let X be set ; :: thesis: for B being SetSequence of X
for f being Function st dom f = NAT & ( for n being Element of NAT holds f . n = union { (B . k) where k is Element of NAT : n <= k } ) holds
f is Function of NAT,(bool X)
let B be SetSequence of X; :: thesis: for f being Function st dom f = NAT & ( for n being Element of NAT holds f . n = union { (B . k) where k is Element of NAT : n <= k } ) holds
f is Function of NAT,(bool X)
let f be Function; :: thesis: ( dom f = NAT & ( for n being Element of NAT holds f . n = union { (B . k) where k is Element of NAT : n <= k } ) implies f is Function of NAT,(bool X) )
assume that
A1: dom f = NAT and
A2: for n being Element of NAT holds f . n = union { (B . k) where k is Element of NAT : n <= k } ; :: thesis: f is Function of NAT,(bool X)
rng f c= bool X

proof

let z be set ; :: according to TARSKI:def 3 :: thesis: ( not z in rng f or z in bool X )
assume z in rng f ; :: thesis: z in bool X
then consider x being set such that
A3: x in dom f and
A4: z = f . x by FUNCT_1:def 5;
reconsider n = x as Element of NAT by A1, A3;
set Y = { (B . k) where k is Element of NAT : n <= k } ;
set y = union { (B . k) where k is Element of NAT : n <= k } ;

now

let z be set ; :: thesis: ( z in union { (B . k) where k is Element of NAT : n <= k } implies z in X )
assume z in union { (B . k) where k is Element of NAT : n <= k } ; :: thesis: z in X
then ex Z being set st
( z in Z & Z in { (B . k) where k is Element of NAT : n <= k } ) by TARSKI:def 4;
then consider Z1 being set such that
A5: Z1 in { (B . k) where k is Element of NAT : n <= k } and
A6: z in Z1 ;
ex k1 being Element of NAT st
( Z1 = B . k1 & n <= k1 ) by A5;
hence z in X by A6; :: thesis: verum

end;

then A7: union { (B . k) where k is Element of NAT : n <= k } is Subset of X by TARSKI:def 3;
z = union { (B . k) where k is Element of NAT : n <= k } by A2, A4;
hence z in bool X by A7; :: thesis: verum

end;

hence f is Function of NAT,(bool X) by A1, FUNCT_2:4; :: thesis: verum