= Solution
Suppose the <incline> is generated by $x_1,\ldots,x_r$. Repeated <distributivity>, followed by <commutativity> and additive <idempotence>, expresses every element as a finite nonempty sum
$$
f_A=\sum_{\alpha\in A}x_1^{\alpha_1}\cdots x_r^{\alpha_r},\qquad A\subseteq\mathbb N^r\setminus\{0\}\text{ finite and nonempty}.
$$
No multiplicative identity is assumed: a zero exponent means that the corresponding generator is omitted, and every displayed <monomial> has at least one factor. Different supports may represent the same element; uniqueness is unnecessary.
Associate to $A$ its upward-closed exponent set
$$
U_A=\{\beta\in\mathbb N^r:\exists\alpha\in A\ (\alpha\le\beta\text{ coordinatewise})\}.
$$
If $U_A\supseteq U_B$, each <monomial> of $f_B$ is either equal to, or a multiplicative extension of, a <monomial> of $f_A$. It is absorbed by that earlier <monomial>, and hence by the full sum $f_A$. Adding the terms of $f_B$ one at a time gives
$$
\boxed{U_A\supseteq U_B\ \Longrightarrow\ f_A+f_B=f_A\ \Longrightarrow\ f_A\preceq f_B.}
$$
For equal exponent vectors the first absorption is additive <idempotence>; for strictly larger vectors it is the incline's multiplicative absorption. Thus the issue is to prove that exponent upsets are <well-quasi-ordered> by reverse inclusion, not merely to apply <Dickson's lemma> to individual exponents.
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