\\
                          = & 0.
        \end{split}$$

Suppose, without loss of generality, that $\alpha \leq \beta \leq \gamma$. The socle degrees of $R/I$ are $\alpha + b + c -3$, $a+\beta+c-3$, and $a+b+\gamma -3$, with the first removed if $\alpha = 0$. The following argument shows that $\alpha + b + c - 3$ is at least $s+1$, however, with a simple changing of names it can be used to show that each of the socle degrees is at least $s+1$.

As we are considering the socle degree $\alpha + b + c - 3$, we may assume $\alpha \geq 1$. Notice that $\alpha + b + c - 3 = 2A + B + C + 2M + \alpha - 3$, which is at least $s+1 = A+B+C+M-1$ exactly when $A+M+\alpha \geq 2$. If $A+M\geq 1$, then we are done. Suppose $A+M = 0$, then $A = M = 0$ and $b+c = \alpha+\beta+\gamma$. Moreover, since $b > \beta$ and $c > \gamma$, then $\alpha+\beta+\gamma = b+c \geq \beta+\gamma + 2$. Thus $\alpha \geq 2$.

An immediate consequence of the previous lemma is that exactly one map need be considered for each algebra in order to determine the presence of the 