Does $ \mathfrak{A} \models (\exists x) \phi $ imply that $ \phi[x / t] $ for some term $ t $. Second approach?

Question

If $ \mathfrak{A} \models (\exists x) \phi $, does it mean that $ \mathfrak{A} \models \phi[x / t] $ for some term $ t $?

$ \mathfrak{A} \models (\exists x) \phi $. It means that there exists $a \in A$ that $(\mathfrak{A}, s[x|a]) \models \phi$ So, the answer for problem is: Generally not, it depends. If the term $t$ is such closed formula that $[t]_{\mathfrak{A}} = a$ the $\mathfrak{A} \models \phi[x/t] $ is satisfied. Otherwise, not.

Answer 1

I think Mauro is answering a slightly different question than the one you're asking. There are examples of structures which "don't have enough terms" (structures and theories which do have enough terms are said to have the witness property or existence property). For example, any structure in a language which only has relation symbols! There are no terms at all in such a language (besides free variables), but of course we have plenty of true existential sentences - e.g. "$\exists x(x=x)$."

Mauro's point is that any structure $\mathfrak{A}$ has a canonical expansion $\mathfrak{A}_{+}$ where we add a new constant symbol for each element of the domain, and this expansion does indeed have the witness property. And it is frequently useful, in trying to understand $\mathfrak{A}$, to study $\mathfrak{A}_+$. However, the two are indeed different structures, and the general answer to you question is: no, the witness property does not always hold.

Answer 2

We can see :

• Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 80 :

A structure for a first-order language will tell us :

1. What collection of things the universal quantifier symbol ($∀$) refers to, and

2. What the other parameters (the predicate and function symbols) denote.

Formally, a structure $\mathfrak A$ for our given first-order language is a function whose domain is the set of parameters [i.e. the symbols of the language] and such that

1. $\mathfrak A$ assigns to the quantifier symbol $∀$ a nonempty set $|\mathfrak A|$ called the universe (or domain) of $\mathfrak A$.

2. $\mathfrak A$ assigns to each $n$-place predicate symbol $P$ an $n$-ary relation $P^{$\mathfrak A} \subseteq |\mathfrak A|^n$; i.e., $P^{$\mathfrak A}$ is a set of $n$-tuples of members of the universe.

3. $\mathfrak A$ assigns to each constant symbol $c$ a member $c^{$\mathfrak A}$ of the universe $|\mathfrak A|$.

4. $|\mathfrak A|$ assigns to each $n$-place function symbol $f$ an $n$-ary operation $f^{$\mathfrak A}$ on $|\mathfrak A|$; i.e., $f^{$\mathfrak A} : |\mathfrak A|^n \to |\mathfrak A|$.

Thus : YES, if we "enlarge" the language adding more constant symbols, we modifiy the structure because we have to modify the point 3 above of the mapping.

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Setting apart the terminology, this is very similar to what happens in Computability & Logic, page 117 :

Let us say that in the interpretation $\mathcal M$ the individual $m$ satisfies $F(x)$, and write $\mathcal M \vDash F[m]$, to mean "if we considered the extended language $L \cup \{ c \}$ obtained by adding a new constant $c$ in to our given language $L$, and if among all the extensions of our given interpretation $\mathcal M$ to an interpretation of this extended language we considered the one $\mathcal M^c_m$ that assigns to $c$ the denotation $m$, then $F(c)$ would be true":

$\mathcal M \vDash F[m]$ if and only if $\mathcal M^c_m \vDash F(c)$.

(For definiteness, let us say the constant to be added should be the first constant not in $L$ in our fixed enumeration of the stock of constants.)