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

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.