cap B' " "(becauseX cap Y' = X-Y)`
(iii) `A=A capU`
`=A cap(BcupB')" "(because BcapB'=U)`
`=(A cap B)cup(A capB')`(From distributive law)
`= (AcapB)cupphi" "(because A cap B'=phi)`
`A cap B`
`:. sube B`
(iv) `A' cupB=U`
`rArr A cap(A'cupB)=A capU`
`rArr(A capA')cup(A capB)=A (because A cap U = A)`
`rArr phi cup (A cap B) = A (because A cap A' = phi)`
`rArr A cap B = A`
`rArr A sube B`.