A non-recursively enumerable set W⊆N, where N\W is recursively enumerable and the predicate n∈W expresses a fundamental property of n
Abstract
Let F(x,n) denote the formula ∃ab ∀i⩽n ∃swpq ∀jv ∃eg {(s+w)^2+3w+s=2i ∧ ⟨[j=w ∨ v=q] ∨ [j=3i ∧ v=p+q] ∨ [j=s ∧ (v=p ∨ (i=n ∧ v=q+x))] ∨ [j=3i+1 ∧ v=pq] ⇒ a=v+e+ejb ∧ v+g=b⟩} from J. P. Jones' article in vol. 43 of J. Symbolic Logic. From the results of Jones' article, it follows that the set J={n∈N: ¬F(n,n)} is co-recursively enumerable and not recursively enumerable. We prove that the set W={n∈N: ∃p,q∈N ((2n=(p+q)(p+q+1)+2q) ∧ ∀(x_0,...,x_p)∈N^{p+1} ∃(y_0,...,y_p)∈{0,...,q}^{p+1} ((∀i,j,k∈{0,...,p} (x_j+1=x_k ⇒ y_j+1=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i \cdot x_j=x_k ⇒ y_i \cdot y_j=y_k))))} is co-recursively enumerable and not recursively enumerable. Let β:N^3→N denote Gödel's β function. For x_1,x_2,x_3∈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. We prove that the set W consists of all n∈N such that ∀u,v∈N ∃a,b,p,q∈N ((2n=(p+q)(p+q+1)+2q) ∧ ∀i,j,k∈{0,...,p} ((β(a,b,i)⩽q) ∧ (β(u,v,j)+1=β(u,v,k) ⇒ β(a,b,j)+1=β(a,b,k)) ∧ (β(u,v,i) \cdot β(u,v,j)=β(u,v,k) ⇒ β(a,b,i) \cdot β(a,b,j)=β(a,b,k)))). We express the above formula in Peano arithmetic.
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BibTeX
@misc{tyszka2026nonrecursively,
title = {{A non-recursively enumerable set W⊆N, where N\W is recursively enumerable and the predicate n∈W expresses a fundamental property of n}},
author = {Apoloniusz Tyszka},
howpublished = {HAL},
year = {2026},
month = sep,
url = {https://hal.science/hal-01614087v85},
note = {CC BY-NC-ND 4.0},
}
RIS
TY - UNPB
TI - A non-recursively enumerable set W⊆N, where N\W is recursively enumerable and the predicate n∈W expresses a fundamental property of n
AU - Apoloniusz Tyszka
PY - 2026
DA - 2026/09/16
PB - HAL
UR - https://hal.science/hal-01614087v85
AB - Let F(x,n) denote the formula ∃ab ∀i⩽n ∃swpq ∀jv ∃eg {(s+w)^2+3w+s=2i ∧ ⟨[j=w ∨ v=q] ∨ [j=3i ∧ v=p+q] ∨ [j=s ∧ (v=p ∨ (i=n ∧ v=q+x))] ∨ [j=3i+1 ∧ v=pq] ⇒ a=v+e+ejb ∧ v+g=b⟩} from J. P. Jones' article in vol. 43 of J. Symbolic Logic. From the results of Jones' article, it follows that the set J={n∈N: ¬F(n,n)} is co-recursively enumerable and not recursively enumerable. We prove that the set W={n∈N: ∃p,q∈N ((2n=(p+q)(p+q+1)+2q) ∧ ∀(x_0,...,x_p)∈N^{p+1} ∃(y_0,...,y_p)∈{0,...,q}^{p+1} ((∀i,j,k∈{0,...,p} (x_j+1=x_k ⇒ y_j+1=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i \cdot x_j=x_k ⇒ y_i \cdot y_j=y_k))))} is co-recursively enumerable and not recursively enumerable. Let β:N^3→N denote Gödel's β function. For x_1,x_2,x_3∈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. We prove that the set W consists of all n∈N such that ∀u,v∈N ∃a,b,p,q∈N ((2n=(p+q)(p+q+1)+2q) ∧ ∀i,j,k∈{0,...,p} ((β(a,b,i)⩽q) ∧ (β(u,v,j)+1=β(u,v,k) ⇒ β(a,b,j)+1=β(a,b,k)) ∧ (β(u,v,i) \cdot β(u,v,j)=β(u,v,k) ⇒ β(a,b,i) \cdot β(a,b,j)=β(a,b,k)))). We express the above formula in Peano arithmetic.
ER -