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Suppose t is a term, $f_n, n\in\mathbb{N}$ are function symbols, I have no doubt that $f_1(t),f_1 f_2(t),f_1 f_2 f_3(t)$ are all terms. But what about these:

$f_1 f_2 f_3...f_n(t)$ (note that n is not a specific natural number but a variable)

$f_1 f_2 f_3...f_n...f_{+\infty}(t)$

Are they terms? If not, what are they?

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    $\begingroup$ The subscript notation and the numbers $1$ through $n$ aren't part of the formal language, they're just part of the language we're using to describe it. $\endgroup$ Commented Sep 26 at 15:28
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    $\begingroup$ @TankutBeygu Infinitary Logic does not respond to the question actually because the underlying structure of infinitary logic is basically the same, it only admits "formulas with infinite connectives" and "just" symbols of finite arity. What are you asking is a "Infinity-arity" first order language. $\endgroup$ Commented Sep 26 at 16:06
  • $\begingroup$ @TankutBeygu There are some studies and some results in "Infinity arity" Logic mainly connected with categories. Accessible categories are examples of categories of models of infinitary theories , where infinitary predicates are considered. If you want more, you can reed Adamek, Rosicky "Locally presentable and accessible categories". $\endgroup$ Commented Sep 26 at 16:17
  • $\begingroup$ Variables don't belong to the natural numbers. Each natural number is a constant. n ∈ N, implies that that in any expression which is a term, n gets replaced with a (symbol for a) natural number. So, neither of those symbolic forms you put up are terms, since no natural number in particular gets suggested by 'n'. The first expression behaves as a placeholder for any term with unary functions only. The second might behave as a way to suggest that terms with unary functions only cannot have a longest term, if the set of natural numbers is not well understood. $\endgroup$ Commented Sep 26 at 20:37
  • $\begingroup$ Terms also have a definite length (or number of symbols). When we have "..." we don't have a definite number of symbols. The entire point of using "..." lies in that we can't specify all of the details, because we either have too large of a set of entites for us to write now, or have an infinite set of entities. The use of "..." thus suffices to imply that we don't have a term. The exact definition of a term in first-order logic would probably become clearer if you first considered a formal system with a clearly specified a priori number of functions. $\endgroup$ Commented Sep 26 at 23:46

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A term is either a variable or a constant or an n-ary function symbol applied to n terms.

And a term is a finite expression (string of symbols).

Examples from the language of arithmetic: $x, 0, s(0), s(0) +x, (s(0)+x)+0$.

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A first order language $\mathcal{L}$ is composed by a set of variables (possibly infinite) $\mbox{Var}=\{x_1,\dots,x_n,\dots\}$, a set $C$ of symbols of constant, a set $R$ of symbols of predicate and a set $F$ of symbols of functions and a function $a\colon F\cup R\rightarrow\mathbb{N} $ such that represents the $\textbf{Arity}$ of that symbols and logical connectives like $\neg$ or $\wedge$.

Let $\mathcal{L}$ a first order language, Term$_{\mathcal{L}}$ is definined as the smallest set closed under the following rules:

  1. $\mbox{Var}\subseteq$ Term$_{\mathcal{L}}$
  2. $\mbox{C}\subseteq\mbox{Term}_{\mathcal{L}}$
  3. If $f\in \mbox{F}$ such that $a(f)=n$ and $t_1,\dots,t_n\in \mbox{Term}_{\mathcal{L}}$, then $f(t_1,\dots, t_n)\in\mbox{Term}_{\mathcal{L}}.$

For example, if the language is the classical language of First order Group Theory with the binary predicate of equality, $x\cdot y=z$ is a possible formula involving a binary function symbol ($\cdot$), a binary predicate symbol (=) and three variables and $x\cdot y$ is a possible term. Clearly $x\cdot\cdot y\cdot\cdot$ is not a term in that language.

It should be clear that terms contains only a finite amount of variables.

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