 Solution

The PDF's displayed inequality uses the relation it purports to define and therefore is not itself a complete definition. The absorption order consistent with that inequality and with the requested result isThis is the reverse of the usual natural order of an incline. We specify it explicitly: using instead $x+y=y$ would make the finite-generation conclusion false in general.

Idempotence gives $x+x=x$, proving reflexivity. If $x+y=x$ and $y+z=y$, then associativity givesso the relation is transitive. If also $y+x=y$, commutativity forces $x=y$. Thus it is even a partial order. In this order $x+y⪯x$ follows because $(x+y)+x=x+y$. Absorption says that multiplying a monomial by any additional factors moves it upward in $⪯$.

Let $g_1,\dots ,g_r$ generate the incline. Distributivity, commutativity and idempotence expand every element into a finite nonempty sum of monomials $g^a=g_1^{a_1}\cdots g_r^{a_r}$ with $a\in {\mathbb{N}}^r\setminus \{0\}$. A zero exponent means that factor is omitted; no multiplicative identity is assumed. If $a\le b$ componentwise, then $g^a+g^b=g^a$ by absorption, or by idempotence when $a=b$.

We prove the needed reverse-inclusion well-quasi-ordering of monomial ideals directly. Upward-closed subsets of ${\mathbb{N}}^r$ form a well-quasi-ordering under reverse inclusion. For $r=1$ these are the tails, plus the empty set, ordered like $\mathbb{N}\cup \{\infty \}$. For the induction, write an upset $U\subseteq {\mathbb{N}}^{r+1}$ as increasing slices $U_0\subseteq U_1\subseteq \cdots$ in ${\mathbb{N}}^r$. Their union is an upset with a finite minimal basis by Dickson lemma. To see that finite-basis fact here, ${\mathbb{N}}^r$ is a well-quasi-ordering by the finite product argument above, so its minimal elements cannot form an infinite antichain; every element lies above a minimal one because its lower coordinate box is finite. All basis elements of the union appear by some finite slice, so the slices eventually stabilize.

Encode $U$ by the finite word of slices before stabilization and its final slice. By the finite-word theorem proved in the other alternative, and finite product closure, these encodings are a well-quasi-ordering with slice comparisons given by reverse inclusion. For two comparable encodings, let the earlier pre-stabilization slice $U_k$ match the later slice $V_{f(k)}$, where $f(k)\ge k$. ThenBeyond the earlier stabilization index, its final slice contains the later final slice and hence every later slice. Therefore $U\supseteq V$. This proves the induction, including empty slices and the empty upset.

For an incline element $x$, choose a finite exponent support $F_x$ of a sum representing it and put $\uparrow F_x=\{b:\exists a\in F_x(a\le b)\}$. Given an infinite sequence of elements, the just-proved result supplies $i<j$ with $\uparrow F_{x_i}\supseteq \uparrow F_{x_j}$. Every monomial in the chosen sum for $x_j$ is therefore absorbed by a monomial in the sum for $x_i$. Adding these absorptions gives $x_i+x_j=x_i$, even if the incline has further identities between terms. HenceThe order direction matters. On the positive integers take $x+y=\min (x,y)$ and multiplication to be ordinary addition of integers. This is an incline generated by $1$. Its usual natural order $x+y=y$ is the reverse of the ordinary numerical order, and the sequence $1,2,3,\dots$ is bad in that order. Its reverse absorption order is the ordinary numerical order, consistent with the theorem above.