As to why the existence of closed terms is necessary, a formula with a free variable generally cannot be assigned a truth value that is consistent with all possible valuations. Some valuations could yield $$ while others could yield $$. However, if all of the variables in a formula are bounded so that the formula is a sentence, then it has only one truth value for all possible valuations. Now, one way of turning a formula with free variables into a sentence is to replace each free variable by a closed term (if constant symbols are available).
It has been written in my last post: Does $ \mathfrak{A} \models (\exists x) \phi $ imply that $ \phi[x / t] $ for some term $ t $?
I would like to ask somebody to explain me it. I don't understand what the author mean, especially by one truth.
Let's see how I reasoning about it. Let $\phi = \exists x p(x)$ . Let $\mathfrak{A} = (A, \Sigma^f, \Sigma^r) $ be a model for $\phi$. $A = \{1,2\}, \Sigma^f = \emptyset, \Sigma^r = \{p(1)\}$ Let $\delta_1 (x) = 1, \delta_2 (x) = 2$.
Now, we see that $(\mathfrak{A}, \delta_1) $ satisfies $\phi$ and $(\mathfrak{A}, \delta_2) $ satisfies $\phi$ doesn't. So it does mean that it depends on valutaion.
Please help me understand where I am wrong- I know that you are right, of course.
Some comments abou the comments.
Regarding :
I would like to rephrase it as follows :
Consider the formula of first-order language for arithmetic : $(x=0)$ and interpret it in the domain $\mathbb N$ of natural numbers.
With the variable assignment function $s$ such that $s(x)=0$, the formula is evaluated to $\mathsf T$ while with the function $s'$ such that $s'(x)=1$ the formula is evaluated to $\mathsf F$.
For the existentially quantified formula $\exists x \ (x=0)$, the semantical specification is :
With this specification, obviously the above $s$ satisfy $\exists x \ (x=0)$, but also $s'$ does, because $s'(x|0)$ satisfy it.
Thus, the "trick" of the specification is simply to formalize the fact that, in order to satisfy $\exists x \varphi$ in a certain interpretation, it is enough to find an "object" in the domain such that $\varphi$ holds of it [irrespective of the fact that we have a name for it or not : see Computability & Logic].
We can easily prove that, for a closed formula $\varphi$ (a sentence), an interpretation $\mathfrak A$ satisfies $\varphi$ with every function $s$ from $\text {Var}$ into $|\mathfrak A|$, or $\mathfrak A$ does not satisfy $\varphi$ with any such function.
This is the meaning of :