[proofplan] We prove the contrapositive by assuming that $S$ is nonempty. The image $M(S) \subset \mathbb{N}$ is then a nonempty set of natural numbers, so it has a least element. Choosing an element of $S$ whose measure realizes this least value contradicts the assumed descent property. [/proofplan] [step:Choose a solution of minimal measure] Assume, 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 $\mathbb{N}$, the set $A$ has a least element; denote it by $m_0 \in A$. Since $m_0 \in M(S)$, there exists $s_0 \in S$ such that \begin{align*} M(s_0) = m_0. \end{align*} Thus $s_0$ is an element of $S$ whose measure is minimal among all values of $M$ on $S$. [/step] [step:Apply the descent hypothesis to contradict minimality] By the descent hypothesis applied to $s_0 \in S$, there exists $s_1 \in S$ such that \begin{align*} M(s_1) < M(s_0). \end{align*} Using $M(s_0) = m_0$, this gives \begin{align*} M(s_1) < m_0. \end{align*} But $s_1 \in S$, so $M(s_1) \in M(S) = A$. This contradicts the fact that $m_0$ is the least element of $A$. [/step] [step:Conclude that no element exists] The contradiction arose from the assumption that $S \neq \varnothing$. Therefore that assumption is false, and hence \begin{align*} S = \varnothing. \end{align*} This proves the Infinite Descent Principle. [/step]