= Solution
Use the reverse absorption order on the <incline>:
$$
\boxed{a\preceq b\quad\Longleftrightarrow\quad a+b=a.}
$$
This is exactly the relation given by the pairs $(x+y,x)$. Indeed, if $a=x+y$ and $b=x$, then $a+b=x+y+x=a$ by additive <idempotence>, <associativity> and <commutativity>. Conversely $a+b=a$ expresses $a$ as $b+a$, so $(a,b)$ is one of these pairs.
Additive <idempotence> gives $a+a=a$, hence reflexivity. If $a+b=a$ and $b+c=b$, then
$$
a+c=(a+b)+c=a+(b+c)=a+b=a,
$$
so the relation is transitive. In fact it is antisymmetric as well: $a+b=a$ and $b+a=b$, together with <commutativity>, imply $a=b$. Thus \b[the relation is a partial order, and therefore a quasi-order]. It is the reverse of the usual join order of an <incline>; keeping this reversal is essential for the <well-quasi-ordering> conclusion.
Both operations are <order-preserving> for this order. If $a\preceq b$, then $(a+c)+(b+c)=a+c$ by additive <idempotence>, while <distributivity> gives $ac+bc=(a+b)c=ac$. The multiplicative absorption identity also says
$$
a\preceq ab.
$$
Thus multiplying a <monomial> by another factor moves upward in the reverse absorption order.
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