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In defining the language, according to my logic book, if $f$ is n-ray function and $\alpha_1, \alpha_2, \cdots, \alpha_n$ are terms, then $f(\alpha_1, \alpha_2, \cdots, \alpha_n)$ is always a term. Why is there no mention of the domain of a function here? I think we need to think of domain as well as arity of function... What can we do if we want to think about two functions with different domains in the first order language?

amoogae
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    Because $f$ is not a function, it's a function symbol. – Reveillark Mar 27 '19 at 03:38
  • @Reveillark Hmm ... what's the difference? Let $L$ is the first order language, and $J$ is an interpretation. If there is no corresponding elements of $f(\alpha_1, \alpha_2, \cdots, \alpha_n)$ because $(\alpha_1, \alpha_2, \cdots, \alpha_n)$ is not in the domain of $ f $. Then, what is the semantic value $f(\alpha_1, \alpha_2, \cdots, \alpha_n)^J$? – amoogae Mar 27 '19 at 03:56
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    Ok so then we're arguing semantics. To argue semantics, you need a structure, which is a non empty set $A$ together with an interpretation $J$. If $f$ is an $n$-ary function symbol, then $J(f):A^n\to A$ by definition. The function symbol $f$ does not have a domain. – Reveillark Mar 27 '19 at 03:58
  • @Reveillark I understood that. Thank you. But how can we consider function $f_1$ with a domain $A^n$ and function $f_2$ with a domain $B\subsetneq A^n$ both. For example, If we want to think of the square root function in real numbers. – amoogae Mar 27 '19 at 04:11
  • Derek's answer is right. I'll point out that using standard (one-sorted) FOL, you cannot just combine various axiomatizations of common structures into a single foundational system. For example if you want to have PA inbuilt, plus ZFC-like axioms, then your best approach is via multi-sorted FOL, or better still an extension of that where sorts can be created on-the-fly like described here, and then add definitorial expansion. Then every function-symbol you use/introduce has specified input types. – user21820 Mar 27 '19 at 07:27

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The issue is simply that the first-order logic that is usually presented is single-sorted first order logic. In multi-sorted first-order logic, we have a collection of sorts and a function symbol can take an arbitrary (but fixed) list of sorts and produce an arbitrary sort, e.g. $f : (X,Y)\to Z$. In the single-sorted case, we have (unsurprisingly) only one sort and so we only get things like $f : (X,X)\to X$. At this point, the only information we need to specify the is the number of inputs, i.e. the arity. What's usually called the domain of the interpretation (not to be confused with the domain of a function) is the interpretation of this single sort. In multi-sorted first-order logics, you will have a domain per sort.

As Reveillark stated, there's a difference between a function symbol and a function. The scenario you describe in your last comment simply can't happen in single-sorted first-order logic. The interpretation of all function symbols (with the same arity) have the same domain. By definition, those functions must be defined on the entire domain of the interpretation. A structure that interprets the domain as $\mathbb R$ but interprets some unary function symbol as $\mathsf{sqrt}$ is simply not a valid interpretation.

Your options are 1) use a multi-sorted first-order logic, but this very likely won't do what you want, 2) talk about functional predicates that get interpreted as functional (but not necessarily total) relations which may work if all your domains are a subset of another domain but probably isn't want you want in general, or 3) talk about set-theoretic functions in a (presumably single-sorted) first-order logic but then interpretation doesn't come into it. (Or, use some other logic entirely.)

Most of your experience with functions comes from (nominally) set-theoretic functions. You can do many things to such set-theoretic functions because they are, after all, just sets (namely, sets of pairs satisfying some rules), so you can do anything you can do to sets to them. When you were asked to determine the domain of a function like $\frac{1}{1-x}$ before, you were really being asked to describe the set $\{x\mid \exists y.(x,y)\in f\}$. Within the syntax of a first-order logic, the only thing you can do with a function symbol is apply it to some terms. In more general approaches to semantics, i.e. categorical semantics, you don't even need to interpret function symbols as functions.

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