There must be a flaw in the following argument, but I don't see it at the moment. Who can point it out?
In first-order logic, suppose that a structure $\mathfrak{U}$ is a model of the formula ($x = 3$):
$$ \mathfrak{U} \vDash x = 3 $$
Then, by the definition of a model, for any assignment function s it is true that
$$ \mathfrak{U} \vDash (x = 3) [s]$$
So the universe of $\mathfrak{U}$ has only one element, which equals the constant 3. And so the structure is also a model of $\forall x (x = 3)$:
$$ \mathfrak{U} \vDash \forall x (x = 3) $$
We have just seen that any model of $x = 3$ is also a model of $\forall x (x = 3)$. So
$$ x = 3 \vDash \forall x (x = 3) $$
Choose any complete deduction system for first-order logic. Then by completeness we have
$$ x = 3 \vdash \forall x (x = 3) $$
There is a natural deduction rule
$$ \frac{\Gamma, A \vdash B}{\Gamma \vdash A \to B} $$
So we can apply it to obtain
$$ \varnothing \vdash x = 3 \to \forall x (x = 3) $$
or
$$ \vdash x = 3 \to \forall x (x = 3) $$
But that statement seems to be invalid. For example, it's not true in a structure whose universe has more than one element under an assignment that maps x to 3.
What am I missing here?
Best Answer
So, $ x = 3 \models (\forall x \mathop. x = 3)$ is not true, and it is not true for the following reason.
$ \Phi \models \Delta $, where $\Phi$ and $\Delta$ are sets of formulas, means the following:
$$\Phi \models \Delta \;\; \textit{if and only if} \;\; \text{for all $(M, v)$, if $M, v \models \Phi$, then $M, v \models \Delta$} $$
$M, v \models \varphi$ means that $\varphi$ holds in the structure $M$ with the variables interpreted by $v$, a mapping from variable names to their denotations. Note that we are quantifying over $M$ and $v$ at the same time.
The relation that you appear to be using in place of $\models$ is well-defined, and we might write it $\Phi \models^g \Delta$. I have seen it called the global consequence relation of first-order logic, but it is not commonly used and certainly isn't the default. I'm only mentioning the global consequence relation at all to stress the importance of running back to the definitions in logic. I first heard about the idea of a global consequence relation from sequitur's answer to a question I asked about modal logic.
$$ \Phi \models^g \Delta \;\;\textit{if and only if}\;\; \text{for all $M$ such that $M \models \Phi$, it holds that $M \models \Delta$} $$
In the above definition $M \models \Phi$ holds if and only if $M, v \models \Phi$ for all $v$.
In the family of proof systems that use the $\vdash$ notation, $\vdash$ is the syntactic analogue of $\models$ not $\models^g$.
Also, as you have noticed, it is indeed not true that $x = 3 \models^g (\forall x \mathop. x = 3)$ implies $\models^g (x = 3) \to (\forall x \mathop. x = 3)$.
The equivalent of implication introduction in some hypothetical proof system in $\models^g$-land would give us $\models^g (\forall x \mathop. x = 3) \to (\forall x \mathop. x = 3)$. Basically, because the free variables are quantified over early, we have to protect statements that contain them.