theorem Th34:
  p^{} = p & {}^p = p
proof
  dom(p^{}) = Seg len (p^{}) by Def3
    .= Seg (len p + len {}) by Th22
    .= dom p by Def3;
  hence p^{} = p by Def7;
A2: dom({}^p) = Seg len ({}^p) by Def3
    .= Seg (len {} + len p) by Th22
    .= dom p by Def3;
  for k st k in dom p holds p.k = ({}^p).k
  proof
    let k;
    assume
A3: k in dom p;
    thus ({}^p).k =({}^p).(len {} + k) .=p.k by A3,Def7;
  end;
  hence {}^p=p by A2;
end;
