theorem Th19:
  k < n implies Px(a,k) < Px(a,n)
proof
  assume
A1: k < n;
  then reconsider nk=n-k as Nat by NAT_1:21;
A2: nk<>0 by A1;
  defpred P[Nat] means $1 >0 implies Px(a,k) < Px(a,k+$1);
A3: P[0];
A4: for i be Nat holds P[i] implies P[i+1]
  proof
    let i be Nat;
    assume
A5:   P[i];
A6:   Px(a,k+i+1) = Px(a,k+i)*a+Py(a,k+i)*(a^2-'1) by HILB10_1:6;
    Px(a,k+i) >= Px(a,0) by HILB10_1:10;
    then Px(a,k+i) >= 1 by HILB10_1:3;
    then
A7:   Px(a,k+i)+Px(a,k+i) >= Px(a,k+i)+1 by XREAL_1:7;
    Px(a,k+i)*a >= Px(a,k+i)*2 by NAT_2:29,XREAL_1:64;
    then Px(a,k+i+1) >= Px(a,k+i)*2+0 by A6,XREAL_1:7;
    then
A8:   Px(a,k+i+1) >= Px(a,k+i)+1 by A7,XXREAL_0:2;
    then
A9:   Px(a,k+i+1) > Px(a,k+i) by NAT_1:13;
    per cases;
    suppose i=0;
      hence thesis by A8,NAT_1:13;
    end;
    suppose i >0;
      hence thesis by A5,A9,XXREAL_0:2;
    end;
  end;
  P[n1] from NAT_1:sch 2(A3,A4);
  then Px(a,k) < Px(a,k+nk) by A2;
  hence thesis;
end;
