[proofplan] The result is the sentence case of Łoś's Theorem. The general theorem compares satisfaction of an $L$-formula in the ultraproduct with satisfaction on a $\mathcal U$-large set of coordinates, relative to coordinate representatives of the parameters. Since a sentence has no free variables, there are no parameters to choose, and the coordinate truth set is exactly $\{i \in I : M_i \models \sigma\}$. The theory statement follows by applying the sentence case to each axiom of $T$. [/proofplan] [step:Apply Łoś's Theorem with no free variables] Let $\sigma$ be an $L$-sentence. By Łoś's Theorem for ultraproducts (citing a result not yet in the wiki: Łoś's Theorem), for every $L$-formula $\varphi(x_1,\dots,x_n)$ and every choice of elements $a_1,\dots,a_n \in \prod_{i \in I} M_i / \mathcal U$ represented by coordinate functions \begin{align*} \tilde a_k: I &\to \bigsqcup_{i \in I} M_i, \end{align*} with $\tilde a_k(i) \in M_i$, one has \begin{align*} \prod_{i \in I} M_i / \mathcal U \models \varphi(a_1,\dots,a_n) \quad \Longleftrightarrow \quad \{i \in I : M_i \models \varphi(\tilde a_1(i),\dots,\tilde a_n(i))\} \in \mathcal U. \end{align*} Apply this theorem to the formula $\varphi = \sigma$. Since $\sigma$ is a sentence, it has $n = 0$ free variables, so no representatives or parameters occur. The displayed equivalence therefore becomes \begin{align*} \prod_{i \in I} M_i / \mathcal U \models \sigma \quad \Longleftrightarrow \quad \{i \in I : M_i \models \sigma\} \in \mathcal U. \end{align*} This proves the asserted equivalence for each $L$-sentence $\sigma$. [/step] [step:Pass from one sentence to all axioms of the theory] Let $T$ be a set of $L$-sentences, and suppose there exists $J \in \mathcal U$ such that $M_i \models T$ for every $i \in J$. To prove \begin{align*} \prod_{i \in I} M_i / \mathcal U \models T, \end{align*} it suffices to prove that the ultraproduct satisfies every sentence $\tau \in T$. Fix $\tau \in T$. Since $M_i \models T$ for every $i \in J$, we have $M_i \models \tau$ for every $i \in J$. Hence \begin{align*} J \subset \{i \in I : M_i \models \tau\}. \end{align*} Because $\mathcal U$ is a filter, it is upward closed under inclusion. Since $J \in \mathcal U$, this gives \begin{align*} \{i \in I : M_i \models \tau\} \in \mathcal U. \end{align*} Applying the sentence case already proved to $\tau$, we obtain \begin{align*} \prod_{i \in I} M_i / \mathcal U \models \tau. \end{align*} As $\tau \in T$ was arbitrary, the ultraproduct satisfies every sentence of $T$, and therefore \begin{align*} \prod_{i \in I} M_i / \mathcal U \models T. \end{align*} [/step]