Problems understanding: $\Sigma \vdash \theta \text{ iff }\Sigma \vdash \forall x \theta$

Question

I'm trying to make sense of a theorem I came across recently:

$$\Sigma \vdash \theta \text{ iff }\Sigma \vdash \forall x \theta$$

Say $\theta$ is $(x=1)$, then we have $\Sigma \vdash (x=1)\text{ iff }\Sigma \vdash \forall x (x=1)$.

Well, $\Sigma \vdash \theta \Rightarrow \Sigma \models \theta$, which means for every structure $(\Bbb A)$ with universe $(A)$ and every assignment function $(s)$: $$(\forall \Bbb A)(\forall s)(\Bbb A \models \Sigma[s] \Rightarrow \Bbb A \models \theta[s])$$ and $$(\forall \Bbb A)(\forall s)(\forall a \in A)(\Bbb A \models \theta[s(x|a)])$$ where $s(x|a)(v) = \{s(v) \text{ if v is not x}, a \text{ if v is x}\}$.

But then doesn't this mean that with ANY structure and ANY universe associated with it and EVERY assignment function that every $x$ is $1$?

You have $\Sigma\vDash \theta$ means every structure that satisfy $\Sigma$ also satisfy $\theta$, so yes, it does mean that "with ANY structure and ANY universe associated with it(I assume you mean $\mathbb A\vDash\Sigma$ in this part) and EVERY assignment function that every $x$ is $1$" — ℋolo, Dec 29 '18 at 16:21
See Universal generalization : the theorem holds provided that $x$ is not free in $\Sigma$. — Mauro ALLEGRANZA, Dec 29 '18 at 16:24
The intuition is : if $\theta$ holds, without any assumptions on $x$ [this is the gist of the proviso : "$x$ not free in $\Sigma$"], for $x$ whatever, then it holds of every $x$. — Mauro ALLEGRANZA, Dec 29 '18 at 16:26
See the post Why does universal generalization work? and also the post How does Universal Generalization agree with facts? — Mauro ALLEGRANZA, Dec 29 '18 at 18:00
See also I don't get how “universal generalization” works? — Mauro ALLEGRANZA, Dec 29 '18 at 18:03
I think I see now. It's ASSUMING that the structure satisfies $\Sigma$ then this is true, not just any structure. — Oliver G, Dec 29 '18 at 21:42
Yes, correct. $\theta$ is true in every structure that satisfies the "axioms" in $\Sigma$. — Mauro ALLEGRANZA, Dec 30 '18 at 08:37
Consider as $\Sigma$ the set of first-order Peano axioms. The proviso on $x$ not free prevents deriving the fallacious $\forall x (x=1)$ simply because we cannot prove $(x=1)$ (the logic is sound and it is not true that $\mathbb N \vDash (x=1)$). — Mauro ALLEGRANZA, Dec 30 '18 at 08:39

Problems understanding: $\Sigma \vdash \theta \text{ iff }\Sigma \vdash \forall x \theta$

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