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 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;
