# satisfiability in a structure implies satisfiability in a substructure?

My level: I've studied mathematics and now work through Hebert Enderton's book "An introduction to mathematical logic", second edition, in my free time. Relevant pages: 135-142, specifically 140

Suppose you have a first-order language that doesn't include the equality symbol and you add countably many constant symbols to this original language. Suppose $\Gamma$ is a set of formulas of the original and hence the new language. Let $\Delta$ be a set of formulas of the new language that is maximally consistent. By "maximally" I mean that for each formula $\phi$ of the new language either $\phi$ or $\neg\phi$ is in $\Delta$ (but not both). Let $\mathfrak{A}$ be the structure for the new language, such that the universe is the set of terms of the new language and $P^{\mathfrak {A}}(t_1, \ldots , t_n)$ iff $Pt_{1}\ldots t_n \in \Delta$ for a predicate symbol $P$ and $f^{\mathfrak{A}}(t_{1}, \ldots, t_n)=ft_{1}\ldots t_n$ for function symbols $f$ and $c^{\mathfrak {A}}=c$. By the way $\Delta =\{\phi\colon \models_{\mathfrak {A}}\phi [s]\}$, where $s$ is the identity function on the set of variables.

I want to show that when $\mathfrak{A}$ satisfies $\Gamma$ with the identity function then $\mathfrak{B}$ satisfies $\Gamma$ with the identity function. Here $\mathfrak{B}$ is the structure for the original language that results from $\mathfrak{A}$ by restricting it to the original language.

Initially I thought this would be easily done by induction but for deferral days now I didn't manage to do it. The problem is that in the induction step I couldn't show that $\models_{\mathfrak {A}}\gamma[s]$ iff $\models_{\mathfrak{B}}\gamma[s]$ (s the identity function on the set of variables) the case of the quantifier symbol. In this case I can only show that from the first assumption follows the latter, which might seem sufficient fit my purpose. But when $\neg$ is involved it might not be so.

The proof by induction that the structure $\mathfrak {A}$ satisfying $\Delta$ with function $s$ is "already there" : step 4, page 137-on.

Being $\Gamma \subseteq \Delta$, this structure satisfy also $\Gamma$ with $s$.

Your error, I presume, is in thinking in term of "sustructure" $\mathfrak {B}$ such that $|\mathfrak {B}| \subseteq |\mathfrak {A}|$.

This is not the correct interpretation of [page 140] :

we need only restrict $\mathfrak {A}$ to the original language to obtain a structure that satisfies every member of $\Gamma$ with the identity function [$s$].

The structure for $\Gamma$ has the same domain $|\mathfrak {A}|$; what we have to "restrict" is the interpretation, because the original language (that of $\Gamma$) has no "added" Henkin-constant $c_i$; thus, we need no "reference" $c_i^{\mathfrak A}$ for them.

But this amounts to "removing" the un-needed "interpreting-relation" for the constants of the language, and not the "objects" themselves (which are the stuff of the structure).

The domain $|\mathfrak {A}|$ is "made of" the terms and we still need them also for $\Gamma$.

• Now I've got it - your answer was most helpful! You were right, I assumed that $\mathfrak{B}$ should be the set of the original terms (no Henkin-constants a you call them there) and therefore my assuming $\mathfrak{B} \subset \mathfrak{A}$ resulted in me having trouble with $\forall$ in the induction step.. – Pi o r Aug 15 '14 at 18:39