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This question builds on top of my former question in which I asked whether constant symbols are necessary in first-order theories. In that question I gave a rough idea on how to get rid of these symbols, which was not completely thought through. I will try to fix this now and would be glad if someone can verify this approach.

Lets say we have a first-order theory $T$

  • over a language $L$ with (possibly countably infinite many) constant symbols $c_i$.
  • we choose (possibly countably infinite many) predicates $\varphi_i(x_1,...,x_{n_i})$ so that $\varphi_i(c_{\sigma_i(1)},...,c_{\sigma_i(n_i)})$ makes up an axiom system of $T$. The $\sigma(i)$ will choose the constant symbols that are present in the $i$-th axiom and $n_i$ (possibly zero) will denote how many are in there.

I want to construct a theory $T'$ over a language $L'$ which got all the constant symbols removed. Still, $T$ and $T'$ should have the exact same models (in some sense). I will do this by replacing the axioms $\varphi_i$

I will need some shorthand notations:

  • I write $\vec x$ for a sequence of some of the variables $x_1,x_2,...$ where the exact names are derived from the place of use of this sequence. E.g. in $(\forall \vec x)\varphi_i(\vec x)$ the sequence will consist of the variables $x_{\sigma_i(1)},...,x_{\sigma_i(n_i)}$.
  • I write $\vec x=\vec y$ for the statement that the two sequences agree element-wise on the variables that are contained in both of them. If a variable $x_j$ is not in both sequences then $\vec x=\vec y$ does not contain $x_j$.
  • I write $\Delta(\vec x)$ for the statement, that all the variables of $\vec x$ are different. So $\Delta(\vec x)$ contains the statement $x_i\not= x_j$ for any pair of distinct variables $x_i,x_j\in\vec x$ conjugated together.

Now I am going to replace the axiom $\varphi_i$ by

$$(A_i)\qquad (\exists \vec x)[\Delta(\vec x)\wedge \varphi_i(\vec x)\quad\wedge\quad(\forall \vec y)\left[\Delta(\vec y)\wedge \varphi_i(\vec y)\rightarrow \vec x=\vec y]\right],$$

and for any pair $\varphi_i, \varphi_j$ of former axioms I will introduce the axiom

$$(A_{ij})\qquad (\forall \vec x \forall \vec y)[\varphi_i(\vec x)\wedge\varphi_j(\vec y)\rightarrow \vec x= \vec y] $$

In the special case that we have constant symbols in $T$, but no axiom contains any of them, I will introduce the axiom $(\exists x)[x=x]$ to ensure that there is at least one element in the model.

In my opinion, this reflects all the properties that the constants inherently brought into the theory. Is this correct? Does this give me an equivalent theory (in some sense, see note 2)?

NOTE:

I do not think we should get rid of contant symbols. My motivation is purely academic in the sense, that I want to know whether there is always an equivalent theory without constant symbols, like there is always an equivalent theory replacing function symbols and constant symbols by relation symbols.

As Giorgio and Rene pointed out in there answers to my former question, there are legitimate and useful applications for constant symbols. I do not doubt this.

NOTE 2:

I am not sure how to directly compare the (whole "space" of) models of these theories as they are defined over different languages. So the first on is an $L$-structure, the second one an $L'$-structure. Is this done by comparing the categories of structures created by these two theories?

M. Winter
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    The most natural way to compare the class of models of $T'$ with the class of models of $T$, when $T'$ is the reduct of $T$ to a smaller language, is to look at (semantic) strong conservativity: can every model of $T'$ be expanded to a model of $T$? This is the sense in which I and Apostolos provide negative answers. There are other comparison methods one might use, but this is most direct one and seems most in line with what you are asking. – Noah Schweber May 02 '17 at 14:20

3 Answers3

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No, constants are needed:

Assume that $\mathcal{L}=\{R,(c_n)_{n\in\omega}\}$ where $R$ is a binary relation and let us consider the following theory $T=\{(R(c_0,c_n))_{n\in\omega},(c_n\neq c_m)_{n\neq m}\}.$ I claim that there is no theory $T'$ of $\mathcal{L}^-=\{R\}$ such that a model satisfies $T$ if and only if its $\mathcal{L}^-$-reduct satisfies $T'$.

Indeed, assume to the contrary that such a $T'$ exists. Let $M=(\mathbb{N}, >)$ be the natural numbers with $R$ defined as the reverse order.

We have that $T\models T'$, that is every model of $T$ satisfies every formula of $T'$. For every formula of $\varphi\in T'$ there exists a finite number of formulas from $T$, let's call it $U_\varphi$, such that $U_\varphi\models\varphi$. This is because if this was not the case, then for every $U$, a finite subset of $T$, we would have that $U\cup\{\lnot\varphi\}$ has a model. This implies by compactness that $T\cup\{\lnot\varphi\}$ has a model, which contradicts the fact that $T\models\varphi$.

Finally, notice that any finite set of formulas of $T$ is satisfied in $M$. Hence for any $\varphi\in T'$, we have that $M\models U_\varphi$. This and the fact that $U_\varphi\models\varphi$ implies that $M\models\varphi$. Since this is true for every $\varphi\in T'$, we have that $M\models T'$.

On the other hand, $M$ can not be expanded to $\mathcal{L}$ so that it satisfies $T$. This concludes the proof.

As for your question regarding the size of the language. If the language contains only finite many constants and finite many relations, then you can write in a single formula exactly how the constants relate w.r.t. the relations and hence you can get rid of the constants.

On the other hand, if you have infinite relations then you can do something similar with what I did above: Take $\mathcal{L}=\{c_1,c_2,(R_n)_{n\in\omega}\}$, the theory $T=\{(R_n(c_1,c_2))_{n\in\omega}\}$ and consider the model $M=\{\mathbb{N}, <_n\}$ where $k<_n\ell$ if and only if $n\leq \ell$ and $k<\ell$. Using these you can show exactly what I showed above.

As for functions, this is something that at some point was present in Noah's answer, but it seems removed now, so I add it here: If you have one constant $c$ and one function $f$, then you can define countable many closed terms in the form of $f(f(c))$, $f(f(f(c)))$, etc. Then you can write the same theory as the one I wrote in my original answer by using these terms instead of constants.

As Alex points out, the whole idea behind these proofs is that you can have a constant satisfy infinite many relations, but there is no sentence in first-order logic that can express this: The constant binds an object and you can describe infinite many properties of that particular object. This is closely related to the notion of saturation in model theory.

Finally, yes, if you have a theory such that a formula $\varphi$ exists such that $\forall x(\varphi(x)\to(x=c))$ you can do away without $c$, as long as this formula $\varphi$ does not involve $c$.

Apostolos
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  • +1. This is a better answer than mine - note to the OP that there is no need to make the language uncountable! – Noah Schweber May 02 '17 at 14:13
  • @NoahSchweber & Apostolos. While I am still in the process of understanding the proof, maybe you can answer this: are constants equally important if we restrict us to only using finitely many constant symbols in $\mathcal L$? – M. Winter May 02 '17 at 15:22
  • @M.Winter at first glance I would say that with a finite language you don't need constants since you can describe how the constants interact with one sentence. But if you have infinite relations and functions then probably a similar example exists. That is for every relation there exists some elements of the model that look like the constants but there doesn't exist elements that look like the constants for all relations. This, I think, has to do with the concept of saturation in model theory. Also if you have any question about the proof, please let me know. – Apostolos May 02 '17 at 18:25
  • @Apostolos I'm still struggeling because I am not so fluent in this model theoretic language. How exactly does the statement of the compactness theorem (a theory has a model iff any finit subtheory has a model) lead to the following two implications: 1. for every $\varphi\in T'$ there is a $U_\varphi\subset T$ with $U_\varphi\vDash \varphi$. 2. $M$ is a model of $T'$. Isn't it that the compacntess theorem only says "there is a model", but not that it is $M$? Maybe you can include more steps in your answer so that I can understand the details. I would be very glad. – M. Winter May 03 '17 at 09:54
  • @Apostolos Another side question: If $c$ is a constant symbol of my language and my theory proves for some formula $\varphi(x)$ that $\varphi(c)$ and $(\forall x)[\varphi(x)\rightarrow x=c]$, then it should be possible to remove $c$ from the language as it is uniquely determined by $\varphi$, right? And this will work for infinitely many constants as well as in the presence of arbitrary many relation/function symbols. Especially, this should apply to PA where $0$ is (as far as I know) provably the only individual satisfying $(\forall n)[Sn\not= 0]$. – M. Winter May 03 '17 at 10:38
  • @M.Winter Compactness implies that if a theory $T\models\phi$ then there exists $U$, a finite subset of $T$, such that $U\models\phi$. To show this assume otherwise, then for every $U$ finite subset of $T$, we have that there exists a model that satisfies $U\cup{\lnot\phi}$. This means that $T\cup{\lnot\phi}$ is satisfiable (because every finite subset of it is), which is a contradiction. (cont.) – Apostolos May 03 '17 at 10:44
  • @M.Winter For your second question: we have that $U_\varphi\models\varphi$, for every $\varphi\in T'$. This means that every model of $U_\varphi$ is also a model of $\varphi$. Now we have that $M$ is a model of every finite subset of $T$, hence it is a model of $U_\varphi$, whis means that $M\models\varphi$, for every $\varphi\in T'$. This means that $M\models T'$. – Apostolos May 03 '17 at 10:46
  • @Apostolos Ok, I understand this. Then why can't I apply the same reasoning to prove that $M\vDash T$? Is this because for any $U_\varphi$, the model $M$ satisfyling $U_\varphi$ is actually always a slightly different model because we have to interpret the $c_i$ in a different way? – M. Winter May 03 '17 at 10:59
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    @M.Winter Yes, the reason that you cannot do the same to show that $M\models T$ is that for every finite subset $U$ of $T$ you define a different interpretation of the atoms in $M$ in order to make $M\models U$. These different interpretations for atoms do not affect the fact that $M\models T'$ because $T'$ does not involve constants. – Apostolos May 03 '17 at 13:22
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I want to make a point that I think hasn't been clearly articulated yet in the discussion.

Let $L$ be a language with no constant symbols, and let $L'$ be $L$ together with some constant symbols (finitely or infinitely many).

Let's say we have an $L'$-sentence $\varphi$, which mentions the constant symbols $c_1,\dots,c_n$ and no others. Then replacing each constant symbol $c_i$ with the fresh variable $x_i$ everywhere it appears in $\varphi$, we get an $L$-formula $\widehat{\varphi}(\overline{x})$. And the $L'$-sentence $\varphi$ is equivalent to the $L$-sentence $\exists \overline{x}\, \widehat{\varphi}(\overline{x})$, in the sense that an $L$-structure satisfies $\exists \overline{x}\, \widehat{\varphi}(\overline{x})$ if and only if it is a reduct of an $L'$-structure satisfying $\varphi$.

This is all well and good for a single sentence, but you run into some trouble when you try to do this for multiple sentences at once. When $\varphi$ and $\psi$ have some constant symbols in common, an $L'$-structure satisfying both $\varphi$ and $\psi$ is not the same as an $L$-structure satisfying both $\exists\overline{x}\,\widehat{\varphi}(\overline{x})$ and $\exists\overline{x}\,\widehat{\psi}(\overline{x})$. The reason is that the tuple witnessing the existential quantifiers in $\exists\overline{x}\,\widehat{\varphi}(\overline{x})$ is not required to be the same as the tuple witnessing the existential quantifiers in $\exists\overline{x}\,\widehat{\psi}(\overline{x})$.

OK, but we can fix this if we only care about finitely many sentences (i.e. a finitely axiomatizable theory). A model for the $L'$-theory $\{\varphi_1,\dots,\varphi_n\}$ is the same as a model for the $L'$-sentence $\bigwedge_{i=1}^m \varphi_i$, and this sentence is equivalent (in the sense above) to the $L$-sentence $\exists\overline{x}\,\bigwedge_{i=1}^m \widehat{\varphi_i}(\overline{x})$. What we did here was ensure that every instance of a each constant was handled by the same existential quantifier.

But most theories we care about are infinite. And you can't use the trick above for an infinite theory (at least not in first-order logic), because you can't form the infinite conjunction. The situation is even worse when you have infinitely many constant symbols, because you can't form an infinite block of existential quantifiers.

Alex Kruckman
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No, there is no way to just get rid of constants.

Consider the language $L$ consisting of uncountably many constant symbols $c_i$ and no non-constant symbols. Let $T$ be the set of axioms asserting $c_i\not=c_j$ whenever $i\not=j$. Then every model of $T$ is uncountable.

However, by Lowenheim-Skolem there is no theory in the empty language such that every model of that theory is uncountable. So to get an equivalent "constant-free" theory, we would need to add something to the language in addition to deleting the constant symbols.

Another construction, this time in a countable language, closely related to but not quite the same as Apostolos' answer:

Take the language $\{c_i: i\in\mathbb{N}\}\cup \{R\}$ where $R$ is a binary relation symbol. Let $T$ be the set of axioms asserting that $R$ is cycle-free and that the $c_i$ form a strictly $R$-descending sequence: $c_0Rc_1Rc_2...$

Now for any theory $T'$ in the language $\{R\}$ containing the axioms asserting that there are no cycles we can construct a model of $T'$ with no infinite $R$-descending sequences (this is a good exercise). Such a model cannot be expanded to a model of $T$.

This example is really really close to Apostolos'; I'm including it, however, because avoiding ill-foundedness is a useful technique elsewhere in logic (as is avoiding well-foundedness, but that's much easier via compactness).

Noah Schweber
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