but $\sigma_2 \not \subseteq c$, or $\sigma_2 \subseteq c$ but $\sigma_1 \not \subseteq c$, cannot be in $C$ (this would contradict the fact that $U_{\sigma_1} = U_{\sigma_2}$), and so $p_c = \prod_{i \in c} x_i \prod_{j \notin c}(1+ x_j) \in J_C$ for every such $c$. Letting $C_1 = \{ c \subseteq [n] \mid \sigma_1 \subseteq c$ but $\sigma_2 \not \subseteq c \}$, $C_2 =\{ c \subseteq [n] \mid \sigma_2 \subseteq c$ but $\sigma_1 \not \subseteq c \}$, and $f = \sum_{c \in C_1 \cup C_2} p_c$, we claim that $f = x_{\sigma_1} + x_{\sigma_2}$ and that $f \in J_C$ as desired.

To see this, first note that $f \in J_C$ because each $p_c$ is in $J_C$ by the remarks above. To prove that $f = x_{\sigma_1}+x_{\sigma_2}$, we will consider $x_\sigma$ for all $\sigma \subseteq [n]$ and determine whether $x_{\sigma}$ appears as a summand in $f$.

First consider $x_\sigma$, where $\sigma_1 \not \subseteq \sigma$ and $\sigma_2 \not \subseteq \sigma$. Then $x_\sigma$ is not a term in the pseudo-monomial $p_c$ for any $c \in C_1 \cup C_2$ since all terms in such a pseudo-monomial must be of the form $x_{\sigma_1}x_{\sigma'}$ or $x_{\sigma_2}x_{\sigma'}$ for some $\sigma' \subseteq [n]$. So $x_{\sigma}$ is not a 