For any 3 sets A, B and C, prove that :
(i) `A-(B cupC)=(A-B)cap(A-C)`
(ii) `A-(BcapC)=(A-B)cup(A-C)`
(iii) `A cap(B-C)= (A cap B) - (A cap C)`
(iv) `(A cupB)-C=(A-C)cup(B-C)`
(v) `A cap (B DeltaC)=(A capB)Delta(A capC)`.

Explanation

(i) `A-(BcupC)`
`=Acap(BcupC)'" "(becauseX-Y=X capY')`
`=Acap(B'capC')=(AcapB')cap(A capC)`
`=(A-B)cap(A-C)`.
(ii) `A-(BcapC)`
`=Acap(BcapC)'`
`=A cap(B'cupC')`
`= (A cap B')cup (A cap C')` (From distributive law)
`= (A-B) cup (A-C)`.
(iii) `A cap(B-C)`
`=A cap(BcapC')" "(becauseX-Y=XcapY')`
`=(AcapB)capC'`(From associative law)

`= phicup[(AcapB)capC']`
`=[(AcapB)capA']cup[(AcapB)capC']`
`=(A capB)cap (A' cup C')` (From distributive law)
`=(AcapB)cap(AcapC)'`
`=(AcapB)-(A capC)" "(because XcapY'=X-Y)`
(iv) `(A cupB)-C`
`=(A cup B)capC'" "(X-Y=X capY')`
`=(A capC')cup (B cap C')` (From distributive law)
`= (A-C)cup(B-C)" "(becauseX capY'=X-Y)`
(v) `A cap(B DeltaC)`
`=A cap[(B-C)cup(C-B)]`
`[Acap(B-C)]cup[Acap(C-B)]` (From distributive law)
`= [(A capB)-(AcapC)]cup[(AcapC)-(AcapB)]`
`=(A capB)Delta(AcapC)`.