reserve k, l, m, n, i, j for Nat,
  K, N for non empty Subset of NAT,
  Ke, Ne, Me for Subset of NAT,
  X,Y for set;

theorem Th14:
  Ne c= Segm n & n>0 implies min* Ne <= n-1
proof
  assume that
A1: Ne c= Segm n and
A2: n>0;
  now
    per cases;
    suppose
      Ne is non empty;
      hence thesis by A1,Th12;
    end;
    suppose
      Ne is empty;
      then min* Ne =0 by NAT_1:def 1;
      hence thesis by A2,NAT_1:20;
    end;
  end;
  hence thesis;
end;
