theorem Th21: p 'or' q in tau X implies p in tau X & q in tau X
  proof
    assume p 'or' q in tau X;
    then (('not' p) '&&' ('not' q)) in tau X by Th19;
    then 'not' p in tau X & 'not' q in tau X by Th20;
    hence thesis by Th19;
  end;
