of the $\cal S$ Cancellation Lemma.]{} We proceed as in the proof of the ${\cal K}_c$ Cancellation Lemma of [@DP03] until we reach the case that $f=\varphi_0\cirk f'$ and $g=\varphi_0\cirk g'$. Then we must ensure that $\varphi_0$ in $\varphi_0\cirk f'$ and $\varphi_0$ in $\varphi_0\cirk g'$ are tied in $\delta(\psi(f))$, which is equal to $\delta(\psi(g))$, to circles encompassing the same circular forms. It is always possible to achieve that. We conclude that $\delta(\psi(f'))\cong_{\cal L} \delta(\psi(g'))$. This is because $\delta(\psi(\varphi_0\cirk f'))$ and $\delta(\psi(\varphi_0\cirk
g'))$ are both $\cal L$-equivalent to $\delta(c_1^\alpha)$, for some $\alpha\geq 1$, while $\delta(\psi(f'))$ and $\delta(\psi(g'))$ must both be $\cal L$-equivalent to the same $\delta(b_1^{\beta}c_1^{\gamma})$, for $\beta$ and $\gamma$ lesser than $\alpha$. We conclude that ${f'\equiv g'}$, and since $f'$ and $g'$ are of type $0\str 2$, by the $\cal L$ Cancellation Lemma of [@DP03], we have that $f'=g'$ in $\cal S$. From that we obtain that $f=g$ in $\cal S$.

As $\cal S$, the isomorphic category ${\cal S}^{\ast t}$ is the disjoint union of two categories ${\cal S}_{\cal A}^{\ast t}$ and ${\cal S}_{\cal B}^{\ast t}$, isomorphic respectively to $\cal
S_A$ and $\cal S_B$. If ${\langle [f], n, m\rangle}$ is in ${\cal
S}_{\cal A}^{\ast t}$, then $n$ and $m$ are 