reserve A for non trivial Nat,
        B,C,n,m,k for Nat,
        e for Nat;
reserve a for non trivial Nat;

theorem Th10:
  n <= a implies ex T be _Theta st
    Py(a,n+1) = (2*a) |^n *(1+T*(n/a))
proof
  assume
A1: n <=a;
  per cases;
  suppose
A2:   n=0;
    then Py(a,n+1) = Px(a,0) + Py(a,0)*a & Px(a,0) =1 & Py(a,0) =0
      by HILB10_1:3,6;
    then Py(a,n+1) = 1 = (2*a) |^n *(1+0*(n/a)) by A2;
    hence thesis;
  end;
  suppose
A3:   n >0;
A4:   -1 is theta;
    2*n <= 2*a by A1,XREAL_1:64;
    then 1/ (2*a) <= 1/(2*n) by A3,XREAL_1:118;
    then consider T be _Theta such that
A5:   (1 + (-1) *(1/(2*a)))|^n = 1+T*2*n*(1/(2*a)) by A4,Th9;
A6:   T*2*n*(1/(2*a)) = T * ((2*n)* (1/(2*a)))
      .= T * ( (2*n*1) / (2*a)) by XCMPLX_1:74
      .= T*(n/a) by XCMPLX_1:91;
    2*a-1 = 2*a*1 - (2*a)*(1/(2*a)) by XCMPLX_1:87;
    then (2*a-1)|^n = ( (2*a) *(1 - 1/(2*a)))|^n
      .= (2*a)|^n * (1+T*(n/a)) by A6,A5,NEWTON:7;
    then (2*a)|^n + (2*a)|^n * (T*(n/a)) <= Py(a,n+1)
      <= (2*a) |^ n +0 by HILB10_1:17;
    then (2*a)|^n * (T*(n/a)) <= Py(a,n+1) - (2*a)|^n <= (2*a)|^n * (0*(n/a))
      by XREAL_1:19,20;
    then T*(n/a) <= (Py(a,n+1) - (2*a)|^n) / ((2*a)|^n) <= 0*(n/a)
      by XREAL_1:77;
    then consider T1 be _Theta such that
A7:   T1* (n/a) = (Py(a,n+1) - (2*a)|^n) / ((2*a)|^n) by Th4;
    take T1;
    T1* (n/a) * ((2*a)|^n) = Py(a,n+1) - (2*a)|^n by A7,XCMPLX_1:87;
    hence thesis;
  end;
end;
