Let $\Sigma $ be a signature. Let $A$ be a universum of $\mathfrak{A}$. We will say that the structure $\mathfrak{A}$ over $\Sigma$ has property $F$ if for every two terms $t,s$ ( with one free variable $x$) set $S$ of such $a \in A$ that $$(\mathfrak{A}, x:a) \models t(x) = s(x) $$ is finite or $S = A$
Show that:
If $f \in \Sigma $ where $f$ is a one-argument function symbol then there does't exist the such set of sentences $\Delta$ over $\Sigma$ that $\mathfrak{A} \models \Delta \iff \mathfrak{A} \text{ has property F}$
If $\Sigma $ contains only constants symbols and relationships symbols ( in another words: it doesn't contain function symbols) then there exists the such set of sentences $\Delta$ over $\Sigma$ that $\mathfrak{A} \models \Delta \iff \mathfrak{A} \text{ has property F}$
2. Because $s, t$ are terms and there are no function symbols we can see it is always true that $S$ is finite. It seems that every structure without function has $F$ property.
1. Let assume that there eixsts the such $\Delta$ over $\Sigma = \{ f \} $ where $f$ is one-arugment functional symbol that $(\mathfrak{A}, x:a) \models \Delta \iff \mathfrak{A} $ has property $F$.
Let's extend signature $\Sigma$. $\Sigma' = \{ f \} \cup \{ c_i | i \in \mathbb{N}\}$
( later, we can interpret $c_i$ as just $i $ from $\mathbb{N}$ in model)
Let $\Delta' = \Delta \cup \Gamma$
$\Gamma = \{ f(c_i) = c_{i+1} | i \in \mathbb{N} \text{ and is even}\} \cup \{f(c_i) = c_i | i \in \mathbb{N} \text { and is odd} \} $
Let's take finite subset $\Delta_0 \subset \Delta'$.
It is satisfable. It is easy to point model:
Let $x = \max{\{i | (f(c_i) = s) \in \Delta_0 }\}$ Now, we can take a model $M = (N, f) $ where $N = \{i | i \le x\}$ and $f$ is defined: $$f(n) = n \text{ n is odd} \\ f(n) = n + 1 \text{ n is even}$$
It is easy to check that $M$ has the $F$ property and satisfies $\Delta_0 \cap \Gamma$ So, by Compactness theorem $\Delta' $ is not satisfable. But, it is not possible that $\Delta$ is satisfied while $\Gamma$ is satisfied. Contradiciton.