The answer is Yes. This answer is based on the idea of Clemens Grabmayer, which makes the observation that addition $+$ is definable from multiplication $\cdot$ and successor. The idea generalizes to the following: **Theorem.** Suppose that relation $R$ is implicitly definable in model $M$, that $S$ is implicitly definable in the expansion $\langle M,R\rangle$, and that $R$ is explicitly definable in $\langle M,S\rangle$. Then $S$ is implicitly definable in $M$. **Proof.** Suppose that $R$ is the unique relation fulfilling sentence $\varphi(\dot R)$ in $M$, in the language expanded with predicate $\dot R$. Suppose $S$ is the unique relation fulfilling sentence $\psi(R,\dot S)$ in $\langle M,R\rangle$. And suppose that $R$ is definable by formula $\theta(x,S)$ in $\langle M,S\rangle$, in that $Rx\leftrightarrow\theta(x,S)$. Let $\Phi(\dot S)$ be the sentence asserting: - the relation defined by $\theta(x,\dot S)$ fulfills property $\varphi$, and - $\psi("\theta(x,\dot S)",\dot S)$ holds, that is, the assertion $\psi(\dot R,\dot S)$ holds where $\dot R$ is interpreted by the relation defined by $\theta(x,\dot S)$. I claim that this is an implicit definition of $S$ in $M$. The reason is that whatever relation $\Phi$ does define has the property that the relation extracted from it via $\theta(x,\dot S)$ will have to be $R$, since it fulfills the implicit definition of $R$ given by $\varphi$. And further, since $\Phi$ asserts that $\psi$ is fulfilled by $\dot S$ relative to that relation, it follows that $\dot S$ must be $S$. $\Box$ The corollary is that: **Corollary.** Multiplication is implicitly definable from successor. **Proof.** Addition is implicitly definable in $\langle\mathbb{N},S,0\rangle$, and multiplication is implicictly definable over addition $\langle\mathbb{N},S,0,+\rangle$, and by the Boolos/Jeffrey observation, addition is explicitly definable from multiplication and successor. So we are in the case of the theorem. $\Box$ The theorem falls short of proving that the property of being implicitly-definable-over is transitive. That seems to be false, in light of counterexamples discussed in the comments.