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Are constant symbols really necessary in first-order theories? And I am not talking about replacing them with unary predicates like here, but instead I would like to remove any sign of their existence from the language and only hint to their existence using the axioms.

Take for example the Peano axioms. Is the zero symbol $0$ really necessary? Assume $\varphi(x)$ is an unary predicate and $\varphi(0)$ is an axiom of PA containing the symbol for the zero. Couldn't I just replace it by the following axiom?

$$(\exists x)\varphi(x) \quad\wedge\quad (\forall x \forall y)[\varphi(x)\wedge\varphi(y)\rightarrow x=y].$$

The first part states that there is an individual satisfying this predicate, the second parts states that it is unique. I can think about ways to generalize this to several and even countably infinite many constant symbols.

If it is about simplicity, then why doesn't ZFC uses a constant symbol for the empty set? Just for historical reasons?

I posted follow up questions here and here.

M. Winter
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    First of all: What do you have against constants ? That's constantism. Second all the constants you mention are definable. There are theories with constants not representing definable elements. What do you say to the theory of infinitely many different constants $c_n\neq c_m$ for $n\neq m$. – Rene Schipperus May 01 '17 at 19:43
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    @ReneSchipperus I think you should incorporate that example into your answer - it's really important (or rather, an example of a really important class of constructions) and should be explicitly mentioned. – Noah Schweber May 01 '17 at 20:19
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    You can formalize the Empty Set axiom without a constant: $\exists x \ \forall y \ \lnot (y \in x)$ and by Extensionality prove that it is unique. – Mauro ALLEGRANZA May 01 '17 at 20:30
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    Having done this, what are the benefits of not introducing a "name" for the unique set such that .... ? – Mauro ALLEGRANZA May 01 '17 at 20:31
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    @MauroALLEGRANZA The question is what distinguishes ZFC from PA in this point, so that we introduce $0$ in PA, but not $\emptyset$ in ZFC? I know that we don't need a constant in ZFC and I know that $0$ comes in handy in PA. But this should be more a question about possibilities. – M. Winter May 02 '17 at 07:48
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    @ReneSchipperus I think I can handle this, at least when I restrict myself to countable languages. If the constants should not satisfy any further axioms, then I can just add axioms $a_n$ for each $n\in\Bbb N$ where $a_n$ says that any mode should have more than $n$ elements. I am pretty sure I can do something similar if there are additional axioms involving the constantes. Maybe I will post a follow up question with my idea and I would be glad if you people can check it. – M. Winter May 02 '17 at 07:56
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    The theory of an infinite set which you describe and the theory of infinitely many constants are very different. One has a unique constable model, the other has infinitely many countable models. – Rene Schipperus May 02 '17 at 08:27
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    I think that the difference is: if we try to dispense with $0$ in PA, using $\exists x \ \forall y \ \lnot (x=S(y))$, we cannot prove its uniqueness. Thus, we have to assume the existence of the unique $x$ such that, and this amounts to using from the beginning the constant $0$. – Mauro ALLEGRANZA May 02 '17 at 08:40
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    @ReneSchipperus I am not an expert here. Can you please give me an example of a structure which is a model of only one of them? I cannot think about one myself. – M. Winter May 02 '17 at 08:46

2 Answers2

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The trick you outlined in the question has some drawbacks as the example of natural numbers already shows: consider the two formulas $$\varphi_1(z) \equiv \forall x\ s(x) \ne z$$ $$\varphi_2(z) \equiv \forall x\ x+z = x$$ we know that $\{\varphi_1(0),\varphi_2(0)\}$ is a sub-theory of PA, if you try to replace this two axioms in the way you described you get the theory with axioms $$\exists z\ \varphi_1(x) \land \forall y\ \varphi_1(y) \rightarrow y=z$$ $$\exists z\ \varphi_2(x) \land \forall y\ \varphi_1(y) \rightarrow y=z$$ that is not equivalent the previous one, that's because you could have models of this theory having two distinct elements satisfying separately the two axioms.

You could solve this problem by taking the the axiom $$\exists z \varphi_1(z) \land \varphi_2(z) \land \dots$$ but this doesn't help when you have infinitely many formulas $\varphi_i(z)$.

While I don't have a proof at the moment, I'm afraid that you cannot eliminate constants from the language unless you add unary-predicate symbols to the language with some axioms that allow to obtain an equivalent theory.

On the other hand constants are incredibly useful because they allow to build Henkin theories that are theories $T$'s such that for each formula $\varphi(x)$ if $T \vdash \exists \varphi(x)$ then there's a constant $c$ such that $T \vdash \varphi(c)$.

These theories are the key instrument for proving the completeness theorem: indeed it's easy to show that for an Henkin consistent theory the term model, that is the structure build out of the terms quotiented up to provable equality, is actually a model of the theory.

But that's not all of it, lots of constructions in model theory make use of constants. For instance for every structure $M$, in a given language $L$, it's possible to define the elementary diagram which is a theory in a language $L'$ obtained by $L$ adding a constant for each element in $M$, the models of this theory are basically the same as the elementary extensions of $M$.

There are other applications of constants inside logic but I hope that these examples gave you an idea of why constants matter.

Giorgio Mossa
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Constants affect the model theoretic properties of the theories. There are many examples of this, to give a simple one:

You say for example that the structures (or their theories) $$(\mathbb{N},<,0)$$ and $$(\mathbb{N},<)$$ are similar. But the set $(\{1,2,\cdots\},<)$ is a substructure of the second model, but not a substructure of the first.

Another example occurs in ring theory, we often require homomorphisms to preserve the unit element, this is only the case (in general) if $1$ is included as a constant.

Rene Schipperus
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    But isn't it isomorphic to a substructure of $(\Bbb N,<)$ which is essentially the same as being a substructure itself? – M. Winter May 02 '17 at 08:57
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    Or is it not a substructure because it is defined over another language. Is there terminology for substructures that are defined over a sub-language? – M. Winter May 02 '17 at 10:28