 \ b'(0) = b(0)+0 .$$ Note that $a(b') = a(b) e_0$ and $b'$ has type B0. We say that $b'$ is obtained from $b$ by an extension of type 0 (on the right). If $(B \setminus \{ b \}) \cup \{ b' \}$ is a [bracket system]{}, then replacement of $b \in B$ with $b'$ in $B$ is called an [*extension* ]{} operation over $B$ of type 0.  Assume that $b \in B$ is a 