the Type 6 relation held, then it would be true that $$x_1+x_2+1 \in I(C).$$ However, $$(1+x_1)(1+x_2) = 1+x_1+x_2+x_1x_2 \in J_C \subseteq I(C).$$ Adding expressions (1) and (2) would then imply that $f:=x_1x_2 \in I(C)$. However, for a point $p \in U_1 \cap U_2$, $f(c(p)) = 1\cdot 1 = 1$ and thus $f \notin I(C)$, a contradiction. So $x_1+x_2+1 \notin I(C)$.

Now we are ready to show that none of converses of the Type 4-6 relations hold.

For each of the following relations, there exists a collection of sets $\mathcal{U} = \{U_i\}^n_{i=1}$ in a stimulus space $X$ (with corresponding neural code $C = C(\mathcal{U})$) such that the relation does [**not**]{} hold:

-   $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U^c_i) = U_{\sigma_2} \cap (\cap_{j \in \tau_2} U^c_j) \Rightarrow x_{\sigma_1} \prod_{i \in \tau_1} (1+x_i) + x_{\sigma_2} \prod_{j \in \tau_2} (1+x_j) \in J_C$ for any subsets $\sigma_1$, $\sigma_2$, $\tau_1$, $\tau_2 \subseteq [n]$.

-   For all $m \geq 2$ indices $1 \leq i_i < i_2 < ... < i_m \leq n$,

    1.  $m$ is even and $U_{i_k} \subseteq \cup_{j \in [m] \setminus  \{k\}} U_{i_j}$ for all $k = 1,...,m \Rightarrow x_{i_1} + ... + x_{i_m} \in J_C$, and

    2.  $m$ is odd, $U_{i_k} \subseteq \cup_{j \in 