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

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 \( \Bigl\{n\in\mathbb{N} : \exists p,q\in\mathbb{N}\;\Bigl((2n=(p+q)(p+q+1)+2q)\;\wedge\ \) \( \forall (x_0,\ldots,x_p)\in\mathbb{N}^{p+1}\;\exists (y_0,\ldots,y_p)\in\{0,\ldots,q\}^{p+1}\ \) \( \bigl((\forall j,k\in\{0,\ldots,p\}\;(x_j+1=x_k \Rightarrow y_j+1=y_k))\;\wedge\ \) \( (\forall i,j,k\in\{0,\ldots,p\}\;(x_i\cdot x_j=x_k \Rightarrow y_i\cdot y_j=y_k))\bigr)\Bigr)\Bigr\}\ \).