4.8. More recursion tools

We first defined recursion as a particular case of the interpretation of a draft in an algebra, whose proof of existence was a little hard (its use of the conditional operator may be removed by adding the proper justifications). Here we shall review diverse variants of this construction, which can be useful in diverse contexts.

Well-founded recursion

Let RX² be a well-founded relation, and denote its reverse curried form R(x) as [x], so that ∀x,yX, xRyx∈[y].
As first seen for drafts where [x] = Im lx, the well-foundedness of R can be equivalently written

AX, (∀xX, [x] ⊂ AxA) ⇒ A = X
AX, AX ⇒ ∃xX\A, [x] ⊂ A
AX, A≠∅ ⇒ ∃xA, A∩[x] = ∅

Then, for any set E and any Φ : ∐xX E[x]E, there exists a unique h : XE such that

xX, h(x) = Φ(x,h|[x]).

The interpretation of a ground draft (X, tGr ψX) in an algebra E, was the particular case of Φ defined as

xX, ∀uE[x], Φ(x,u) = σ(x)E(ulx) = φE(LuX(x)))

The same proof works in the general case, after a simple re-writing left as an exercise for the reader.

Graphs of morphisms

A different generalization will be inspired from the fact the wanted morphism was defined by its graph.

Let L an algebraic language, E, F two L-systems, G = E×F the product system, f : EF and H = Gr fG. Then

(E injective ∧ HG(LH)) ⇒ f ∈ MorL(E,F)
(E surjective ∧ f ∈ MorL(E,F)) ⇒ HG(LH)
(F serial ∧ G(LH) ⊂ H)) ⇒ f ∈ MorL(E,F)
(F functional ∧ f ∈ MorL(E,F)) ⇒ G(LH) ⊂ H

The proofs are easy and left as exercises to the reader. The last formula provides another proof of the stability of equalizers: The last two together give : if F is an L-algebra then ∀fFE, f ∈ MorL(E,F) ⇔ Gr f ∈ SubL(E×F).
This can also be deduced from previous results when both E,F are L-algebras and thus G too: Now the search for morphisms from a system E to an algebra F, is reduced to that of subsets H∈ SubL(E×F) which would be graphs of functions with domain E. Hence the idea of a different view on the construction of H, as an intersection instead of a union (generalizing the proof from Halmos's Naive Set Theory): we shall prove that, when E is minimal, these conditions imply H = MinL(E×F), and that E being a ground draft implies the converse: MinL(E×F) is functional with domain E.

Generalized algebraic recursion


Let us modify our hypothesis to handle a more general case, very similar to the case of well-founded recursion, but still using an algebraic language (for compatibility with previous concepts).

Let L an algebraic language, and E, G two L-systems. We shall forget F and the hypothesis G = E×F, to only remember its first projection π ∈ MorL(G,E).
The hypothesis that F was an algebra, is re-expressed by the formula of the fullness of π modified by replacing ∃ by ∃! :

xLG, ∀yE, (Lπ(x),y) ∈ E ⇒ ∃!zG, π(z) = y ∧ (x,z)∈G

letting the structure G of G be expressed by a function Φ : KG where

K = {(x,y)∈ LG×E | (Lf(x),y) ∈ E}
∀(x,y)∈K, π(Φ(x,y)) = y
G = {(x, Φ(x,y)) | (x,y) ∈ K}

As π is full, π[MinL G] = MinL E. Thus if E is minimal then π[MinL G] = E, so that

H∈ SubL(G), MinL GH ∴ (Inj π|HH = MinL G)

Then, for any HG, the following hypotheses imply the stability of B = {yE | !zH, π(z) = y}.
Indeed ∀uLB, !xLH, Lπ(x) = u and the conclusion easily follows.
Finally since MinL G is surjective, when E is a ground term algebra and G is expressible by Φ as above, π|MinL G is injective with image E. ∎

An easy further generalization comes by only using the second half of this proof: while keeping E a ground term algebra, the hypothesis on G is weakened to the use of ! only, fitting a Φ with domain a subset of K. Then MinL G is the only surjective stable subset of G, and the graph of a function with domain a subset of E, called the function defined by a partial recursion, whose exact formulation will be left to the reader.

The use of the coproduct

The above construction can be dualized, once expressed in terms of categories : Both are concretely related, since E×F ⊂ (FE)2, and the minimal stable subset of E×F is part of the minimal congruence of FE:

MinL(E×F) = E×F ∩ MinL(FE)2E×F ∩ ≃FE

with equality when E is a functional draft and F is a partial algebra (while MinL(FE)2 is not generally an equivalence relation when F is not injective).

The below will be revised later
Countability of the set of finite number sequences

Also the set ℕ(ℕ) = {u∈ ℕ | ∃k∈ ℕ, ∀n>k, un=0} is countable. There are different ways to see it:

Rebuilding recursion

Let us work in a first-order theory just assumed to have 3 types ℕ, E and HE, the language and axioms of arithmetic for ℕ with the schema of induction over expressible subsets of ℕ, and the axiom that H provides "all finite sequences"

n∈ℕ, ∀uH, ∀yE, ∃vH, vn=y ∧ ∀k<n, vk = uk

The following existential result has a simple proof by induction making no use of uniqueness:

R⊂ℕ×E2, ∀n∈ℕ, (∀i<n, ∀xE, ∃yE, R(i,x,y)) ⇒ ∀aE, ∃uH, u0=a ∧ ∀i<n, R(i,ui,ui+1).

Its simplicity only goes for this case, only generalizable to conditions of the form R(n,(uk)k<n,un), while the case of algebraic terms is also true but more difficult to prove.
As a particular case, comes the finite choice theorem written as

R⊂ℕ×E, ∀n∈ℕ, (∀i<n, ∃yE, R(i,y)) ⇒ ∃uH, ∀i<n, R(i,ui).

With fEℕ×E the restriction of such u to numbers ≤ n is also unique by induction, from which the xE such that x0=a ∧ ∀n∈ℕ, xn+1 = f(n,xn), can be defined by its graph

{(n,un) | (n,u)∈ℕ×H, u0=a ∧ ∀i<n, ui+1 = f(i,ui)}

As H contains any finite sequence expressible in the theory, this definition of recursion turns out to be "the recursion which the theory can express" independently of the particular choice of H.
The construction of ℕ(ℕ) from the last section, and similarly simple candidate expressions of H as a countable set, assumes recursion, thus cannot be used to define recursion by the above method. To actually provide a definition of recursion that does not assume it, requires an independent expression of an enumerated H, which is harder but finally possible. Such a construction was achieved by Godel as part of his work to prove the incompleteness theorem.

Covering of stable subsets by terms

For any L'-system (E,E), any XE and any x∈〈XL in E there exists a draft (D,D) where xDE, DE and VDX.
It makes no difference to fix VD=X, or X=∅ by adding it as a set of constants to the language.

Proof, fixing X=∅ to simplify.
Let us show the stability of the union B of the D such that there exists a ground draft (D,D) with DE.
Let (s,x,y)∈E such that Im xB. By finite choice there exists a tuple of ground drafts (Di,Di)i<ns such that ∀i<ns, xiDiDiE.
Let D = ⋃i<ns Di.
Let us check that Ψ = (Dz ↦ Ψi(z) for the smallest i such that zDi), defines a ground draft.
For any nonempty AD, take the smallest i such that ADi ≠ ∅.
zADi, ∅ = ADi∩ Im li,z = A ∩ Im li,z because Im li,zDi.
Ψ(z) = Ψi(z) because i is also smallest such that zDi, thus lz = li,z.
We conclude ∃zA, A ∩ Im lz = ∅, thus Ψ defines a ground draft on D.
Then either yD, or adding it there with (s,x,y) gives a ground draft that fits.∎

Thus independently of the axiom of infinity (existence of term algebras) anyway all elements of a minimal algebra are values of ground terms.


Set theory and foundations of mathematics
1. First foundations of mathematics
2. Set theory
3. Algebra 1
4. Arithmetic and first-order foundations
4.1. Algebraic terms
4.2. Quotient systems
4.3. Term algebras
4.4. Integers and recursion
4.5. Presburger Arithmetic
4.6. Finiteness
4.7. Countability and Completeness
4.8. More recursion tools
4.9. Non-standard models of Arithmetic
4.10. Developing theories : definitions
4.11. Constructions
4.A. The Berry paradox
5. Second-order foundations
6. Foundations of Geometry