a and b are sets, prove a intersection b = empty set if and only if power set of a intersection with the power set of b = empty set.

I will use * for intersection, 0 for the empty set.
let P(s) be the power set of s, that is the set of all subsets of s.
Suppose a*b = 0. let s be in P(a) * P(b)
if s != 0, then there exists x in s.
since s is in P(a), s subset of a, therefore x is in a
since s is in P(b), s subset of b, therefore x is in b.
this contradicts that a*b = 0
therefore s = 0. so the only member of P(a) * P(b) = 0.
now suppose P(a) * P(b) = 0, let x be in a*b
then {x} is in P(a) and in P(b), which contraticts that P(a) * P(b) = 0
therefpre a*b = 0
let P(s) be the power set of s, that is the set of all subsets of s.
Suppose a*b = 0. let s be in P(a) * P(b)
if s != 0, then there exists x in s.
since s is in P(a), s subset of a, therefore x is in a
since s is in P(b), s subset of b, therefore x is in b.
this contradicts that a*b = 0
therefore s = 0. so the only member of P(a) * P(b) = 0.
now suppose P(a) * P(b) = 0, let x be in a*b
then {x} is in P(a) and in P(b), which contraticts that P(a) * P(b) = 0
therefpre a*b = 0