$\newcommand{\Th}{\mathsf{Th}}$ $\newcommand{\M}{\mathfrak{M}}$ $\newcommand{\N}{\mathfrak{N}}$ $\newcommand{\L}{\mathscr{L}}$
I'm trying to prove equivalent statements of a complete theory by cyclic implication. However, I'm unable to prove $(4) \implies (5)$ or equivalently $(4) \implies (3)$ as $(3) \implies (5)$ is obvious.
Let $T$ be a consistent $\L$-theory. The following are equivalent:
- $T$ is complete ;
- The deductive closure of $T$ is a maximal consistent theory ;
- $\forall \M \models T, T = \Th(\M)$ ;
- $\exists \M \models T, T = \Th(\M)$ ;
- $\forall \M, \N \models T, \M \equiv \N$
- for any $\L$-sentence $\varphi$, $T \not\models \varphi \implies T \models \neg \varphi$.
($T$ is complete $\iff$ for all sentence $\varphi$, $T \models \varphi$ or $T \models \neg \varphi$.)
Draft for $(4) \implies (5)$:
Let $\M \models T$ s.t. $T = \Th(\M)$. Then for $\N_1, \N_2 \models T$, there is $\Th(\N_1) \supseteq \Th(\M)$ and $\Th(\N_2) \supseteq \Th(\M)$.
I have no idea how to prove the inclusion $\subseteq$.
Let $M$ be as in 4.(I don't like the curly stuff)
The important point is that $T = Th(M)$ is complete. This is because for any $\varphi$ sentence, either $M \vDash \varphi$ or $M \vDash \neg \varphi$, so either $\varphi \in T$ or $\neg \varphi \in T$.
Well then given any $N \vDash T$. Let $\varphi$ be any sentence. We then have:
$M \vDash \varphi \iff \varphi \in T \iff N \vDash \varphi$
The first $\iff$ follows from $T = Th(M)$.
The second $\implies$ follows from $N \vDash T$.
The second $\impliedby$ is by completeness of $T$. Either $\varphi \in T$ or $\neg \varphi \in T$. Since $N \vDash T$ and $N \vDash \varphi$, clearly $\neg \varphi \not \in T$ (else $N \vDash \varphi$ and $N \vDash \neg \varphi$, and so since $\neg \varphi \not \in T$,we have $\varphi \in T$).
Thus for any $N \vDash T$, $N \equiv M$. So given $N, N' \vDash T$, $N \equiv M \equiv N' \rightarrow N \equiv N'$