reserve r,r1,r2, s,x for Real,
  i for Integer;

theorem
  2*PI*i <= r & r <= PI+2*PI*i implies sin r >= 0
proof
  assume
A1: 2*PI*i <= r & r <= PI+2*PI*i;
  per cases by A1,XXREAL_0:1;
  suppose
    2*PI*i < r & r < PI+2*PI*i;
    hence thesis by Th11;
  end;
  suppose
A2: 2*PI*i = r;
    sin(0+2*PI*i) = sin 0 by COMPLEX2:8;
    hence thesis by A2,SIN_COS:31;
  end;
  suppose
    r = PI+2*PI*i;
    hence thesis by COMPLEX2:8,SIN_COS:77;
  end;
end;
