 \ b'(0) = b(0)+{\varepsilon}_0 , \ b'(0) = b(0) .\end{aligned}$$ 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.  On the other hand, if $| b(0) | = b(0)-0$