theorem Th2:
  seq(a,0) = {}
proof
  hereby
    set x =the  Element of seq(a,0);
    assume
A1: seq(a,0) <> {};
    then reconsider x as Element of NAT by TARSKI:def 3;
    1+a <= x & x <= 0+a by A1,Th1;
    hence contradiction by NAT_1:13;
  end;
end;
