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