There Exists a Non-Recursively Enumerable Set {n ∈ N : φ(n)} Such That the Formula φ(n) Is Short and Can Be Easily Translated into a First-Order Formula Which Uses Only + and ·

Apoloniusz Tyszka

doi:10.20944/preprints202508.0363.v9

Submitted:

07 April 2026

Posted:

08 April 2026

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Abstract

We prove that the set \( T=\Bigl\{n\in\mathbb{N}: \exists p,q\in\mathbb{N}\;\Bigl((2n=(p+q)(p+q+1)+2q)\;\wedge\ \) \( \forall (x_0,\ldots,x_p)\in\mathbb{N}^{p+1}\;\exists (y_0,\ldots,y_p)\in\{0,\ldots,q\}^{p+1}\ \) \( \bigl((\forall k\in\{0,\ldots,p\}\;(1=x_k \Rightarrow 1=y_k))\;\wedge\ \) \( (\forall i,j,k\in\{0,\ldots,p\}\;(x_i+x_j=x_k \Rightarrow y_i+y_j=y_k))\;\wedge\ \) \( (\forall i,j,k\in\{0,\ldots,p\}\;(x_i\cdot x_j=x_k \Rightarrow y_i\cdot y_j=y_k))\bigr)\Bigr)\Bigr\}\ \) is not recursively enumerable. By using Gödel's \( \beta \) function, we prove that the formula that defines the set T can be easily translated into a first-order formula which uses only + and \( \cdot \). The same properties has the set \( \Bigl\{n\in\mathbb{N} : \exists p,q\in\mathbb{N}\;\Bigl((2n=(p+q)(p+q+1)+2q)\;\wedge\ \) \( \forall (x_0,\ldots,x_p)\in\mathbb{N}^{p+1}\;\exists (y_0,\ldots,y_p)\in\{0,\ldots,q\}^{p+1}\ \) \( \bigl((\forall j,k\in\{0,\ldots,p\}\;(x_j+1=x_k \Rightarrow y_j+1=y_k))\;\wedge\ \) \( (\forall i,j,k\in\{0,\ldots,p\}\;(x_i\cdot x_j=x_k \Rightarrow y_i\cdot y_j=y_k))\bigr)\Bigr)\Bigr\}\ \).

Keywords:

computable function

;

eventual domination

;

Gödel's β function

;

limit-computable function

;

recursively enumerable set

;

undecidable decision problem

Subject:

Computer Science and Mathematics - Logic

MSC: 03D25

This article is a shortened version of the article [6]. Semi-algorithms differ from algorithms, as they may not terminate.

Definition 1.

(cf. ([4], pp. 233–235)). A computation in the limit of a function

f : N \to N

is a semi-algorithm which takes as input a non-negative integer n and for every

m \in N

prints a non-negative integer

ξ (n, m)

such that

lim_{m \to \infty} ξ (n, m) = f (n)

.

By Definition 1, a function

f : N \to N

is computable in the limit when there exists an infinite computation which takes as input a non-negative integer n and prints a non-negative integer on each iteration and prints

f (n)

on each sufficiently high iteration.

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.

([3], 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 [5]. The term "dominated" in the title of [5] means "eventually dominated".

Theorem 2.

([5]). Flowchart 1 shows a semi-algorithm which computes

f (n)

in the limit.

Flowchart 1

A semi-algorithm which computes

f (n)

in the limit

Definition 2.

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)

.

Flowchart 2 shows a simpler semi-algorithm which computes

f (n)

in the limit.

Flowchart 2

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 3.

For every

n, m \in N

, Flowcharts 1 and 2 print the same number.

Proof.

It follows from Lemmas 1 and 2. □

Corollary 1.

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 4.

For every

n \in N

,

f (n)

is 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}

.

Proof.

It follows from Theorem 2 and Corollary 1. □

Theorem 5.

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 Theorem 4. □

Lemma 3.

([2]). The function

N^{2} ∋ (p, q) \to \frac{1}{2} (p + q) (p + q + 1) + q \in N

is bijective.

Theorem 6.

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

\exists p, q \in N ((2 n = (p + q) (p + q + 1) + 2 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 5 and Lemma 3. □

Let

T = {n \in N : \exists p, q \in N ((2 n = (p + q) (p + q + 1) + 2 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 7.

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

.

Flowchart 3

A semi-algorithm which takes as input

n \in N

and terminates if and only if

n \in N ∖ T

□

Theorem 8.

The set T is not recursively enumerable.

Proof.

It follows from Theorems 6 and 7. □

Let

β : N^{3} \to N

denote Gödel’s

β

function, see [1]. For

x_{1}, x_{2}, x_{3} \in N

,

β (x_{1}, x_{2}, x_{3})

equals the remainder after integer division of

x_{1}

by

1 + (x_{3} + 1) \cdot x_{2}

.

Lemma 4.

([1]). If

(d_{0}, \dots, d_{p}) \in N^{p + 1}

, then

\exists b, c \in N \forall l \in {0, \dots, p} β (b, c, l) = d_{l}

.

Theorem 9.

The formula that defines the set T can be easily translated into a first-order formula which uses only + and ·.

Proof.

By Lemma 4, the set T consists of all

n \in N

such that

\forall u, v \in N \exists a, b, p, q \in N ((2 n = (p + q) (p + q + 1) + 2 q) \land

\forall i, j, k \in {0, \dots, p} ((1 = β (u, v, k) \Rightarrow 1 = min (β (a, b, k), q)) \land

(β (u, v, i) + β (u, v, j) = β (u, v, k) \Rightarrow min (β (a, b, i), q) + min (β (a, b, j), q) = min (β (a, b, k), q)) \land

(β (u, v, i) \cdot β (u, v, j) = β (u, v, k) \Rightarrow min (β (a, b, i), q) \cdot min (β (a, b, j), q) = min (β (a, b, k), q))))

The above formula can be easily translated into a first-order formula which uses only + and ·. □

A more sophisticated proof shows that the set

W = {n \in N : \exists p, q \in N ((2 n = (p + q) (p + q + 1) + 2 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 [6]. By using Gödel’s

β

function, the formula that defines the set W can be easily translated into a first-order formula which uses only + and ·.

References

Gödel’s β function. Available online: https://en.wikipedia.org/wiki/G%C3%B6del%27s_%CE%B2_function.
Pairing function. Available online: https://en.wikipedia.org/wiki/Pairing_function.
Royer, J. S.; Case, J. Subrecursive Programming Systems: Complexity and Succinctness; Birkhäuser: Boston, 1994. [Google Scholar]
Soare, R. I. Interactive computing and relativized computability. In Computability: Turing, Gödel, Church and beyond; Copeland, B. J., Posy, C. J., Shagrir, O., Eds.; MIT Press: Cambridge, MA, 2013; pp. 203–260. [Google Scholar]
Tyszka, A. 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; Daras, N. J., Rassias, M. Th., Eds.; Springer: Cham, 2015; pp. 577–590. [Google Scholar] [CrossRef]
Tyszka, A. There exists a non-recursively enumerable subset of N. which has a short definition in terms of arithmetic. Available online: https://arxiv.org/abs/1309.2605.

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There Exists a Non-Recursively Enumerable Set {n ∈ N : φ(n)} Such That the Formula φ(n) Is Short and Can Be Easily Translated into a First-Order Formula Which Uses Only + and ·

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