[m] \setminus  \{k\}} U_{i_j}$ for all $k = 1,...,m$, and $\cap_{k=1}^m U_{i_k} = \emptyset \Rightarrow x_{i_1} + ... + x_{i_m} \in J_C$.

-   $ \cup^m_{k=1} U_{i_k} = X \Rightarrow x_{i_1} + ... + x_{i_m} + 1 \in J_C$ for all $m \geq 2$ indices $1 \leq i_i < i_2 < ... < i_m \leq n$.

We present a counterexample of each of the three statements, beginning with the converse of the Type 4 relation. The counterexamples for the Type 5 and 6 relations are the same as in the proof of Lemma 3.1. As we will see, we can use the same counterexamples for these two relations because $J_C \subseteq I(C)$.

[*Converse of Type 4:*]{} Consider the code $C=\{\emptyset,1,2,12\}$ on 2 neurons pictured in Figure \[fig:T4C\]. Also let $\sigma_1 = \tau_1 = 1$ and $\sigma_2 = \tau_2=2$. Then $U_1  \cap U_1^c = \emptyset = U_2 \cap U_2^c$, but $J_C = \langle 0 \rangle$, so $x_1(1+x_1) + x_2(1+x_2) \notin J_C$.

[*Converse of Type 5:*]{} Again let $C = \{\emptyset, 12, 13, 14, 123 \}$. The realization of this code is shown in Figure \[fig:T5C\]. From the proof of Lemma 3.1, we know that $x_1 + x_2 + x_3 + x_4 \notin 