theorem Th30:
  rng q c= rng(p^q)
proof
    let x be object;
    assume x in rng q;
    then consider y being object such that
A1: y in dom q and
A2: x=q.y by FUNCT_1:def 3;
    reconsider k=y as Element of NAT by A1;
    len p + k in dom(p^q) & (p^q).(len p + k) = q.k by A1,Def7,Th28;
    hence x in rng(p^q) by A2,FUNCT_1:def 3;
end;
