 reserve n,s for Nat;

theorem Th27:
  s >= 2 implies Polygon (s,n) is natural
  proof
    assume s >= 2; then
A1: s - 2 >= 2 - 2 by XREAL_1:9;
    per cases;
    suppose n = 0;
      hence thesis;
    end;
    suppose n > 0; then
      n >= 0 + 1 by NAT_1:13; then
      n - 1 >= 1 - 1 by XREAL_1:9; then
      (n * (s - 2) * (n - 1)) / 2 + n >= 0 by A1;
      hence Polygon (s,n) in NAT by INT_1:3;
    end;
  end;
