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This is a continuation of a problem I asked yesterday seen here: Prove this proposition concerning a theory with ∀∃-axiomatization

The setting is reproduced below for easy reference.

Setting

A theory $\pmb{T}$ has a $\forall\exists$-axiomatization if it can be axiomatized by sentences of the form $$\forall v_1\ldots \forall v_n \exists w_1 \ldots \exists w_n ~~ \phi(\bar{v},\bar{w})$$ where $\phi$ is a quantifier free $\mathcal{L}$-formula.

Furthermore, suppose whenever $(\mathcal{M}_i : i \in \mathbb{I})$ is a chain of models of $\pmb{T}$, then $$\mathcal{M} = \bigcup \mathcal{M}_i \models \pmb{T}.$$ Let $\Gamma = \{ \phi : \phi \text{ is a $\forall\exists$-sentence and $\pmb{T} \models \phi$}\}$. Let $\mathcal{M} \models \Gamma$.

We also showed $\mathcal{N} \models \pmb{T}$ such that if $\psi$ is an $\exists\forall$-sentence and $\mathcal{M} \models \psi$, then $\mathcal{N} \models \psi$.

Finally there is $\mathcal{N}' \supseteq \mathcal{M} $ with $\mathcal{N}' \equiv \mathcal{N}$.

Now I would like to show there is an $\mathcal{M}' \supseteq \mathcal{N}'$ such that $\mathcal{M} \prec \mathcal{M}'$.

My Attempt

Update: since I read the question wrong, my proof is also wrong. Please disregard.

We create $\mathcal{M}'$ from $\mathcal{N}'$ by removing all constants of $\mathcal{N}$ from the universe of $\mathcal{N}'$, and let each $f^{\mathcal{M}'}_n = f^{\mathcal{N}'} | \mathcal{M}'^n$, and $R^{\mathcal{M}'} \subseteq R^{\mathcal{N}'} \cap \mathcal{M}'^n$. Clearly $\mathcal{M}'$ is a substructure of $\mathcal{N}'$. Now we can create some function $j : \mathcal{M} \rightarrow \mathcal{M}'$ that interprets each constant $c_{\mathcal{M}}$ in $\mathcal{M}$ as $c_{\mathcal{M}'}$ in $\mathcal{M}'$. Because $\mathcal{M}'$ contains an isomoporphic copy of $\mathcal{M}$, for each $\bar{a} \in \mathbb{M}$ there is a correponding $j(\bar{a}) \in \mathbb{M}'$. Finally, for all $\bar{a} \in \mathbb{M}$ and $\mathcal{L}$-formulas $\phi(\bar{x})$, we have

$$\mathcal{M} \models \phi(\bar{a}) \iff \mathcal{M}' \models \phi(j(\bar{a})).$$ Where $\mathcal{M}' \models \phi(j(\bar{a})) \Rightarrow \mathcal{M} \models \phi(\bar{a})$ because $M'$ contains a copy of $\mathcal{M}$. And $\mathcal{M} \models \phi(\bar{a}) \Rightarrow \mathcal{M}' \models \phi(j(\bar{a}))$ because ...

My Problems

  1. I would like to confirm my construction of $\mathcal{M}'$ is correct. To be honest it does not look correct to me. Isn't my $\mathcal{M}'$ just an isomorphic copy of $\mathcal{M}$?

  2. Suppose my construction is correct, I am having trouble proving why $$\mathcal{M} \models \phi(\bar{a}) \Rightarrow \mathcal{M}' \models \phi(j(\bar{a}))$$

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I must have misunderstood something, or perhaps you have omitted some assumption. The statement you want to prove seems to be false.

Consider a language with a binary relation symbol $\lt$. Let $\sf T$ be the trivial theory, axiomatized by the empty set of sentences; so $T$ is the set of all logically valid sentences, and $\Gamma$ is the set of all logically valid $\forall\exists$ sentences.

Let $N=\{0,1,2,3,\dots\}$, the set of all nonnegative integers, and let $\mathcal N'=\mathcal N=(N,\lt)$. Let $\mathcal M=(M,\lt)$ where $M=N\setminus\{0\}=\{1,2,3,\dots\}$, the set of all positive integers. Then $\mathcal N'\supseteq\mathcal M$, and $\mathcal N'\equiv\mathcal N\equiv\mathcal M$ since $\mathcal N'=\mathcal N\cong\mathcal M$.

Suppose there is $\mathcal M'\supseteq\mathcal N'$ such that $\mathcal M\prec\mathcal M'$. Then $1$ must be the least element of $\mathcal M'$, because $1$ is the least element of $\mathcal M$ and $\mathcal M\prec\mathcal M'$. But $1$ cannot be the least element of $\mathcal M'$, because $\mathcal M'\supseteq N'$ and $1$ is not the least element of $\mathcal N'$.

What am I missing?

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