If A = {b, c, e, g, h}, B = {a, c, d, g, i} and C = {a, d, e, g, h}, then show that `A-(BcapC)=(A-B)cup(A-C)`.

Explanation

Let `x in (A-B) cup (A-C) `
` rArr x in ( A-B) and x in (A-c)`
`rArr ( x in A and x in B ) and ( x in A and x in C)`
`rArr x in A and x in B) and ( x in A and notin C)`
`rArr x in A and ( x notin B and x notin C) `
`rArr x in A and x notin ( B cup C) `
`rArr x in A - ( B cup C)`
` rArr ( A-B) cap ( A-C) sub A - ( B cup C)`
Now let
`y in A-(B cup C)`
`rArr y in A and y notin ( B cup C)`
`rArr y in A and ( y notin B and y notin C)`
`rArr y in ( A- B) and y in ( A-C)`
`rArr y in ( A - B ) cap ( A-c)`
`rArr A-( B cup C) sub ( A-B) cap ( A-C)`
From (1) and (2) we get
`A-(B cup C) = ( A - B ) cap (A-C)