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/(sin(x))^2) & sin(x)<>0 & sin(x+h)<>0
  implies fD(f,h).x = 16*cos((2*x+h)/2)*sin((-h)/2)*cos((-h)/2)*sin((2*x+h)/2)
  /((cos(2*x+h)-cos(h))^2)
proof
  assume that
A1:for x holds f.x=1/(sin(x))^2 and
A2:sin(x)<>0 & sin(x+h)<>0;
  fD(f,h).x = f.(x+h) - f.x by DIFF_1:3
    .= 1/(sin(x+h))^2 - f.x by A1
    .= 1/(sin(x+h))^2 - 1/(sin(x))^2 by A1
    .= (1*(sin(x))^2-1*(sin(x+h))^2)/((sin(x+h))^2*(sin(x))^2)
                                                by A2,XCMPLX_1:130
    .= ((sin(x))^2-(sin(x+h))^2)/((sin(x+h)*sin(x))^2)
    .= ((sin(x))^2-(sin(x+h))^2)/((-(1/2)*(cos((x+h)+x)-cos((x+h)-x)))^2)
                                                               by SIN_COS4:29
    .= ((sin(x))^2-(sin(x+h))^2)/((1/4)*(cos(2*x+h)-cos(h))^2)
    .= ((sin(x))^2-(sin(x+h))^2)/(1/4)/((cos(2*x+h)-cos(h))^2) by XCMPLX_1:78
    .= 4*((((sin(x))-(sin(x+h)))*((sin(x))+(sin(x+h))))
       /((cos(2*x+h)-cos(h))^2))
    .= 4*(((2*(cos((x+(x+h))/2)*sin((x-(x+h))/2)))*((sin(x))+(sin(x+h))))
       /((cos(2*x+h)-cos(h))^2)) by SIN_COS4:16
    .= 4*(((2*(cos((2*x+h)/2)*sin((-h)/2)))*(2*(cos((-h)/2)*sin((2*x+h)/2))))
       /((cos(2*x+h)-cos(h))^2)) by SIN_COS4:15
    .= 16*cos((2*x+h)/2)*sin((-h)/2)*cos((-h)/2)*sin((2*x+h)/2)
       /((cos(2*x+h)-cos(h))^2);
  hence thesis;
end;
