
theorem Th4:
  for X,N,I being non empty set for s being Function of X,I for c
being Function of X,N st c is one-to-one for n being Element of I holds (N-->n)
  +*(s*c") is Function of N,I
proof
  let X,N,I be non empty set;
  let s be Function of X,I;
  let c be Function of X,N;
  assume c is one-to-one;
  then
A1: dom (c") = rng c by FUNCT_1:33;
  let n be Element of I;
  set f = N-->n, g = s*c";
A2: dom g c= dom (c") by RELAT_1:25;
  rng g c= rng s by RELAT_1:26;
  then rng g c= I by XBOOLE_1:1;
  then
A3: rng f \/ rng g c= I by XBOOLE_1:8;
A5: rng (f+*g) c= rng f \/ rng g by FUNCT_4:17;
  dom (f+*g) = dom f \/ dom g by FUNCT_4:def 1;
  then dom (f+*g) = N by A2,A1,XBOOLE_1:1,12;
  hence thesis by A5,A3,FUNCT_2:2,XBOOLE_1:1;
end;
