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I have been asked to prove that DAG,the theory of non-trivial torsion-free divisible abelian groups, has quantifier elimination and deduce from that that it is complete.

What are the conditions to conclude completeness from quantifier elimination?

Also, I have been looking at different sources for the proof of quantifier elimination and they always take L={0,+,-} for the language of groups. Does it make a big difference if I only take L={0,+}?

Thanks!

Rodrigo de Azevedo
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2 Answers2

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Let me add a bit to Primo Petri's answer.

Let $T$ be a consistent theory with quantifier elimination. The following are equivalent:

  1. $T$ is complete.
  2. For every quantifier-free sentence $\varphi$, $T\models \varphi$ or $T\models \lnot \varphi$.
  3. For any models $M\models T$ and $N\models T$, the substructures generated by the empty set in $M$ and $N$ are isomorphic.
  4. There is a structure $A$ such that for any model $M\models T$, $A$ embeds in $M$.

Let's prove this.

$(1)\implies(2)$: Since $T$ is complete, it decides the truth of every sentence, and in particular every quantifier-free sentence.

$(2)\implies (3)$: For any model $M\models T$, the elements of the substructure generated by the empty set, $\langle \varnothing\rangle_M$, are exactly the evaluations of terms with no free variables. Then the isomorphism type of this structure is entirely described by quantifier-free sentences of the form $f(t_1,\dots,t_n) = t_{n+1}$ and $(\lnot) R(t_1,\dots,t_n)$, where $t_1,\dots,t_{n+1}$ are terms with no free variables, $R$ is an $n$-ary relation symbol, and $f$ is an $n$-ary function symbol. Since $T$ decides the truth value of every quantifier-free sentence, it completely describes the isomorphism type of $\langle \varnothing \rangle_M$. Thus $\langle\varnothing\rangle_M\cong \langle \varnothing \rangle_N$ for any model $N\models T$.

$(3)\implies (4)$: Let $N\models T$, and let $A = \langle \varnothing\rangle_N$. Then for any $M\models T$, $A\cong \langle \varnothing\rangle_M$, so $A$ embeds in $M$.

$(4)\implies (1)$: Let $M$ and $N$ be models of $T$. It suffices to show that if $\varphi$ is a sentence such that $M\models \varphi$, then also $N\models \varphi$. By quantifier elimination, $\varphi$ is equivalent in models of $T$ to a quantifier-free sentence $\psi$. So $M\models \psi$. Since $A$ embeds in $M$ and $\psi$ is quantifier-free, $A\models \psi$. Similiarly, $N\models \psi$. Since $N\models T$, $N\models \varphi$, as desired.

We can apply criterion 4 to show that DAG is complete: the trivial group embeds in every model.

Let me note that there's a subtlety involving languages with no constant symbols. To make the equivalent statements above make sense in this case, we have to use the correct conventions for first-order logic. Namely, we have to allow empty structures, and we have to include the symbols $\top$ (true) and $\bot$ (false) in our logic. This fixes two issues:

  1. If our language $L$ has no constant symbols, and we don't include $\top$ and $\bot$, then there are no quantifier-free $L$-sentences at all. If $T$ has quantifier elimination, every sentence should be equivalent to a quantifier-free sentence, but this is impossible if there are no quantifier-free sentences!
  2. If our language $L$ has no constant symbols, then the substructure $\langle \varnothing\rangle_M$ generated by the empty set is empty, and hence not a substructure (if we use the convention that structures must be non-empty).
Alex Kruckman
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  • +1 but I do not understand 2. Why discharge the empty structure? Anren't two empty structures isomorphic (equal)?. – Primo Petri Feb 07 '22 at 15:33
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    @PrimoPetri My point is that condition 3 doesn't even make sense if we don't allow empty structures (there is no "substructure generated by the empty set" when there are no constant symbols). I'm advocating for including empty structures in our semantics to fix this problem. [Also, I think one should include $0$-ary relation symbols or "proposition symbols" in first-order logic. If we do this, then we can have multiple empty structures which are non-isomorphic, because they disagree on the truth values of the proposition symbols. But this is not really relevant to the question at hand.] – Alex Kruckman Feb 07 '22 at 16:06
  • @PrimoPetri Also, sorry for stealing your green check mark! That wasn't my intention. – Alex Kruckman Feb 07 '22 at 16:07
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    As you've stolen the green check mark :), you owe us an answer to the second part of the question. Is there an easy way to see that (if at all true??) DAG has QE in the language ${0,+}$ other than going through the proof? – Primo Petri Feb 07 '22 at 16:30
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    @PrimoPetri Here's a syntactic argument: Let $\varphi$ be a formula in the language $L = {0,+}$. Since DAG has QE in the language $L' = {0,+,-}$, $\varphi$ is equivalent to a quantifier-free $L'$-formula $\psi$. We want to show that $\psi$ is equivalent to a quantifier-free $L$-formula. To do this, it suffices to show that every atomic $L'$-formula is equivalent to an atomic $L$-formula. Note every $L'$-term is equivalent (modulo the abelian group axioms) to $0$ or a sum of some variables and some negated variables. Every atomic $L'$-formula is equivalent to an equality between $L'$-terms. – Alex Kruckman Feb 07 '22 at 17:04
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    To finish, just add the negated variables on each side of the equality over to the other side. – Alex Kruckman Feb 07 '22 at 17:04
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    A semantic argument using a semantic equivalent to QE will amount to showing that if $M$ and $N$ are abelian groups and $A$ is a common sub-monoid (substructure in the language ${0,+}$), then the subgroup of $M$ generated by $A$ is isomorphic to the subgroup of $N$ generated by $A$. – Alex Kruckman Feb 07 '22 at 17:06
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A theory with QE is complete iff it decides all sentences in $L_{\rm qf}$.

Examples:

  • the theory of dense linear orders and the theory of the random graph are complete because $L_{\rm qf}$ is empty (the language has no constants).

  • the theory of divisible abelian groups (DAG) is complete because all sentences in $L_{\rm qf}$ are equivalent, modulo the theory of groups, to $n0=m0$

  • the theory of algebraically closed fields is not complete because it does not decide the caracteristic of the field.

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As for the second question:

The natural language to formalize groups is $\{0,+,-\}$. For example, in this language the model theoretic notions of homomorphism and substructure coincide with the algebraic notions of group homomorphism and subgroup.

However, after a quick check it seems that the proof of QE for DAG goes through also when the language is $\{0,+\}$. You might have checked it already.

Primo Petri
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