I've read @TauMu's question about the set of functions $\mathbb N\rightarrow\mathbb N$ generated from the identity map by repeatedly applying exponentiation of two already accepted functions. @TauMu asks if such a set is well-ordered with respect to *eventual domination*.

It seems to me that it would be hard, perhaps impossible, to obtain a set which is not well-ordered regardless of the choice of a binary operation $\mathbb N\times\mathbb N\rightarrow\mathbb N$. Here is a patient formulation of my more general question:

Given an operation $\tau : \mathbb N^2\rightarrow\mathbb N$, let $\bigcirc^{\tau}\subseteq\mathbb N^{\mathbb N}$ be the smallest set such that it has the identity map $I_{\mathbb N}$ as its element, $I_\mathbb N\in\bigcirc^\tau$, and $h:=\tau\circ(f\triangle g)\in\bigcirc^{\tau}$ for every $f\ g\in\bigcirc^{\tau}$.

**REMARK** (*an explanation of the notation above*)
$$\forall_{n\in\mathbb N}\quad h(n) := \tau(f(n)\ g(n))$$

Finally,

**QUESTION** (edited twice after the 1st and 2nd comment by *@Joseph Van Name*) Does there exists an operation $\tau:\mathbb N^2\rightarrow \mathbb N$ dominating the identity in each variable (see below), and such that $\bigcirc^{\tau}$ contains an infinite strictly decreasing sequence with respect to the relation of *eventual domination*?

By $\tau$ dominating the identity in each variable I mean that:

$$\forall_{k\ n\in\mathbb N}\quad \tau(k\ \ n)\ \ \ge\ \ \max(k\ \ \ n)$$