TheoremBase

The formula says that q is a natural number with zero and that some bijection runs from the segment [q] onto S; with it, the expressions #C_i and #S define maps by the scheme for maps given by formulas.

Proof

Predicativity. Let ss and ff be set variables distinct from each other and from qq and SS. Let seg(s,q)\mathrm{seg}(s,q) be the predicative formula of Induction over Finite Sets, and Predicative Formulas for the Segments [n] and for Finiteness (i), with ss and qq for its set variables and YY and RR for its class variables, so that for every set ss and every n∈N0n\in\mathbb{N}_{0}, seg(s,n)\mathrm{seg}(s,n) holds if and only if s=[n]s=[n]. Let β(f,s,S)\beta(f,s,S) be the predicative formula of Predicativity under Renaming of Set Variables by Variables, of Formulas with Ordered Pairs, and of Formulas for Relations, Functions, Maps, Injectivity, Surjectivity and Bijectivity (x), with ff, ss and SS for its variables FF, AA and BB, as Predicativity under Renaming of Set Variables by Variables, of Formulas with Ordered Pairs, and of Formulas for Relations, Functions, Maps, Injectivity, Surjectivity and Bijectivity (i) permits; it holds if and only if ff is a bijection from ss to SS. Put

κ(q,S)  =  q∈X∧∃s ∃f (seg(s,q)∧β(f,s,S)).\kappa(q,S)\;=\;q\in X\wedge\exists s\,\exists f\,\bigl(\mathrm{seg}(s,q)\wedge\beta(f,s,S)\bigr).

It is predicative, since it adds to predicative formulas only connectives and quantifiers over set variables, and its free variables are among qq, SS, XX, YY and RR.

Let nn and BB be sets. If κ\kappa holds for them, then n∈N0n\in\mathbb{N}_{0}, and there are a set ss with s=[n]s=[n] and a bijection ff from [n][n] onto BB; so BB is finite by Finite Sets (i), and n=#Bn=\#B by The Number of Elements of a Finite Set (i). Conversely, if BB is finite and n=#Bn=\#B, then n∈N0n\in\mathbb{N}_{0} and there is a bijection ff from [n][n] onto BB by The Number of Elements of a Finite Set (i); with s=[n]s=[n], which satisfies seg(s,n)\mathrm{seg}(s,n), κ\kappa holds. This proves (iii).

In particular, for a finite set BB exactly one set qq satisfies κ(q,B)\kappa(q,B), namely #B\#B, the length being unique by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (ii). Hence Predicativity of Expressions with Abbreviations, and Symbol-Free Forms of the Basic Defined Set Symbols (ii), with θ=κ\theta=\kappa and with γ(S)=fin(S)\gamma(S)=\mathrm{fin}(S), the predicative formula of Induction over Finite Sets, and Predicative Formulas for the Segments [n] and for Finiteness (ii) (whose class variables are assigned N0\mathbb{N}_{0}, N\mathbb{N} and ≤\le as here), lets #S\#S be eliminated through κ\kappa from every predicative formula, for every finite set SS.

Counting a family. For i∈Ii\in I, the value CiC_{i} is a defined set symbol used properly (Functions, Values of a Function, and Functions from One Class to Another (ii)), and #Ci\#C_{i} is used properly by this elimination, because CiC_{i} is finite; moreover #Ci∈N0\#C_{i}\in\mathbb{N}_{0} by The Number of Elements of a Finite Set (i). Hence Maps and Relations Given by Formulas (i), applied to the expression #Ci\#C_{i} with the parameters CC, N0\mathbb{N}_{0}, N\mathbb{N} and ≤\le, gives exactly one map c:I→N0c:I\to\mathbb{N}_{0} with c(i)=#Cic(i)=\#C_{i} for every i∈Ii\in I. This proves (i).

Counting subsets. Every S∈P(A)S\in\mathcal{P}(A) is a subset of AA by The Union Set and the Power Set of a Set (ii), hence finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iv), so #S\#S is used properly by the same elimination and lies in N0\mathbb{N}_{0}. As before, Maps and Relations Given by Formulas (i), applied to the expression #S\#S on the set P(A)\mathcal{P}(A), gives exactly one map P(A)→N0\mathcal{P}(A)\to\mathbb{N}_{0} with value #S\#S at every S∈P(A)S\in\mathcal{P}(A). This proves (ii).

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