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

theorem
  2*PI*i <= r & r < 2*PI+2*PI*i & 2*PI*i <= s & s < 2*PI+2*PI*i & sin r
  = sin s & cos r = cos s implies r = s
proof
  assume that
A1: 2*PI*i <= r and
A2: r < 2*PI+2*PI*i & 2*PI*i <= s and
A3: s < 2*PI+2*PI*i and
A4: sin r = sin s & cos r = cos s;
A5: cos(r-s) = (cos r)*(cos s)+(sin r)*(sin s) by COMPLEX2:3
    .= 1 by A4,SIN_COS:29;
A6: sin(r-s) = (sin r)*(cos s)-(cos r)*(sin s) by COMPLEX2:3
    .= 0 by A4;
A7: cos (s-r) = (cos r)*(cos s)+(sin r)*(sin s) by COMPLEX2:3
    .= 1 by A4,SIN_COS:29;
A8: sin (s-r) = (sin s)*(cos r)-(cos s)*(sin r) by COMPLEX2:3
    .= 0 by A4;
  per cases by XXREAL_0:1;
  suppose
A9: r > s;
    r+2*PI*i < 2*PI+2*PI*i+s by A2,XREAL_1:8;
    then r+2*PI*i-2*PI*i < 2*PI+2*PI*i+s-2*PI*i by XREAL_1:9;
    then r < 2*PI+s;
    then
A10: r-s < 2*PI by XREAL_1:19;
    r > s+0 by A9;
    then 0 <= r-s by XREAL_1:20;
    then r-s = 0 or r-s = PI by A6,A10,COMPTRIG:17;
    hence thesis by A5,SIN_COS:77;
  end;
  suppose
A11: r < s;
    s+2*PI*i < 2*PI+2*PI*i+r by A1,A3,XREAL_1:8;
    then s+2*PI*i-2*PI*i < 2*PI+2*PI*i+r-2*PI*i by XREAL_1:9;
    then s < 2*PI+r;
    then
A12: s-r < 2*PI by XREAL_1:19;
    s > r+0 by A11;
    then 0 <= s-r by XREAL_1:20;
    then s-r = 0 or s-r = PI by A8,A12,COMPTRIG:17;
    hence thesis by A7,SIN_COS:77;
  end;
  suppose
    r = s;
    hence thesis;
  end;
end;
