reserve
  a for natural Number,
  k,l,m,n,k1,b,c,i for Nat,
  x,y,z,y1,y2 for object,
  X,Y for set,
  f,g for Function;
reserve p,q,r,s,t for FinSequence;
reserve D for set;

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;
