[proofplan] We argue by contradiction. If $S$ were nonempty, then the image $M(S)\subset\mathbb{N}$ would be a nonempty subset of the natural numbers. The well-ordering property of $\mathbb{N}$ gives a least value attained by $M$, but the descent hypothesis produces an element of $S$ with a strictly smaller value, contradicting minimality. [/proofplan] [step:Assume a nonempty counterexample and choose a minimal value of $M$] Suppose, for contradiction, that $S\neq\varnothing$. Define the image set \begin{align*} A:=M(S)=\{M(s):s\in S\}\subset\mathbb{N}. \end{align*} Since $S\neq\varnothing$, the set $A$ is nonempty. By the well-ordering property of $\mathbb{N}$, there exists $m_0\in A$ such that \begin{align*} m_0\leq m \end{align*} for every $m\in A$. By the definition of $A=M(S)$, choose $s_0\in S$ such that \begin{align*} M(s_0)=m_0. \end{align*} [/step] [step:Apply the descent hypothesis to contradict minimality] Applying the hypothesis to $s_0\in S$, there exists $s_1\in S$ such that \begin{align*} M(s_1)<M(s_0). \end{align*} Since $s_1\in S$, we have $M(s_1)\in A$. Using $M(s_0)=m_0$, this gives \begin{align*} M(s_1)<m_0, \end{align*} which contradicts the choice of $m_0$ as the least element of $A$. [/step] [step:Conclude that the set is empty] The assumption $S\neq\varnothing$ leads to a contradiction. Therefore $S=\varnothing$, as claimed. [/step]