First of all, we need to prove === Lemma === Following the definitions on Definition:Ultraproduct :$\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$ {{iff}} :$\left\{ i : \left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right) \right \} \in \mathcal U$ === Proof === Let :$I_k := \left\{ i \in I : m_{k, i} = m'_{k, i} \right\}$ :$I^* :=  \left\{ i : \left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right) \right \}= \displaystyle \bigcap^n_{k = 1} I_k $ Suppose :$\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$ for $k = 1, \dotsc, n$ we have :$I_k \in \mathcal U$ for $k = 1, \dotsc, n$ Since $\mathcal U$ is closed under intersection :$I^* \in \mathcal U$ On the other hand, suppose  :$I^* \in \mathcal U$ Since $\mathcal U$ is upward-closed :$I_k \in \mathcal U$ for $k = 1, \dotsc, n$ Therefore :$\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$ {{qed}} === Proposition 1 === ''The definition of $f^\mathcal M$ is consistent.''  i.e. for $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$ :$\left(f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right)\right)_\mathcal U = \left(f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right)\right)_\mathcal U$ === Proof === Firstly note that:  :$\{i : f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right) = f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right) \} \supseteq  \{i : \left(m_{1, i}, \dots, m_{n, i}\right) = \left(m'_{1, i}, \dots, m'_{n, i}\right) \} $ and by $\mathcal U$ is an ultrafilter on $I$, we have :$\{i : \left(m_{1, i}, \dots, m_{n, i}\right) = \left(m'_{1, i}, \dots, m'_{n, i}\right) \} \in \mathcal U$ implies :$\{i : f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right) = f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right) \} \in \mathcal U$ Therefore,  :$\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$, by #Lemma, which is equvalent to $\{i : \left(m_{1, i}, \dots, m_{n, i}\right) = \left(m'_{1, i}, \dots, m'_{n, i}\right) \} \in \mathcal U$ implies :$\left(f^{\mathcal M_i}\left(m_{1, i}, \dots, m_{n, i}\right)\right)_\mathcal U = \left(f^{\mathcal M_i}\left(m'_{1, i}, \dots, m'_{n, i}\right)\right)_\mathcal U$ {{qed}} === Proposition 2 === ''The definition of $R^\mathcal M$ is consistent.''  i.e. for $\left(m_{k, i}\right)_\mathcal U = \left(m'_{k, i}\right)_\mathcal U$, $k = 1, \dotsc, n$ :$\left\{i \in I: \left({m_{1, i}, \dots, m_{n, i} }\right) \in R^\mathcal M_i\right\} \in \mathcal U$ {{iff}} :$\left\{i \in I: \left({m'_{1, i}, \dots, m'_{n, i} }\right) \in R^\mathcal M_i\right\} \in \mathcal U$ === Proof === Let :$S := \left\{i \in I: \left({m_{1, i}, \dots, m_{n, i} }\right) \in R^\mathcal M_i\right\}$ :$S' := \left\{i \in I: \left({m'_{1, i}, \dots, m'_{n, i} }\right) \in R^\mathcal M_i\right\}$ :$I^* :=  \left\{ i : \left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right) \right \}$ :$T := I^* \cap S$ :$T' := I^* \cap S'$ As #Lemma implies :$I^* \in \mathcal U$ therefore :$S \in \mathcal U$ implies $T \in \mathcal U$ Note that :$\left({m_{1, i}, \dots, m_{n, i} }\right) = \left({m'_{1, i}, \dots, m'_{n, i} }\right)$ for $i \in I^*$ we have :$T = T'$ Hence :$T' \in \mathcal U$ and :$S' \in \mathcal U$ since $S' \supseteq T'$ So far we have proved :$S \in \mathcal U$ implies $S' \in \mathcal U$ By symmetry,  :$S' \in \mathcal U$ implies $S \in \mathcal U$ {{qed}} Category:Model Theory pxaqnmgw5qct26nsoe4koay2w3aghan