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Contrary to my initial expectation, 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:

• $\varphi(\theta(x,\dot S))$, that is, 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 interpretation is given to $\dot S$, it will have 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. $\Box$

A more striking instance might be:

Corollary. First-order arithmetic truth for the standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1<\rangle$ is implicitly definable just from successor $\langle\mathbb{N},S,0\rangle$.

Proof. The truth predicate is uniquely characterized on the standard model $\mathbb{N}$ by fulfilling the Tarski recursion, and so it is implicitly definable in $\langle\mathbb{N},+,\cdot\rangle$. But both addition and multiplication are definable from the truth predicate, and they themselves are implicitly definable from successor. So by the theorem, truth is implicitly definable from successor. $\Box$

And one can of course iterate this by forming the predicate for truth-about-truth, and truth-about-truth-about-truth and so on, proceeding transfinitely up the hierarchy for quite some way.

But lastly, let me mention that 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.

Contrary to my initial expectation, 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:

• $\varphi(\theta(x,\dot S))$, that is, 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 interpretation is given to $\dot S$, it will have 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. $\Box$

A more striking instance might be:

Corollary. First-order arithmetic truth for the standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1<\rangle$ is implicitly definable just from successor $\langle\mathbb{N},S,0\rangle$.

Proof. I intend to use the trinary truth predicate $\text{Tr}(\varphi,x,y,z)$, holding when $\mathbb{N}\models\varphi[x,y,z]$. This truth predicate is uniquely characterized on the standard model $\mathbb{N}$ by fulfilling the Tarski recursion, and so it is implicitly definable in $\langle\mathbb{N},+,\cdot\rangle$. But both addition and multiplication are definable from the truth predicate (this is why we use the trinary version, since with just successor we don't initially have any coding, but once we get $+$ and $\times$, then the usual coding kicks in), and they themselves are implicitly definable from successor. So by the theorem, truth is implicitly definable from successor. $\Box$

And one can of course iterate this by forming the predicate for truth-about-truth, and truth-about-truth-about-truth and so on, proceeding transfinitely up the hierarchy for quite some way.

But lastly, let me mention that 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.

Contrary to my initial expectation, 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:

• $\varphi(\theta(x,\dot S))$, that is, 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 interpretation is given to $\dot S$, it will have 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.

Contrary to my initial expectation, 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:

• $\varphi(\theta(x,\dot S))$, that is, 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 interpretation is given to $\dot S$, it will have 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. $\Box$

A more striking instance might be:

Corollary. First-order arithmetic truth for the standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1<\rangle$ is implicitly definable just from successor $\langle\mathbb{N},S,0\rangle$.

Proof. The truth predicate is uniquely characterized on the standard model $\mathbb{N}$ by fulfilling the Tarski recursion, and so it is implicitly definable in $\langle\mathbb{N},+,\cdot\rangle$. But both addition and multiplication are definable from the truth predicate, and they themselves are implicitly definable from successor. So by the theorem, truth is implicitly definable from successor. $\Box$

And one can of course iterate this by forming the predicate for truth-about-truth, and truth-about-truth-about-truth and so on, proceeding transfinitely up the hierarchy for quite some way.

But lastly, let me mention that 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.

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.

Contrary to my initial expectation, 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:

• $\varphi(\theta(x,\dot S))$, that is, 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 interpretation is given to $\dot S$, it will have 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.