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It is usually said given set of variables, terms of language are defined recursively. But for recursive definition on a set, we need a function p which assigns to each function from a section of integers to T an element of T'.
(Munkers Page 54 math.ucsb.edu/~bigelow/145/munkres.pdf )

But how we defining these function in order to generate terms

Sushil
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1 Answers1

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For a formal treatment, you can see Kenneth Kunen, The Foundations of Mathematics (2009), page 45 :

Theorem 1.9.2 (Primitive Recursion on $ON$) Suppose that $\forall s \ \exists!y \ \varphi(s,y)$, and define $G(s)$ to be the unique $y$ such that $\varphi(s,y)$. Then we can define a formula $\psi$ for which the following are provable:

  1. $\forall x \ \exists!y \ \psi(x,y)$, so $\psi$ defines a function $F$, where $F(x)$ is the $y$ such that $\psi(x,y)$.

  2. $\forall \xi \in ON \ [F(\xi) = G(F \upharpoonright \xi)]$.

See page 91 :

Definition 11.4.1 A lexicon for Polish notation is a pair $(\mathcal W, \alpha)$ where $\mathcal W$ is a set of symbols and $\alpha : \mathcal W \to \omega$. Let $\mathcal W_n = \{s \in \mathcal W : \alpha(s) = n \}$. We say that the symbols in $\mathcal W_n$ have arity $n$. $\mathcal W^{< \omega}$ denotes the set of all finite sequences of symbols in $\mathcal W$. The (well-formed) expressions of $(\mathcal W, \alpha)$ are all sequences constructed by the following rule:

(§) If $s \in \mathcal W_n$ and $\tau_i$ is an expression for each $i < n$, then $s \tau_1 \ldots \tau_{n-1}$ is an expression.

In the "standard" applications, most of the $\mathcal W_n$ are empty. For example, we can let $\mathcal W = \{ x, y, z, !, +, . \}$, with $\mathcal W_0 = \{ x, y, z \}, \mathcal W_1 = \{ ! \}, \mathcal W_2 = \{ +, . \}$ [and thus : $\alpha(+)= \alpha(.)=2$, because $+, .$ have arity $2$]; the rest are empty. Then the following shows nine expressions of this lexicon:

$$x \quad y \quad z \quad +xy \quad .yz \quad +xx \quad +x.yz \quad .+xyz \quad !.+xyz$$

See page 92 :

A symbol is really just a set, since everything is a set. We assume always that $\mathcal W \cap \mathcal W^{< \omega} = \emptyset$, and (§) uses the terminology of concatenation. We should also distinguish between the symbol $x \in \mathcal W_0$ and the expression of length $1$ consisting of $x$, which is set-theoretically $(x) = \{ \langle 0, x \rangle \} \in \mathcal W_1 = \mathcal W^{ \{ \emptyset \} }$. Note that the rule (§) is a recursive definition of the notion of "expression". To justify this definition using Theorem 1.9.2, recursively define $F : \omega \to \mathcal P(\mathcal W^{< \omega})$, where $F(n)$ is the set of expressions of length $\le n$.

  • Oh you using formula in principle of recursion and then using principle of recursion, you are defining well formed formula. Is there any problem? – Sushil Jul 11 '15 at 11:38
  • @Sushil - of course, you can use the formulation adopted by Munkers (page 54) : in any case, what we get is the unique function $F$ (or $h$, for Munkers) that maps $0$ (or $1$) into the set of symbols of the alphabet, and $n$ into $F(n)$ (or $h(n)$), i.e. the set of expressions of length $≤ n$. – Mauro ALLEGRANZA Jul 11 '15 at 11:56
  • But we need a function which maps each section of natural no. not just each natural no. – Sushil Jul 11 '15 at 12:14
  • Moreover to define section we need axiom of specification which needs wffs, again same kind of loop. We defining wff using some application of wffs – Sushil Jul 11 '15 at 12:15
  • @Sushil - see above quotation form Kunen : he speaks of ordinals; in set theory a finite ordinal $n$ is precisely the set ${ 0, 1, \ldots, n-1 }$. – Mauro ALLEGRANZA Jul 11 '15 at 12:27
  • It only defines every natural no. contains all successor ordinals before it. But to define(or prove) n={0,1,…,n−1} we need induction. Am I right? – Sushil Jul 11 '15 at 12:52
  • @Sushil - you can compare with dtldarek's answer to your previous post : the answer is the same. – Mauro ALLEGRANZA Jul 11 '15 at 13:36
  • But there also I am facing some problem as I mentioned in last comment there. Please see in previous post – Sushil Jul 11 '15 at 13:39