\setminus \sigma_1 | \geq 1$ and $k = |\sigma \setminus \sigma_2| \geq 1$ [^1]. However, if we count all such $c$ and $c'$, we will also count twice the $2^l$ codewords $c''$ such that $\sigma_1\cup \sigma_2 \subseteq c'' \subseteq \sigma$. (Here $l = |\sigma \setminus (\sigma_1 \cup \sigma_2)| \geq 0$.) These codewords $c''$ are not in $C_1 \cup C_2$ and therefore must be excluded twice from our count of codewords in $C_1 \cup C_2$ contained in $\sigma$. So there are $2^m + 2^k - 2(2^l)$ codewords contained in $\sigma$ that are in $C_1 \cup C_2$. This implies that $x_\sigma$ will be added $2^m + 2^k - 2(2^l)$ times in $f$ and therefore will cancel out since $2^m + 2^k - 2(2^l)$ is even.

Thus, the only terms that appear in $f$ are $x_{\sigma_1}$ and $x_{\sigma_2}$, so $f = x_{\sigma_1} + x_{\sigma_2} \in J_C$ as desired.

[*Case 2 ($\tau_1 \cup \tau_2 \neq \emptyset$):*]{} In the case where $\tau_1 \cup \tau_2 \neq \emptyset$ and $U_{\sigma_1} \cap (\cap_{i \in \tau_1} U_i^c) = U_{\sigma_2} \cap (\cap_{j \in \tau_2} U_j^c) $, we can apply the same reasoning used in the previous case if we first define a new stimulus space $\{U_i'\}$ (with receptive field code 