 , \ b'(0) = b(0) .$$ Note that $a(b') = e_0 a(b)e_0$. We say that $b'$ is obtained from $b$ by an extension of type 0. If $(B \setminus \{ b \}) \cup \{ b' \}$ is a [pseudobracket system]{}, then replacement of $b \in B$ with $b'$ in $B$ is called an [*extension* ]{} operation over $B$ of type 0.  *Mergers*.  Suppose that $b_0, b_0$ are distinct 