)c_{ij}(T) + B_{ij}c_{ik}(S)b_{ji}(T) = C_{jk}c_{ij}(S)c_{jk}(T) + B_{kj}c_{ik}(S)b_{jk}(T)
        \end{align*}
        By this last equality, we have illustrated that $Y_{12}(i,j,k)=0$.
        
        Given our conditions, all of the requisite equations are satisfied. Hence, the $n$-label Yang-Baxter equation has a nontrivial solution if and only if they all hold.
    \end{proof}

\subsection{When are all R-vertex weights nonzero?}

    It is noteworthy that, assuming the $S$ and $T$-vertex weights are nonzero, in any nontrivial solution $C_{ij}\ne 0$ for all $i$, $j$. Nonetheless, there are cases where some $R$-vertex weights can be 0. In this section, we explore some such cases. To begin, we analyze some of the conditions in Theorem \ref{nlabelCond}.

    \begin{lem} \label{condLem}
        The final three conditions in Theorem \ref{nlabelCond} are equivalent to:
        \begin{center}
            $\beta_{ij} = \left(\displaystyle\frac{\gamma_{ij}}{\gamma_{ik}} - \gamma_{kj}\right) \displaystyle\displaystyle\frac{b_{ij}(S)c_{jk}(T)}{c_{ik}(S)c_{ji}(T)}$ \hspace{2em}
            %
            $\beta_{ij} = \left(\tau_{ik}\displaystyle\frac{\gamma_{ij}}{\gamma_{ik}} - \gamma_{kj}\displaystyle\frac{\tau_{ij}}{\tau_{kj}}\right) \frac{b_{ij}(S)c_{kj}(T)}{c_{ki}(S)c_{ij}(T)}$ \vspace{0.5em}
            
            $\beta_{ij}b_{ji}(T)\displaystyle\frac{\tau_{ji}}{\gamma_{ji}} - \beta_{kj}b_{jk}(T)\frac{\tau_{jk}}{\gamma_{jk}}  =  \left(\frac{1}{\gamma_{jk}} - \frac{\gamma_{kj}}{\gamma_{ij}\gamma_{ji}}\right) \frac{c_{ij}(S)c_{jk}(T)}{c_{ik}(S)}$
        \end{center}
    \end{lem}

    \begin{proof}
        Throughout this proof, we consistently refer to the identity $\frac{b_{ij}(T)}{b_{ij}(S)} = \frac{b_{kj}(T)}{b_{kj}(S)}$, one of the conditions in Theorem \ref{nlabelCond}. It is essential to expressing these conditions using $\gamma$ and $\tau$.
    
        The first condition from Theorem \ref{nlabelCond} that we simplify is $\gamma_{ik}c_{jk}(S)b_{ij}(T) + \beta_{ij}\gamma_{ik}c_{ik}(S)c_{ji}(T) = \gamma_{ij}b_{ij}(S)c_{jk}(T)$. We solve for $\beta_{ij}$ assuming it holds:
        \begin{align*}
            \beta_{ij}\gamma_{ik}c_{ik}(S)c_{ji}(T) &= \gamma_{ij}b_{ij}(S)c_{jk}(T) - \gamma_{ik}c_{jk}(S)b_{ij}(T) \\&= \left(\gamma_{ij} - \gamma_{ik}\frac{c_{jk}(S)b_{ij}(T)}{b_{ij}(S)c_{jk}(T)}\right)b_{ij}(S)c_{jk}(T) \\
            &= \left(\gamma_{ij} - \gamma_{ik}\frac{c_{jk}(S)b_{kj}(T)}{b_{kj}(S)c_{jk}(T)}\right)b_{ij}(S)c_{jk}(T) \\&= (\gamma_{ij} - \gamma_{ik}\gamma_{kj})b_{ij}(S)c_{jk}(T),
        \end{align*}
        assuming $\frac{b_{ij}(T)}{b_{ij}(S)} = \frac{b_{kj}(T)}{b_{kj}(S)}$. Dividing by $\gamma_{ik}c_{ik}(S)c_{ji}(T)$,
        $$\beta_{ij}=\left(\frac{\gamma_{ij}}{\gamma_{ik}} - \gamma_{kj}\right) \frac{b_{ij}(S)c_{jk}(T)}{c_{ik}(S)c_{ji}(T)}$$
        
        The second condition is $\gamma_{ij}b_{ij}(S)c_{kj}(T) = \tau_{ki}\tau_{ij}\gamma_{ik}c_{kj}(S)b_{ij}(T) + \tau_{ki}\beta_{ij}\gamma_{ik}c_{ki}(S)c_{ij}(T)$. Rearranging and factoring,
        \begin{align*}
            \beta_{ij}\gamma_{ik}c_{ki}(S)c_{ij}(T) &= \tau_{ik}\gamma_{ij}b_{ij}(S)c_{kj}(T) - \tau_{ij}\gamma_{ik}c_{kj}(S)b_{ij}(T) \\ 
            = \left(\tau_{ik}\gamma_{ij} - \tau_{ij}\gamma_{ik}\frac{b_{ij}(T)c_{kj}(S)}{b_{ij}(S)c_{kj}(T)}\right)b_{ij}(S)c_{kj}(T)
            &= \left(\tau_{ik}\gamma_{ij} - \tau_{ij}\gamma_{ik}\frac{b_{kj}(T)c_{kj}(S)}{b_{kj}(S)c_{kj}(T)}\right)b_{ij}(S)c_{kj}(T) \\
            &= (\tau_{ik}\gamma_{ij} - \tau_{ij}\gamma_{ik}\gamma_{kj}/\tau_{kj})b_{ij}(S)c_{kj}(T)
        \end{align*}
        where, as before, we assume $\frac{b_{ij}(T)}{b_{ij}(S)} = \frac{b_{kj}(T)}{b_{kj}(S)}$. Dividing by $\gamma_{ik}c_{ki}(S)c_{ij}(T)$, $$\beta_{ij} = \left(\tau_{ik}\frac{\gamma_{ij}}{\gamma_{ik}} - \gamma_{kj}\frac{\tau_{ij}}{\tau_{kj}}\right) \frac{b_{ij}(S)c_{kj}(T)}{c_{ki}(S)c_{ij}(T)}$$
        
        Last, we handle the condition $\gamma_{jk}c_{jk}(S)c_{ij}(T) + \beta_{ij}\gamma_{jk}c_{ik}(S)b_{ji}(T) = \tau_{ij}\gamma_{ji}c_{ij}(S)c_{jk}(T) + \tau_{ij}\tau_{jk}\beta_{kj}\gamma_{ji}c_{ik}(S)b_{jk}(T)$. Rearranging, we see that
        \begin{align*}
            &\beta_{ij}\gamma_{jk}c_{ik}(S)b_{ji}(T) - \tau_{ij}\tau_{jk}\beta_{kj}\gamma_{ji}c_{ik}(S)b_{jk}(T) = \tau_{ij}\gamma_{ji}c_{ij}(S)c_{jk}(T) - \gamma_{jk}c_{jk}(S)c_{ij}(T) \\
            &= \left(\tau_{ij}\gamma_{ji} - \gamma_{jk}\frac{c_{jk}(S)c_{ij}(T)}{c_{ij}(S)c_{jk}(T)}\right)c_{ij}(S)c_{jk}(T) \\
            &= \left(\tau_{ij}\gamma_{ji} - \gamma_{jk}\frac{c_{jk}(S)c_{ij}(T)b_{ij}(S)b_{ij}(T)}{c_{ij}(S)c_{jk}(T)b_{ij}(S)b_{ij}(T)}\right)c_{ij}(S)c_{jk}(T) \\
            &= \left(\tau_{ij}\gamma_{ji} - \gamma_{jk}\tau_{ij}/\gamma_{ij}\frac{c_{jk}(S)b_{kj}(T)}{c_{jk}(T)b_{kj}(S)}\right)c_{ij}(S)c_{jk}(T) = (\tau_{ij}\gamma_{ji} - \gamma_{jk}\gamma_{kj}\tau_{ij}/\gamma_{ij})c_{ij}(S)c_{jk}(T)
        \end{align*}
        Again, we used that $\frac{b_{ij}(T)}{b_{ij}(S)} = \frac{b_{kj}(T)}{b_{kj}(S)}$ to realize the presence of the factor $\gamma_{kj}$. Dividing by $\gamma_{ji}\gamma_{jk}\tau_{ij}c_{ik}(S)$ reveals
        