[proofplan] We argue by contradiction using the well-foundedness of $R$. If some element fails $P$, the set of counterexamples has an $R$-minimal element. Minimality forces every $R$-predecessor of that element to satisfy $P$, and the assumed induction step then gives $P$ for the minimal counterexample itself, a contradiction. [/proofplan] [step:Form the set of counterexamples] Define the set of counterexamples $C \subset A$ by \begin{align*} C := \{a \in A : P(a) \text{ is false}\}. \end{align*} We prove that $C = \varnothing$. Suppose, for contradiction, that $C \neq \varnothing$. [/step] [step:Choose an $R$-minimal counterexample by well-foundedness] Since $R$ is well-founded on $A$ and $C$ is a nonempty subset of $A$, there exists an element $a_0 \in C$ such that no element of $C$ is an $R$-predecessor of $a_0$. Equivalently, \begin{align*} \forall b \in C,\ (b,a_0) \notin R. \end{align*} Because $a_0 \in C$, the predicate $P(a_0)$ is false. [/step] [step:Show that every predecessor of the minimal counterexample satisfies $P$] Let $b \in A$ satisfy $(b,a_0) \in R$. If $P(b)$ were false, then $b \in C$ by the definition of $C$. This would contradict the $R$-minimality condition \begin{align*} \forall c \in C,\ (c,a_0) \notin R. \end{align*} Therefore $P(b)$ is true. Since $b \in A$ with $(b,a_0) \in R$ was arbitrary, we have \begin{align*} \forall b \in A,\ (b,a_0) \in R \implies P(b). \end{align*} [/step] [step:Apply the induction hypothesis at the minimal counterexample] The assumed induction step applies to the element $a_0 \in A$. From \begin{align*} \forall b \in A,\ (b,a_0) \in R \implies P(b), \end{align*} it follows that $P(a_0)$ is true. This contradicts the fact that $a_0 \in C$, which means that $P(a_0)$ is false. [/step] [step:Conclude that there are no counterexamples] The contradiction shows that the assumption $C \neq \varnothing$ is false. Hence $C = \varnothing$. By the definition of $C$, there is no element $a \in A$ for which $P(a)$ is false. Therefore $P(a)$ holds for every $a \in A$. [/step]