indices , so the implication is vacuously true.

When $n=2$, in order for the left-hand side of the implication to be true, we must have $m=2$, $U_1 \subseteq U_2$, and $U_2 \subseteq U_1$, which implies that $U_1 = U_2$. This means that neither 1 nor 2 is in the associated code $C$, so $x_1(1+x_2), x_2(1+x_1) \in J_C$. Therefore, $x_1(1+x_2)+ x_2(1+x_1) = x_1+x_2 \in J_C$

Finally, when $n=3$, we can have $m=2$ or $m=3$. If $m=2$, then $U_{i_1}=U_{i_2}$ as in the $n=2$ case. If we let $i_3$ be the remaining element of $\{1,2,3\}$, then $i_1$, $i_2$, $ i_1i_3$, $i_2i_3 \notin C$, which implies that the following pseudo-monomials are in $J_C$: $$\begin{aligned}
x_{i_1}(1+x_{i_2})(1+x_{i_3}) &= x_{i_1} + x_{i_1}x_{i_2} + x_{i_1}x_{i_3} + x_{i_1}x_{i_2}x_{i_3} \\
x_{i_2}(1+x_{i_1}) (1+x_{i_3}) &=  x_{i_2} + x_{i_1}x_{i_2} + x_{i_2}x_{i_3} + x_{i_1}x_{i_2}x_{i_3} \\
 x_{i_1}x_{i_3}(1+x_{i_2}) &= x_{i_1}x_{i_3} + x_{i_1}x_{i_2}x_{i_3}\\ 
 x_{i_2}x_{i_3}(1+x_{i_3}) &=  x_{i_2}x_{i_3} + x_{i_1}x_{i_2}x_{i_3}\end{aligned}$$ Adding these four terms together gives us $x_{i_1}+x_{i_2} \in J_C$.

If $m=3$, similar reasoning applies, but we also must assume that $\cap_{k=1}^3 U_{i_k} = \emptyset$. We know that the codewords $1$, $2$, $3 \notin C$ since $U_{i_k} \subseteq \cup_{j \in [3] \setminus  \{k\}} U_{i_j}$ for all $k = 1,2,3$, and we know that $123 \notin C$ because $\cap_{k=1}^3 U_{i_k} 