theorem
  x in rng p implies rng(p |-- x) c= rng p
proof
  assume
A1: x in rng p;
  let y be object;
  assume y in rng(p |-- x);
  then consider z being object such that
A2: z in dom(p |-- x) and
A3: (p |-- x).z = y by FUNCT_1:def 3;
  reconsider z as Element of NAT by A2;
  y = p.(z + x..p) & z + x..p in dom p by A1,A2,A3,Def6,Th43;
  hence thesis by FUNCT_1:def 3;
end;
