end, let $C = \{\emptyset, 12, 13, 14, 123 \}$. The corresponding stimulus space is shown in Figure \[fig:T5C\]. From the figure, we can see that $U_{i_k} \subseteq \cup_{j \in [4]\setminus \{k\}} U_{i_j}$, where $i_1=1,...,i_4=4$. However, if $f := x_1+x_2+x_3+x_4$, then for any point $p \in (U_1 \cap U_2 \cap U_3) \setminus U_4$, $f(c(p)) = (x_1 + x_2 + x_3 + x_4)(c(p)) = 1+1+1+0 \equiv 1 \mod 2$, so $f=x_1+x_2+x_3+x_4 \notin I(C)$.

There are also counterexamples for the second implication. For instance, the code $C' = \{\emptyset, 12, 13, 14, 123, 145\}$ (shown in Figure \[fig:Counter5Odd\]) with $i_1= 1, ..., i_5=5$ satisfies $U_{i_k} \subseteq \cup_{j \in [5]\setminus \{k\}} U_{i_j}$ for all $k=1,...,5$, and $\cap_{k=1}^5 U_{i_k} = \emptyset$. But, again, if $g := x_1 + x_2 + x_3 + x_4+x_5$ and $p \in  (U_1 \cap U_2 \cap U_3) \setminus (U_4 \cup U_5)$, then $g(c(p)) = (x_1 + x_2 + x_3 + x_4+x_5)(c(p)) = 1+1+1+0+0 \equiv 1 \mod 2$. So $g \notin I(C')$.

[*Converse of Type 6:*]{} Now consider the code $C = \{1,2,12\}$ on 2 neurons pictured in Figure \[fig:T6C\]. We compute that $$J_C = \langle (1+x_1)(1+x_2) \rangle,$$ and we can see that $\cup_{i=1}^2 U_i = X$. If the converse of 