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

theorem
  PI/2+2*PI*i < r & r <= 2*PI+2*PI*i implies sin r < 1
proof
  assume that
A1: PI/2+T(i) < r & r <= 2*PI+T(i) and
A2: sin r >= 1;
A3: PI/2+T(i)-T(i) < r-T(i) & r-T(i) <= 2*PI+T(i)-T(i) by A1,XREAL_1:9;
A4: sin r <= 1 by Th4;
  sin(r-T(i)) = sin(r+T(-i)) .= sin r by COMPLEX2:8
    .= 1 by A2,A4,XXREAL_0:1;
  hence thesis by A3,Th33;
end;
