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This is really a comment, but it's too long:

I'm not entirely sure what you're asking, but I think there are a couple things worth pointing out:

• In terms of what a model $M$ "knows" about a structure Ackermann-coded into one of its subsets $A$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?

• In terms of Parikh's model, I think you're overestimating its relevance to Ackermann coding. Any model of PA has a unique definable map satisfying the recursive properties of exponentiation. (Indeed, PA proves "there is a unique exponentiation function" in the following sense: PA proves "for all $n$, there is a unique map from $\{0, 1, . . . , n\}^2$ to $\mathbb{N}$ which satisfies [recursive properties]". Note that such a map is a finite object, so can be talked about directly in PA, even though the whole exponentiation map can't.) Ackermann coding works with respect to this definable map. Parikh showed that we can have undefinable maps that also satisfy the properties of exponentiation. But these maps aren't relevant to Ackermann coding.

Based on the comments, I think it's worth clarifying some points about the Ackermann interpretation.

The Ackermann interpretation is definable: it is a formula $\varphi(x, y)$, and - given a model $M\models PA$ - it gives us an interpreted structure $Ack(M)=(M, \varphi^M)$. This is true regardless of what $M$ is.

Now, some facts about $\varphi$ are provable in $PA$, and these facts in turn tell us things about the theory of the structure $Ack(M)$ associated to a model of $PA$. For example, $PA$ proves $$\forall x_1, x_2[\forall y[\varphi(y, x_1)\iff\varphi(y, x_2)]\iff x_1=x_2],$$ and this tells us that $Ack(M)$ satisfies extensionality. In fact, every axiom of $ZF-Inf+\neg Inf$ (call that theory "$T$" for simplicity), when appropriately phrased in terms of "$\varphi$" in place of "$\in$", is a provable sentence of $PA$ (this takes work); so if $M\models PA$, we will always have $Ack(M)\models T$.

In particular, this means we can never "get $\omega$" from the Ackermann interpretation of a model of $PA$. This can in fact be proved pretty easily: show that if $Ack(M)$ thinks $n$ is $\omega$, then $M$ thinks $n$ defines a proper cut in $M$, and this is impossible if $M\models PA$.

This should also help explain how Parikh's phenomenon is irrelevant here: the Ackermann interpretation leans on a specific formula, whereas Parikh shows that undefinable functions with certain properties can exist in some models.

Now let me say some things about definability/undefinability. There are a couple things worth pointing out:

• In terms of what a model $M$ "knows" about the substructure of $Ack(M)$ corresponding to a subset $A\subseteq M$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?

• In terms of Parikh's model, let me recap what I said above in more detail. Any model of PA has a unique definable map satisfying the recursive properties of exponentiation. (Indeed, PA proves "there is a unique exponentiation function" in the following sense: PA proves "for all $n$, there is a unique map from $\{0, 1, . . . , n\}^2$ to $\mathbb{N}$ which satisfies [recursive properties]". Note that such a map is a finite object, so can be talked about directly in PA, even though the whole exponentiation map can't.) Ackermann coding works with respect to this definable map. Parikh showed that we can have undefinable maps that also satisfy the properties of exponentiation. But these maps aren't relevant to Ackermann coding.

This is really a comment, but it's too long:

I'm not entirely sure what you're asking, but I think there are a couple things worth pointing out:

• In terms of what a model $M$ "knows" about a structure Ackermann-coded into one of its subsets $A$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?

• In terms of Parikh's model, I think you're overestimating its relevance to Ackermann coding. Any model of PA has a unique definable map satisfying the recursive properties of exponentiation. (Indeed, PA proves "there is a unique exponentiation function" in the following sense: PA proves "for all $n$, there is a unique map from $\{0, 1, . . . , n\}^2$ to $\mathbb{N}$ which satisfies [recursive properties]". Note that such a map is a finite object, so can be talked about directly in PA, even though the whole exponentiation map can't.) Ackermann coding works with respect to this definable map. Parikh showed that we can have undefinable maps that also satisfy the properties of exponentiation. But these maps aren't relevant to Ackermann coding.

This is really a comment, but it's too long:

I'm not entirely sure what you're asking, but I think there are a couple things worth pointing out:

• In terms of what a model $M$ "knows" about a structure Ackermann-coded into one of its subsets $A$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?

• In terms of Parikh's model, I think you're overestimating its relevance to Ackermann coding. Any model of PA has a unique definable map satisfying the recursive properties of exponentiation. (Indeed, PA proves "there is a unique exponentiation function" in the following sense: PA proves "for all $n$, there is a unique map from $\{0, 1, . . . , n\}^2$ to $\mathbb{N}$ which satisfies [recursive properties]". Note that such a map is a finite object, so can be talked about directly in PA, even though the whole exponentiation map can't.) Ackermann coding works with respect to this definable map. Parikh showed that we can have undefinable maps that also satisfy the properties of exponentiation. But these maps aren't relevant to Ackermann coding.

Both bullet points really fall under the same category: although "strong foundational" theories (like PA and ZF) imply lots of rigidity about the universe in various ways, their nonstandard models have tons of flexibility, undefinably. For example, every nonstandard model $W$ of ZF contains a set model $m$ of ZF, even if it proves that ZF is inconsistent! . . . the trick being, of course, that the model is incorrect about what ZF is, and $m$ fails to satisfy some of the axioms of ZF$^W$, so the sense in which $m$ is a model of ZF is really undefinable. Similarly, Cohen's result that we can have a model of ZF with an automorphism of order $2$ is inherently about the undefinable structure of models, since no model of ZF has any definable automorphisms at all.

Ressayre's substructures and Parikh's exponentiations are examples of this "undefinable flexibility". But because they're appropriately undefinable, they're quite slippery, and don't interact with definable phenomena (like Ackermann coding) in any nice way.

This is really a comment, but it's too long:

I'm not entirely sure what you're asking, but I think there are a couple things worth pointing out:

• In terms of what a model $M$ "knows" about a structure Ackermann-coded into one of its subsets $A$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?

• In terms of Parikh's model, I think you're overestimating its relevance to Ackermann coding. Ackermann coding relies only on being able to prove that binary expansions of natural numbers exist and are unique; this is ultimately only a statement about the map $x\mapsto 2^x$. This is legal, since - in any model $M$ of PA - we have a definable exponentiation map, which PA proves satisfies the recursive properties of exponentiation. The possible existence of other maps which satisfy the recursive properties of exponentiation isn't relevant, since those maps won't be definable: any two definable maps satisfying the recursive properties of exponentiation must be the same, by induction.

This is really a comment, but it's too long:

I'm not entirely sure what you're asking, but I think there are a couple things worth pointing out:

• In terms of what a model $M$ "knows" about a structure Ackermann-coded into one of its subsets $A$, note that this question best makes sense if $M$ knows about $A$ to begin with. But most of the time, the $A$s we'll be interested in won't be definable in $M$: indeed, no model $W$ of ZF can be Ackermann-coded into any nonstandard model $M$ of PA in a definable way (since such an encoding would give evidence in $M$ of the non-standardness of whatever element of $M$ is coding the $\omega$ of $W$). So you have to be very careful when you ask "What does $M$ know?" about such a situation - what exactly do you mean, given that the most basic information about such a configuration will generally be undefinable in $M$?

• In terms of Parikh's model, I think you're overestimating its relevance to Ackermann coding. Any model of PA has a unique definable map satisfying the recursive properties of exponentiation. (Indeed, PA proves "there is a unique exponentiation function" in the following sense: PA proves "for all $n$, there is a unique map from $\{0, 1, . . . , n\}^2$ to $\mathbb{N}$ which satisfies [recursive properties]". Note that such a map is a finite object, so can be talked about directly in PA, even though the whole exponentiation map can't.) Ackermann coding works with respect to this definable map. Parikh showed that we can have undefinable maps that also satisfy the properties of exponentiation. But these maps aren't relevant to Ackermann coding.