theorem Th29:
  rng p c= rng(p^q)
proof
    let x be object;
    assume x in rng p;
    then consider y being object such that
A1: y in dom p and
A2: x=p.y by FUNCT_1:def 3;
    reconsider k=y as Element of NAT by A1;
A3: dom p c= dom(p^q) by Th26;
    (p^q).k=p.k by A1,Def7;
    hence x in rng(p^q) by A1,A2,A3,FUNCT_1:def 3;
end;
