There Exists a Subset of N Which Is Not Recursively Enumerable and Has a Short Description in Terms of Arithmetic

Apoloniusz Tyszka

doi:10.20944/preprints202508.0363.v8

Submitted:

25 February 2026

Posted:

27 February 2026

Read the latest preprint version here

Abstract

We prove that the set {n∈N: ∃p,q∈N ((n=2^p \cdot 3^q) ∧ ∀(x_0,...,x_p)∈N^{p+1} ∃(y_0,...,y_p)∈{0,...,q}^{p+1} ((∀k∈{0,...,p} (1=x_k ⇒ 1=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i+x_j=x_k ⇒ y_i+y_j=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i \cdot x_j=x_k ⇒ y_i \cdot y_j=y_k))))} is not recursively enumerable.

Keywords:

arithmetic of N

;

computable function

;

eventual domination

;

Hilbert's 10th problem

;

limit-computable function

;

recursively enumerable set

;

undecidable decision problem

Subject:

Computer Science and Mathematics - Logic

This article is a shortened version of the article [3]. For

n \in N

, let

E_{n} = {1 = x_{k}, x_{i} + x_{j} = x_{k}, x_{i} \cdot x_{j} = x_{k} : i, j, k \in {0, \dots, n}}

Theorem 1

([1], p. 118). There exists a limit-computable function

f : N \to N

which eventually dominates every computable function

g : N \to N

.

We present an alternative proof of Theorem 1. For

n \in N

,

f (n)

denotes the smallest

b \in N

such that if a system of equations

S \subseteq E_{n}

has a solution in

N^{n + 1}

, then S has a solution in

{0, \dots, b}^{n + 1}

. The function

f : N \to N

is computable in the limit and eventually dominates every computable function

g : N \to N

, see [2]. The term "dominated" in the title of [2] means "eventually dominated". Flowchart 1 shows a semi-algorithm which computes

f (n)

in the limit, see [2].

A semi-algorithm which computes

f (n)

in the limit

Flowchart 2 shows a simpler semi-algorithm which computes

f (n)

in the limit.

A simpler semi-algorithm which computes

f (n)

in the limit

Lemma 1.

For every

n, m \in N

, the number printed by Flowchart 2 does not exceed the number printed by Flowchart 1.

Proof.

For every

(a_{0}, \dots, a_{n}) \in {0, \dots, m}^{n + 1}

,

E_{n} \supseteq {1 = x_{k} : (k \in {0, \dots, n}) \land (1 = a_{k})} \cup

{x_{i} + x_{j} = x_{k} : (i, j, k \in {0, \dots, n}) \land (a_{i} + a_{j} = a_{k})} \cup

{x_{i} \cdot x_{j} = x_{k} : (i, j, k \in {0, \dots, n}) \land (a_{i} \cdot a_{j} = a_{k})}

□

Lemma 2.

For every

n, m \in N

, the number printed by Flowchart 1 does not exceed the number printed by Flowchart 2.

Proof.

Let

n, m \in N

. For every system of equations

S \subseteq E_{n}

, if

(a_{0}, \dots, a_{n}) \in {0, \dots, m}^{n + 1}

and

(a_{0}, \dots, a_{n})

solves S, then

(a_{0}, \dots, a_{n})

solves the following system of equations:

{1 = x_{k} : (k \in {0, \dots, n}) \land (1 = a_{k})} \cup

{x_{i} + x_{j} = x_{k} : (i, j, k \in {0, \dots, n}) \land (a_{i} + a_{j} = a_{k})} \cup

{x_{i} \cdot x_{j} = x_{k} : (i, j, k \in {0, \dots, n}) \land (a_{i} \cdot a_{j} = a_{k})}

□

Theorem 2.

For every

n, m \in N

, Flowcharts 1 and 2 print the same number.

Proof.

It follows from Lemmas 1 and 2. □

Definition 1.

An approximation of a tuple

(x_{0}, \dots, x_{n}) \in N^{n + 1}

is a tuple

(y_{0}, \dots, y_{n}) \in N^{n + 1}

such that

(\forall k \in {0, \dots, n} (1 = x_{k} \Rightarrow 1 = y_{k})) \land

(\forall i, j, k \in {0, \dots, n} (x_{i} + x_{j} = x_{k} \Rightarrow y_{i} + y_{j} = y_{k})) \land

(\forall i, j, k \in {0, \dots, n} (x_{i} \cdot x_{j} = x_{k} \Rightarrow y_{i} \cdot y_{j} = y_{k}))

Observation 1.

For every

n \in N

, there exists a set

A (n) \subseteq N^{n + 1}

such that

card (A (n)) ⩽ 2^{card (E_{n})} = 2^{n + 1 + 2 \cdot {(n + 1)}^{3}}

and every tuple

(x_{0}, \dots, x_{n}) \in N^{n + 1}

possesses an approximation in

A (n)

.

Observation 2.

For every

n \in N

,

f (n)

equals the smallest

b \in N

such that every tuple

(x_{0}, \dots, x_{n}) \in N^{n + 1}

possesses an approximation in

{0, \dots, b}^{n + 1}

.

Observation 3.

For every

n, m \in N

, Flowcharts 1 and 2 print the smallest

b \in {0, \dots, m}

such that every tuple

(x_{0}, \dots, x_{n}) \in {0, \dots, m}^{n + 1}

possesses an approximation in

{0, \dots, b}^{n + 1}

.

Theorem 3.

No algorithm takes as input non-negative integers n and m and decides whether or not

\forall (x_{0}, \dots, x_{n}) \in N^{n + 1} \exists (y_{0}, \dots, y_{n}) \in {0, \dots, m}^{n + 1}

((\forall k \in {0, \dots, n} (1 = x_{k} \Rightarrow 1 = y_{k})) \land

(\forall i, j, k \in {0, \dots, n} (x_{i} + x_{j} = x_{k} \Rightarrow y_{i} + y_{j} = y_{k})) \land

(\forall i, j, k \in {0, \dots, n} (x_{i} \cdot x_{j} = x_{k} \Rightarrow y_{i} \cdot y_{j} = y_{k})))

Proof.

Since the function f is not computable, it follows from Observation 2. □

Theorem 4.

No algorithm takes as input a non-negative integer n and decides whether or not

\exists p, q \in N ((n = 2^{p} \cdot 3^{q}) \land

\forall (x_{0}, \dots, x_{p}) \in N^{p + 1} \exists (y_{0}, \dots, y_{p}) \in {0, \dots, q}^{p + 1}

((\forall k \in {0, \dots, p} (1 = x_{k} \Rightarrow 1 = y_{k})) \land

(\forall i, j, k \in {0, \dots, p} (x_{i} + x_{j} = x_{k} \Rightarrow y_{i} + y_{j} = y_{k})) \land

(\forall i, j, k \in {0, \dots, p} (x_{i} \cdot x_{j} = x_{k} \Rightarrow y_{i} \cdot y_{j} = y_{k}))))

Proof.

It follows from Theorem 3. □

Let

T = {n \in N : \exists p, q \in N ((n = 2^{p} \cdot 3^{q}) \land

\forall (x_{0}, \dots, x_{p}) \in N^{p + 1} \exists (y_{0}, \dots, y_{p}) \in {0, \dots, q}^{p + 1}

((\forall k \in {0, \dots, p} (1 = x_{k} \Rightarrow 1 = y_{k})) \land

(\forall i, j, k \in {0, \dots, p} (x_{i} + x_{j} = x_{k} \Rightarrow y_{i} + y_{j} = y_{k})) \land

(\forall i, j, k \in {0, \dots, p} (x_{i} \cdot x_{j} = x_{k} \Rightarrow y_{i} \cdot y_{j} = y_{k}))))}

Theorem 5.

The set

N ∖ T

is recursively enumerable.

Proof.

For

i \in N

, let

p_{i}

denote the i-th prime number. Flowchart 3 shows a semi-algorithm which takes as input

n \in N

and terminates if and only if

n \in N ∖ T

.

A semi-algorithm which takes as input

n \in N

and terminates if and only if

n \in N ∖ T

□

Theorem 6.

The set T is not recursively enumerable.

Proof.

It follows from Theorems 4 and 5. □

A more sophisticated proof shows that the set

{n \in N : \exists p, q \in N ((n = 2^{p} \cdot 3^{q}) \land

\forall (x_{0}, \dots, x_{p}) \in N^{p + 1} \exists (y_{0}, \dots, y_{p}) \in {0, \dots, q}^{p + 1}

((\forall j, k \in {0, \dots, p} (x_{j} + 1 = x_{k} \Rightarrow y_{j} + 1 = y_{k})) \land

(\forall i, j, k \in {0, \dots, p} (x_{i} \cdot x_{j} = x_{k} \Rightarrow y_{i} \cdot y_{j} = y_{k}))))}

is not recursively enumerable, see [3].

References

J. S. Royer and J. Case, Subrecursive Programming Systems: Complexity and Succinctness, Birkhäuser, Boston, 1994.
A. Tyszka, All functions g: $N$ → $N$ which have a single-fold Diophantine representation are dominated by a limit-computable function f : $N$ ∖{0} → $N$ which is implemented in MuPAD and whose computability is an open problem, in: Computation, cryptography, and network security (eds. N. J. Daras, M. Th. Rassias), Springer, Cham, 2015, 577–590. [CrossRef]
A. Tyszka, Three undecidable decision problems about a non-negative integer n which have a short description in terms of arithmetic, https://philarchive.org/rec/TYSATA and https://ssrn.com/abstract=4710446.

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There Exists a Subset of N Which Is Not Recursively Enumerable and Has a Short Description in Terms of Arithmetic

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