either $\sigma_1=1$ and $\tau_1=\emptyset$, or $\sigma_1 = \tau_1 = \emptyset$. The same is true for $U_{\sigma_2} \cap (\cap_{i \in \tau_2} U^c_i)$, so $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U^c_i) = U_{\sigma_2} \cap (\cap_{j \in \tau_2} U^c_j)$ implies that we have one of the following cases:

-   $\sigma_1 = \sigma_2$ and $\tau_1=\tau_2$

-   $\sigma_1 \neq \sigma_2$ and $\tau_1 = \tau_2 = \{1\}$

-   $\sigma_1 \neq \sigma_2$ and $\tau_1 = \tau_2 = \emptyset$

In case 1, we can use the same reasoning we used when dealing with $C_1$, so it is enough to show that the converse of the Type 4 relation holds in cases 2 and 3. For these cases, we can assume without loss of generality that $\sigma_1 = 1$ and $\sigma_2 = \emptyset$.

In case 2, $$x_{\sigma_1} \prod_{i \in \tau_1} (1+x_i) + x_{\sigma_2} \prod_{j \in \tau_2} (1+x_j)  = x_1(1+x_1) + (1)(1+x_1) = (1+x_1)^2 \in J_C$$ because $1+x_1 \in J_C$.

Similarly, in case 3, $$x_{\sigma_1} \prod_{i \in \tau_1} (1+x_i) + x_{\sigma_2} \prod_{j \in \tau_2} (1+x_j) = x_1(1)+ (1)(1) = 1+x_1 \in J_C.$$ So the implication holds.

[*Converse of Type 5:*]{} We consider the cases where $n=1$, $n=2$, and $n=3$ separately.

In the case where $n=1$, we cannot have $m \geq 2$ 