Let $\Sigma$ will be signature (language). Structure $A$ has property $F$ over $\Sigma$ if for each two terms $s,t$ over $\Sigma$ with one free variable $x$ (both can have only one free variable $x$), set of elements $a\in A$ satisfying $(A, x:a) \models t = s$ is either finite or is $A$. For example, $\langle \mathbb{R}, +, *, 1\rangle$ has a property $F$. Show that
(a) If $\Sigma$ contains only one-argument function $f$ then there is no $\Delta$ over $\Sigma$ such that such that $A\models \Delta$ iff $A$ has property $F$.
(b) If $\Sigma$ contains only relations and constants then exists $\Delta$ over $\Sigma$ such that $\Delta\models A$ iff $A$ has property $F$.
(a)
First of all, what about form of terms ?
$x, f(x), f(f(x)),...$
Let suppose that such $\Delta$ exists. Let $\Delta' = \Delta \cup \{c_i = f(c_{i})|i\in \mathbb{N}\}\cup \{f(d_i) \not = d_i | i\in\mathbb{N}\}$ where $c_i,d_i$ are added constants. Lets consider any finite subset of $\Delta'$. It is satisfied because only finite number of constants $c$ - it is the number of solutions each equation $s=t$. For other arguments we can force that $f(x)\neq x$.
Obvously, $\Delta'$ is not satisfied because there exists infinitely many solutions - thanks to the fact that we have infinitely many constants $c$. Moreover, set of solution is not entire universum - thanks to constans $d$. Now, I can see that it is sufficient only one constant $d$.
(b)First of all, what about form of terms ?
$1, 2, 2, 0, x$ I am not sure if I correctly understand terms, but terms can only contains functions, free variables, and constants (in real 0-arg functions). Relations - we understand difference between function and relation, henec I can't see a sense relation in term. So $t=s$ with only constants:
$3 = 3$, $3 = 5 $
These two cases are ok - in first entire $A$ states solution. In second set of solution is empty, hence finite.
Am I ok ?