reserve n,m for Element of NAT;
reserve h,k,r,r1,r2,x,x0,x1,x2,x3 for Real;
reserve f,f1,f2 for Function of REAL,REAL;
reserve S for Seq_Sequence;

theorem
  (for x holds f.x=1/(cos(x))^2) & cos(x)<>0 & cos(x-h)<>0
  implies bD(f,h).x = (-16)*sin((2*x-h)/2)*sin((-h)/2)*cos((2*x-h)/2)
  *cos((-h)/2)/((cos(2*x-h)+cos(h))^2)
proof
  assume that
A1:for x holds f.x=1/(cos(x))^2 and
A2:cos(x)<>0 & cos(x-h)<>0;
  bD(f,h).x = f.x - f.(x-h) by DIFF_1:4
    .= 1/(cos(x))^2 - f.(x-h) by A1
    .= 1/(cos(x))^2 - 1/(cos(x-h))^2 by A1
    .= (1*(cos(x-h))^2-1*(cos(x))^2)/((cos(x))^2*(cos(x-h))^2)
                                               by A2,XCMPLX_1:130
    .= ((cos(x-h))^2-(cos(x))^2)/((cos(x)*cos(x-h))^2)
    .= ((cos(x-h))^2-(cos(x))^2)/(((1/2)*(cos(x+(x-h))+cos(x-(x-h))))^2)
                                                             by SIN_COS4:32
    .= ((cos(x-h))^2-(cos(x))^2)/((1/4)*(cos(2*x-h)+cos(h))^2)
    .= ((cos(x-h))^2-(cos(x))^2)/(1/4)/((cos(2*x-h)+cos(h))^2) by XCMPLX_1:78
    .= 4*((((cos(x-h))-(cos(x)))*((cos(x-h))+(cos(x))))
       /((cos(2*x-h)+cos(h))^2))
    .= 4*(((-2*(sin(((x-h)+x)/2)*sin(((x-h)-x)/2)))*((cos(x-h))+(cos(x))))
       /((cos(2*x-h)+cos(h))^2)) by SIN_COS4:18
    .= 4*(((-2*(sin((2*x-h)/2)*sin((-h)/2)))*(2*(cos((2*x-h)/2)*cos((-h)/2))))
       /((cos(2*x-h)+cos(h))^2)) by SIN_COS4:17
    .= (-16)*sin((2*x-h)/2)*sin((-h)/2)*cos((2*x-h)/2)*cos((-h)/2)
       /((cos(2*x-h)+cos(h))^2);
  hence thesis;
end;
