Why is it not true that if $\Phi \models \exists x \phi$ then there is a term $t$ with $\Phi \models \phi \frac{t}{x}$?

Question

On p189 in XI Free Models and Logic Programming of Ebbinghaus' Mathematical Logic:

In general, the following statement is false:

(*) If $\Phi \models \exists x \phi$ then there is a term $t$ with $\Phi \models \phi \frac{t}{x}$

We get a counterexample for $S = \{R\}$ with unary $R$, $\Phi = \{\exists x Rx\}$, and $\phi = Rx$.

In the counterexample, why if $\{\exists x Rx\} \models \exists x R x $, then there is no term $t$ with $\{\exists x Rx\} \models R t $?

Thanks.

Answer 1

Well, consider the terms in the language. There are no function or constant symbols, so the only terms are the individual variables themselves. There's nothing special about one variable vs. another, so let's just focus on $x$. Since $\Phi$ consists only of sentences, we have $$\Phi\models Rx\quad\iff\quad\Phi\models\forall x(Rx)$$ (think about the definition of "$\models$" in case we have formulas with free variables on one side or the other). But it's certainly not true that $\Phi\models\forall x(Rx)$: consider for example a structure consisting of two elements where $R$ holds of one but not the other. This structure satisfies $\Phi$ but does not satisfy $\forall x(Rx)$.