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 bD(f,h).x = (-2)*(sin(x-h)-sin(x))/(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;
  bD(f,h).x = f.x - f.(x-h) by DIFF_1:4
    .= 1/sin(x) - f.(x-h) by A1
    .= 1/sin(x) - 1/sin(x-h) by A1
    .= (1*sin(x-h)-1*sin(x))/(sin(x)*sin(x-h)) by A2,XCMPLX_1:130
    .= (sin(x-h)-sin(x))/(-(1/2)*(cos(x+(x-h))-cos(x-(x-h)))) by SIN_COS4:29
    .= (sin(x-h)-sin(x))/((-1/2)*(cos(2*x-h)-cos(h)))
    .= (sin(x-h)-sin(x))/(-1/2)/(cos(2*x-h)-cos(h)) by XCMPLX_1:78
    .= (-2)*((sin(x-h)-sin(x))/(cos(2*x-h)-cos(h)));
  hence thesis;
end;
