Prove complete iff $\mathcal{A} \equiv \mathcal{B}$

Question

I want to prove that a set $\Sigma$ of $\mathcal{L}$-sentences is $\textbf{complete}$ iff when $\mathcal{A}, \mathcal{B} \models \Sigma$, then $\mathcal{A} \equiv \mathcal{B}$.

Starting with the definition of complete, we have:

A set $\Sigma$ of $\mathcal{L}$-sentences is complete if for all $\mathcal{L}$-sentences $\varphi$, either $\Sigma \models \varphi$ or $\Sigma \models (\neg \varphi)$.

---

Seems like such a straightforward proof. My attempt:

$(\Rightarrow)$ Let $\Sigma$ be a set of $\mathcal{L}$-sentences. By the given, we have that $\Sigma$ is complete. Take any $\phi$ to be an $\mathcal{L}$-sentence. Then, we have that either $\Sigma \models \phi$ or $\Sigma \models (\neg \phi)$. Let $\mathcal{A}, \mathcal{B}$ be models such that for all $\phi \in \mathcal{A}, \mathcal{B}$ we have that $\mathcal{A}, \mathcal{B} \models \Sigma$. As $\phi$ satisfies $\Sigma$ then it must be that $\mathcal{A}, \mathcal{B}$ satisfy $\Sigma$, hence for all $\mathcal{L}$-sentences in $\mathcal{A},\mathcal{B}$ we have that $\mathcal{A} \equiv \mathcal{B}$.

Does this make sense?

Answer 1

The structure of your proof is a bit off -- you start speaking about $\phi$ before you introduce $\mathcal A$ and $\mathcal B$, so it is unclear what can depend on what. And it breaks down completely when you begin speaking about "$\phi$ satisfies $\Sigma$", because a formula doesn't satisfy anything; structures do.

In the claim you're trying to prove $\mathcal A$ and $\mathcal B$ are implicitly universally quantified after the "iff" (which is bad style but unfortunately common), so by the structure of your claim, the proof ought to start

($\Rightarrow$). Assume that $\Sigma$ is complete and that $\mathcal A$ and $\mathcal B$ are structures such that $\mathcal A\vDash\Sigma$ and $\mathcal B\vDash\Sigma$. We need to prove that $\mathcal A\equiv\mathcal B$ ...

Now you can start introducing a $\phi$, because the definition of $\mathcal A\equiv\mathcal B$ includes a quantification over $\phi$:

... Let $\phi$ be an arbitrary sentence; we then need to prove that $\mathcal A\vDash\phi \iff \mathcal B\vDash\phi$. ...

By symmetry it is enough to prove one of the directions here, so

... Assume $\mathcal A\vDash \phi$; we want to prove that $\mathcal B\vDash \phi$.

So far we've just unfolded definitions; now we can actually start arguing. The crucial fact is indeed that $\Sigma\vDash \phi$ or $\Sigma\vDash \neg\phi$, but because we have assumed $\mathcal A\vDash \Sigma$ we can't have $\Sigma\vDash\neg\phi$, because then by transitivity $\mathcal A\vDash\neg\phi$ which would contradict our other assumption $\mathcal A\vDash\phi$ (by definition of what $\vDash$ means for structures and negated formulas).

Therefore we must have $\Sigma\vDash\phi$, and from there it is smooth sailing to $\mathcal B\vdash\phi$.

---

The ($\Leftarrow$) direction is more subtle. There you need to assume a $\phi$ before you decide on $\mathcal A$ and $\mathcal B$ -- but on the other hand you get to choose $\mathcal A$ and $\mathcal B$, because they're now quantified on the premise side of the implication.