Proof outline for $\mathcal {T} = {\textsf {EA}}$

Suppose $\sigma (x)$ is a $\Sigma _1$ -formula that defines a set $\Phi $ of sentences of $\mathcal {L}_{{\textsf {PA}}}$ in the standard model of arithmetic. (By Craig’s trick, $\sigma $ can be chosen to be an elementary formula, this does not play a role in this proof, but it will come handy in the proof of Proposition 21, which is based on this one.) Let $\mathsf {True}(y,P)$ be as in Definition 9. The desired scheme template $\tau $ is

$$ \begin{align*} \forall y\left[\mathsf{True}(y,P)\rightarrow \forall x\left[\left(\sigma(x)\wedge \textsf{pdepth}(x)\leq y\right) \rightarrow P(x)\right] \right]. \end{align*} $$

We note that:

(1) ${\textsf {EA}}+S_{\tau } \vdash \Phi $ , because for each $n\in \omega $ the truth predicate for formulae of depth at most n is definable by Theorem 11; and

(2) ${\textsf {EA}}+\Phi \vdash S_{\tau }$ . To see this, suppose to the contrary that some instance $\psi $ of $S_{\tau }$ is not provable in ${\textsf {EA}}+\Phi $ . Then by the completeness theorem of first order logic there is a model $\mathcal {M}$ of ${\textsf {EA}} + \Phi + \lnot \psi $ . Since by the definition of $S_{\tau }$ there is a formula $\varphi (x,v)$ such that $\psi $ is a sentence of the form $\forall v~\tau [\varphi (x,v)/P]$ , we have

$$ \begin{align*}\mathcal{M} \models {\textsf{EA}} + \Phi + \lnot (\forall v~\tau [\varphi (x,v)/P]). \end{align*} $$

Thus $\mathcal {M}$ is a model of ${\textsf {EA}} +\Phi $ in which the sentence $ \exists v ~\lnot \tau [\varphi (x,v)/P])$ holds, i.e., $\mathcal {M}\models \exists v \exists y \theta (v,y),$ where

$$ \begin{align*}\theta(v,y) := \left[\mathsf{True}(y,\varphi(x,v)/P)\land\exists x\left[ \left(\sigma(x)\wedge \textsf{pdepth}(x)\leq y\right) \land \lnot \varphi(x,v)\right] \right].\end{align*} $$

Let a and b be elements in $\mathcal {M}$ such that $\mathcal {M}\models \theta (a,b)$ . The key observation at this point is that b cannot be a standard element since $\mathcal {M}\models \Phi .$ (It is precisely at this step that the argument would have broken down if we had used depth instead instead of pure depth in our formulation of the scheme template $\tau $ .) Together with the fact that $\mathcal {M}\models \mathsf {True}(b,\varphi (x,a)/P)$ , this implies that the formula $\varphi (x,a)$ defines a subset of $\mathcal {M}$ that satisfies Tarski’s compositional clauses for all standard formulae, thus contradicting Tarski’s undefinability of truth theorem.

Convention 18. Simplifying things a little bit, when talking about the structures $\langle \Delta , \leq _{{\textsf {CT}}^-}\rangle $ and $\langle \Delta ^-, \leq _{{\textsf {CT}}^-}\rangle $ , we shall assume that $\Delta $ is replaced by the quotient set ${\Delta }/{ \sim } $ , where $\sim $ is the least equivalence relation that makes $\leq _{{\textsf {CT}}^-}$ antisymmetric, to wit:

$$ \begin{align*} \delta\sim\delta' \text{ iff } \delta\leq_{{\textsf{CT}}^-}\delta' \text{ and } \delta'\leq_{{\textsf{CT}}^-}\delta. \end{align*} $$

Proof We only present the proof for the case of $\Delta $ as it is $(1+\varepsilon )$ -times harder. For showing that both structures are distributive lattices, it is enough to show that given $\delta ,\delta ' \in \Delta $ , one can find elements $\delta \oplus \delta '$ and $\delta \otimes \delta '$ of $\Delta $ such that over ${\textsf {CT}}^-[{\textsf {PA}}]$ we have

(1) $$ \begin{align} T[\delta]\wedge T[\delta'] &\leftrightarrow T[\delta \oplus \delta'], \end{align} $$

(2) $$ \begin{align} T[\delta]\vee T[\delta'] &\leftrightarrow T[\delta\otimes \delta']. \end{align} $$

Indeed, this is because the Lindenbaum algebra of ${\textsf {CT}}^-$ is a distributive lattice. It can be readily seen that if we define

$$\begin{align*}\delta\oplus\delta'(x):= \delta(x)\vee \delta'(x),\end{align*}$$

then $\delta \oplus \delta ' \in \Delta $ and (1) is satisfied. For $(2)$ it is sufficient to define

$$\begin{align*}\delta\otimes\delta'(x):= \exists y,z<x\bigl(\delta(y)\wedge \delta'(z) \wedge x = y \vee z\bigr),\end{align*}$$

where $x=y\vee z$ expresses that x is a disjunction of y and z. To see that (2) holds and $\delta _{{\textsf {PA}}}\leq _{pt} \delta \otimes \delta '$ one simply applies reasoning by cases; the proof of $\delta \otimes \delta '\leq _{pt}\delta _{{\textsf {PA}}}$ is trivial.

Proof Fix $\mathcal {T}$ and $\delta $ as in the assumptions. Let $\delta '$ be the natural elementary definition of the set $S_{\tau }$ , where $\tau $ is the template defined as in the proof of Theorem 12. This works, since in the proof of Theorem 12, the verification of the fact that the deductive closure of ${\textsf {EA}}+ S_{\tau }$ and ${\textsf {EA}} + \Phi $ coincide formalizes smoothly in the subsystem $\mathsf {WKL^{*}_0}$ of second order arithmetic, which is well-known to be a conservative extension of ${\textsf {EA}}$ , as first shown in [Reference Simpson and Smith28]. More explicitly, the verification of ${\textsf {EA}}+ S_{\tau }\vdash \Phi $ requires only the existence of well-behaved partial truth predicates (that can be developed within ${\textsf {EA}}$ , as demonstrated, e.g., in [Reference Beklemishev1, Proposition 2.6]). On the other hand, the verification of ${\textsf {EA}} + \Phi \vdash S_{\tau }$ requires the completeness theorem of first order logic (which is readily available in $\mathsf {WKL^{*}_0}$ ) together with Tarski’s undefinability of truth theorem. Although Tarski’s theorem presupposes the consistency of $\Phi $ , this can be assumed, because if $\Phi $ is inconsistent, so is $S_{\tau }$ by the proof of the first implication, and in such a scenario the two theories clearly coincide. Hence $\delta '$ is indeed proof-theoretically reducible to $\delta $ . However, in this case is trivially equivalent to ${\textsf {CT}}^-[{\textsf {PA}}] + T[\tau ]$ , hence is a conservative extension of $\mathcal {T}$ , due to Theorem 8.

Proof As previously we do the case of a smaller structure, with $\mathsf {Sch}_{{\textsf {PA}}}$ as the universe. Arguing as previously in Proposition 19, it is enough to define $\oplus $ and $\otimes $ such that ${\textsf {CT}}^-[{\textsf {PA}}]$ proves the following for all $\tau ,~\tau ' \in \mathsf {Sch}_{{\textsf {PA}}}$ :

(3) $$ \begin{align} T[\tau]\wedge T[\tau']&\leftrightarrow T[\tau\oplus\tau'], \end{align} $$

(4) $$ \begin{align} T[\tau]\vee T[\tau']&\leftrightarrow T[\tau\otimes\tau']. \end{align} $$

The case of $\oplus $ is trivial. We put

$$\begin{align*}\tau\oplus \tau' := \tau\wedge \tau'.\end{align*}$$

The case of $\otimes $ is (a little bit) harder. We put

$$\begin{align*}\tau\otimes\tau' := \tau \vee (\tau'[Q/P]),\end{align*}$$

where Q is a fresh unary predicate. As remarked earlier (compare footnote 4) thanks to the coding apparatus, $\tau \otimes \tau '$ can be expressed as a scheme with a single unary predicate P. Then we obtain

$$\begin{align*}{\textsf{CT}}^-[{\textsf{EA}}]\vdash T[\tau\otimes\tau'] \equiv \forall \varphi~\forall \psi ~T\bigl(\tau[\varphi/P]\vee \tau'[\psi/Q]\bigr).\end{align*}$$

It is very easy now to check that $(4)$ is satisfied.

Proof Each of the four sets is readily seen to be definable by a $\Pi _{2}$ -formula, so it suffices to show that each is $\Pi _{2}$ -hard, i.e., the complete $\Pi _{2}$ -set $\mathsf {True}_{\Pi _{2}}^{\mathbb {N}}$ consisting of (Gödel numbers of) $\Pi _{2}$ -sentences that are true in the standard model $\mathbb {N}$ of ${\textsf {PA}}$ is many-one reducible (denoted $\leq _{\mathsf {m}} $ ) to each of them. Recall that $\leq _{\mathsf {m}} $ is defined among subsets of $\omega $ via:

$$ \begin{align*} A\leq_{\mathsf{m}} B \text{ iff there is a total recursive function } f \text{ such that: } \forall n\in \omega \ (n\in A\Leftrightarrow f(n)\in B). \end{align*} $$

The proof will be complete once we demonstrate the following four assertions:

1. (i) $\mathsf {True}_{\Pi _{2}}^{\mathbb {N}}\leq _{\mathsf {m}} \mathsf {Sch}_{{\textsf {PA}}}^{-}$ .Footnote ⁹

2. (ii) $\mathsf {Sch}_{\mathsf {PA}}^{-}\leq _{\mathsf {m}} \mathsf {Sch}_{\mathsf {PA}}.$

3. (iii) $\mathsf {Sch}_{\mathsf {PA}}^{-}\leq _{\mathsf {m}} \ \mathsf {Sch}_{ \mathsf {PA}}^{T}.$

4. (iv) $\mathsf {True}_{\Pi _{2}}^{\mathbb {N}}\leq _{\mathsf {m}} \mathsf {Cons}$ .

To prove $(i)$ , suppose $\pi =\forall x\ \exists y\ \delta (x,y)$ is a $\Pi _{2}$ -statement, where $\delta (x,y)$ is $\Delta _{0}$ . We first observe that by $\Sigma _{1}$ -completeness of ${\textsf {PA}}$ :

$(\ast ) \ \pi \in \mathrm {True}_{\Pi _{2}}^{\mathbb {N}}$ iff $ \forall n\in \omega \ \mathrm {PA}\vdash \exists y\ \delta (\underline {n},y)$ .

On the other hand, $R=\{\exists y\ \delta (\underline {n},y):n\in \omega \}$ is a recursive set of sentences, so by Theorem 12 there is $\tau $ such that $\tau \in \mathsf {Sch}_{\mathsf {PA}}^{-}$ iff $\mathsf {PA} \vdash R.$ To finish the proof, it remains to observe that the transition from $\pi $ to the $\Sigma _1$ -formula $\sigma $ that defines R in $\mathbb {N}$ is given by a recursive function g, therefore if f is the total recursive function as in Theorem 12 then we have

$$ \begin{align*} \pi \in \mathsf{True}_{\Pi _{2}}^{\mathbb{N}} \text{ iff } f(g(\pi ))\in \mathsf{Sch}_{\mathrm{PA}}^{-}. \end{align*} $$

The proof of $(ii)$ is based on the observation that $\tau \in \mathsf {Sch}^{-} _{\mathsf {PA}}$ iff $h(\tau )\in \mathsf {Sch}_{\mathsf {PA}}$ , where $h(\tau ):=\tau \wedge \tau _{\mathsf {PA}}$ , and $\tau _{{\textsf {PA}}}$ is defined as follows:

$$ \begin{align*} \tau _{{\textsf{PA}}}:={\textsf{Q}} \wedge [P(0) \wedge \forall x ( P(x) \rightarrow P(S(x))) \rightarrow \forall x P(x)]. \end{align*} $$

To verify $(iii)$ , we claim that $\tau \in \mathsf { Sch}_{\mathsf {PA}}^{-}$ iff $\left ( \tau \wedge \tau _{\mathsf {PA} }\right ) \in \mathsf {Sch}_{\mathsf {PA}}^{T}$ . The implication $ \tau \in \mathsf {Sch}_{\mathsf {PA}}^{-}\Rightarrow \left ( \tau \wedge \tau _{\mathsf {PA}}\right ) \in \mathsf {Sch}_{\mathrm {PA}}^{T}$ follows directly from Theorem 3 (since ${\textsf {PA}}$ proves $S _{\tau \wedge \tau _{{\textsf {PA}}}}$ if $\tau \in \mathrm {Sch}_{\mathsf {PA}}^{-}).$ On the other hand, if $\left ( \tau \wedge \tau _{{\textsf {PA}}}\right ) \in \mathsf {Sch}_{{\textsf {PA}}}^{T}$ , then by the definition of $\mathsf { Sch}_{\mathsf {PA}}^{T}$ , ${\textsf {PA}}$ proves $S _{\tau }$ , so $\tau \in \mathsf {Sch}_{\mathsf {PA}}^{-}$ .

Finally, to establish $(iv)$ suppose $\pi =\forall x\ \exists y\ \delta (x,y)$ is a $\Pi _{2}$ -statement, where $ \delta (x,y)$ is $\Delta _{0}$ . In the proof of part (i) we showed that there are recursive functions f and g such that

$$ \begin{align*}\pi \in \mathsf{True}_{\Pi _{2}}^{\mathbb{N}} \Longleftrightarrow f(g(\pi ))\in \mathsf{Sch}_{\mathsf{PA}}^{-}.\end{align*} $$

Let h be the function that takes a template $\tau $ as input, and outputs the sentence $T[\tau ]\in \mathcal {L}_{T}$ expressing “T is $\tau $ -correct.” Clearly h is a recursive function. Also, it is evident that $\tau \in \mathsf {Sch}_{\mathsf {PA}}^{-}$ iff $T[\tau ]\in \mathsf {Cons}$ (the direction $\Rightarrow $ follows from Theorem 8; and the direction $\Leftarrow $ follows from the relevant definitions). Therefore,

$$ \begin{align*} \pi \in \mathsf{True}_{\Pi _{2}}^{\mathbb{N}} \Longleftrightarrow h\left( f(g(\pi ))\right) \in \mathsf{Cons}.\\[-36pt] \end{align*} $$

Proof By Theorem 6, $\mathsf {CT}_{0}$ can be axiomatized by $\mathsf {CT}^{-}[{\textsf {EA}}]+\mathsf {GRP}({\textsf {PA}})$ . This makes it clear that $\varphi _s$ is provable in $\mathsf {CT}_{0}.$ For the other direction, working in ${\textsf {CT}}^-[{\textsf {EA}}] + \varphi _s$ , suppose $\psi $ is ${\textsf {PA}}$ -provable. Then the scheme given by $\forall x(\psi \vee P(x))$ is PA-provable, so the instance of this scheme in which P is replaced with $x\neq x$ is true, but since $T(\forall x (x=x))$ , we have $T(\psi )$ . Thus, since $\psi $ was arbitrary, ${\textsf {CT}}^-[{\textsf {EA}}]+\varphi _s\vdash \mathsf {GRP}({\textsf {PA}}).$

Convention 29. Working in ${\textsf {CT}}^-[{\textsf {EA}}]$ and having fixed an (possibly nonstandard) arithmetical formula with one free variable $\theta (v)$ , $T{*}\theta (x)$ will abbreviate the formula $T(\theta [\underline {x}/v])$ . Hence $T{*}\theta (x)$ says that x satisfies $\theta $ . This notation was first introduced in [Reference Łełyk and Wcisło19] and is very successful in decreasing the number of brackets and improving readability.

Proof We prove the contrapositive. Fix $\psi $ which is not interpretable in ${\textsf {CT}}^-[{\textsf {PA}}]$ . We modify the Pakhomov–Visser diagonalization from [Reference Pakhomov and Visser22, Theorem 4.1]. Observe that for two finite theories $\alpha $ , $\beta $ , the condition “ $\alpha $ interprets $\beta $ ” is $\Sigma _1$ . Let $\alpha \triangleright \beta $ denote the formalization of this relation. Consider a $\Sigma _1$ -sentence $\varphi = \exists x \varphi '(x)$ , where $\varphi '(x)\in \Delta _0$ such that the following equivalence is provable in ${\textsf {CT}}^-[{\textsf {PA}}]$ :

Similarly to the Pakhomov–Visser argument, we argue that $\varphi $ is false. Suppose not and take the least $n\in \omega $ such that $\varphi '(n)$ holds. Then, in ${\textsf {Q}}$ , $\forall z \leq x\neg \varphi '(z)$ is equivalent to $x< \underline {n}$ , hence the following is provable in ${\textsf {CT}}^-[{\textsf {PA}}]$ :

$$ \begin{align*}\forall \theta(x) \bigl( T\bigl(V_{(\forall z\leq y\neg\varphi'(z),\delta_{{\textsf{PA}}})}[\theta]\bigr)\leftrightarrow T\bigl(V_{(y< \underline{n}, \delta_{{\textsf{PA}}})}[\theta]\bigr)\bigr).\end{align*} $$

We claim that

(*) $$ \begin{align} {\textsf{CT}}^-[{\textsf{PA}}] \vdash \forall \theta(x)~T\bigl(V_{(y< \underline{n}, \delta_{{\textsf{PA}}})}[\theta]\bigr). \end{align} $$

Indeed, working in ${\textsf {CT}}^-[{\textsf {PA}}]$ fix $\theta \in {\textsf {Form}}^{\leq 1}_{\mathcal {L}_{{\textsf {PA}}}}$ . By compositional conditions $T\bigl (V_{(y< \underline {n}, \delta _{{\textsf {PA}}})}[\theta ]\bigr )$ is equivalent to

$$\begin{align*}\bigwedge_{i< n} \big[\bigl(T{*}\text{True}(\underline{i},\theta)\bigr) \rightarrow \forall x \bigl((\delta_{{\textsf{PA}}}(x) \wedge \textsf{pdepth}(x)\leq \underline{i})\rightarrow T{*}\theta(x)\bigr)\big].\end{align*}$$

However, once again by compositional conditions imposed on T, $T{*}\text {True}(\underline {i}, \theta )$ is equivalent to $\text {True}(\underline {i}, T{*}\theta (x))$ , hence to the assertion that $T{*}\theta (x)$ is a compositional truth predicate for formulae of depth at most i. Assuming that this is the case, since i is standard, every induction axiom of pure depth at most i is true in the sense of $T{*}\theta (x)$ . This concludes our proof of (*).

Now, since $\varphi $ is true, it follows that

$$\begin{align*}{\textsf{CT}}^-[{\textsf{PA}}] + \forall \theta(x)~ T\bigl(V_{(y<\underline{n},\delta_{{\textsf{PA}}})}[\theta]\bigr)\text{ interprets } \psi.\end{align*}$$

However, by the above argument it would mean that ${\textsf {CT}}^-[{\textsf {PA}}]$ interprets $\psi $ , contrary to the assumption.

Since $\varphi $ is false, $V_{(\forall z \leq y \neg \varphi '(z),\delta _{{\textsf {PA}}})}[P]$ is a scheme template, such that the scheme associated with it axiomatizes ${\textsf {PA}}$ . Moreover, ${\textsf {CT}}^-[{\textsf {PA}}] + T\bigl [V_{(\forall z\leq y\neg \varphi '(z),\delta _{{\textsf {PA}}})}\bigr ]$ does not interpret $\psi $ .

Proof Fix $\tau $ and apply Corollary 31 to . This is legal, since the latter theory is a finitely axiomatizable extension of ${\textsf {PA}}$ , hence it is not interpretable in ${\textsf {PA}}$ .Footnote ¹⁰ So there is a scheme $\tau "\in \mathsf {Sch}_{{\textsf {PA}}}$ such that does not relatively interpret . Now it is sufficient to take $\tau ':= \tau \otimes \tau "$ , as in the proof of Proposition 23.

Proof Fix arbitrary arithmetical formulae $\psi $ and $\varphi $ with exactly one free variable. Let $\delta _{\psi }$ and $\delta _{\varphi }$ be defined in the bullet point above the current proposition. Without loss of generality, assume that the variable x occurs in $\psi $ . Working in ${\textsf {CT}}^-[{\textsf {PA}}]$ assume that $\forall x \bigl (\varphi (x)\rightarrow \psi (x)\bigr )$ and $T[V_{\delta _{\varphi }}]$ hold. We argue that $T[V_{\delta _{\psi }}]$ holds as well. Fix arbitrary a, $\theta $ , b such that $\mathsf {True}(a,T{*}\theta )$ and $\textsf {pdepth}(\psi (\underline {b}))\leq a$ . It follows that for some standard n, $\textsf {pdepth}(\varphi (\underline {b}))\leq a+n$ , hence there exists a formula $\theta '(x)$ such that

$$\begin{align*}\mathsf{True}(a+n, T{*}\theta').\end{align*}$$

By $T[V_{\delta _{\varphi }}]$ we conclude $T{*}\theta '(\varphi (\underline {b})).$ However, since $\varphi (x)$ is of standard depth, it follows that $\varphi (b)$ holds. Hence $\psi (b)$ holds as well. Since $\psi (b)$ is also of standard depth, we conclude that $T{*}\theta (\psi (\underline {b}))$ , which ends the proof.

Proof We start by showing that provably in ${\textsf {CT}}_0$ all arithmetical partial truth predicates are coextensive, i.e., the following is provable in ${\textsf {CT}}_0$ :

(*) $$ \begin{align} \forall x ~\forall \theta\in{\textsf{Form}}^{\leq 1}_{\mathcal{L}_{{\textsf{PA}}}} \forall \varphi \in \mathsf{Sent}_{\mathcal{L}_{{\textsf{PA}}}} \bigl[(\mathsf{True}(x,T{*}\theta) \land \mathsf{depth}({\varphi})\leq x) \rightarrow \bigl(T{*}\theta(\varphi)\leftrightarrow T(\varphi)\bigr)\bigr]. \end{align} $$

Fix an arbitrary $(\mathcal {M},T)\models {\textsf {CT}}_0$ . For an arbitrary $c\in M$ , let $T_c$ denote the ( $(\mathcal {M},T)$ -definable) restriction of T to all sentences of depth at most c. Then $(\mathcal {M},T_c)\models \mathsf {True}(c,T)$ . However, as proved in [Reference Łełyk and Wcisło20, Fact 32], $(\mathcal {M},T_c)$ satisfies full induction scheme for $\mathcal {L}_T$ . Hence $T_c$ is a fully inductive truth predicate for formulae of depth at most c. Using this we argue that (*) holds in $(\mathcal {M},T)$ . Working in the model, fix an arbitrary a and an arbitrary $\theta \in {\textsf {Form}}^{\leq 1}_{\mathcal {L}_{{\textsf {PA}}}}$ . Assume that the depth of $\theta $ is b and let $c = \max \{a,b\}$ . Assume $\mathsf {True}(a,T{*}\theta )$ , i.e. the formula $T{*}\theta $ is a partial truth predicate for formulae of depth $\leq a$ . Since for every formula $\varphi $ of depth at most c, $T_c(\varphi )$ is equivalent to $T(\varphi )$ , we conclude that $\mathsf {True}(a, T_c{*}\theta )$ holds. Moreover, it is sufficient to show that

$$\begin{align*}\forall x\bigl(T_c{*}\theta(x)\leftrightarrow T_c(x)\bigr).\end{align*}$$

In other words, it is sufficient to prove that

$$\begin{align*}(\mathcal{M},T_c)\models\forall x\bigl(T{*}\theta(x)\leftrightarrow T(x)\bigr).\end{align*}$$

The above can be demonstrated by a routine induction on the build-up of formulae. More precisely, let

$$\begin{align*}\Xi(y):= \forall \varphi\in{\textsf{depth}}(y)\bigl(T{*}\theta(\varphi)\leftrightarrow T(\varphi)\bigr).\end{align*}$$

Then $\Xi (0)$ and $\forall x<a \bigl (\Xi (x)\rightarrow \Xi (x+1)\bigr )$ hold (in $(\mathcal {M}, T_c)$ ), because both $T_c{*}\theta $ and $T_c$ are partial truth predicates for formulae of depth at most a. Since $\Xi (y)$ is a formula of $\mathcal {L}_T$ , in $(\mathcal {M},T_c)$ we have an induction axiom for it, and we can conclude

$$\begin{align*}(\mathcal{M},T_c)\models \forall y \leq a ~ \Xi(y).\end{align*}$$

This completes the proof of (*).

We show that over ${\textsf {CT}}^-[{\textsf {PA}}]$ , $T[\delta ]$ implies $T[V_{\delta }]$ . We fix an arbitrary $\delta $ and working in ${\textsf {CT}}_0$ assume that $\forall x \bigl (\delta (x)\rightarrow T(x)\bigr )$ . We show that T is $V_{\delta }$ -correct, i.e., for every arithmetical formula $\theta $ (possibly nonstandard) $T(V_{\delta }[\theta ])$ holds. By the compositional conditions, $T(V_{\delta }[\theta ])$ is equivalent to

$$\begin{align*}\forall x \left(\mathsf{True}(x,T{*}\theta)\rightarrow \left[\forall y\bigl(\delta(y)\wedge \textsf{pdepth}(y)\leq x)\rightarrow T{*}\theta(y)\bigr)\right]\right).\end{align*}$$

Fix x, assume $\mathsf {True}(x, T{*}\theta )$ and fix an arbitrary y such that $\textsf {pdepth}(y)\leq x$ and $\delta (y)$ . By $\delta $ -correctness $T(y)$ holds, hence y is a formula and since $\textsf {pdepth}(y)\leq x$ , y is a formula of depth at most x. Then, by the previous claim (*) we know that for every $\varphi $ whose depth is at most x, $T{*}\theta (\varphi )$ is equivalent to $T(\varphi )$ . Hence $T{*}\theta (y)$ holds as well.

For the converse direction, we assume T is $V_{\delta }$ -correct. Fix an arbitrary x and assume that $\delta (x)$ holds. In particular x is a formula. Let y be the depth of x and let $\theta $ be any arithmetical truth predicate such that ${\textsf {Pr}}_{{\textsf {PA}}}(\mathsf {True}(\underline {y},\theta ))$ holds. By the Global Reflection in ${\textsf {CT}}_0$ , $T(\mathsf {True}(\underline {y},\theta ))$ holds as well, and this in turn implies, by compositional conditions, $\mathsf {True}(y, T{*}\theta )$ . Consequently, by $V_{\delta }$ -correctness, $T{*}\theta (x)$ holds. Finally, it follows that $T(x)$ holds by our claim (*). This concludes the proof of $\delta $ -correctness and the whole proof.

Proof Suppose that $\preceq $ satisfies the assumptions. Firstly, we build a family of consistent theories $\{\sigma _n\}_{n\in \omega }$ such that the following hold for all $m,n \in \omega $ :

1. 1. If $m\preceq n$ , then ${\textsf {PA}}\vdash \sigma _n\subseteq \sigma _m$ .

2. 2. ${\textsf {CT}}_0\nvdash {\textsf {Con}}_{\sigma _m}$ .

3. 3. If $m\npreceq n$ , then ${\textsf {CT}}_0+{\textsf {Con}}_{\sigma _m}\nvdash {\textsf {Con}}_{\sigma _n}$ .

As shown in [Reference Lindström21, Section 2.3, Theorem 11], there is a $\Pi _1$ -formula $\pi (x)$ that is flexible over $\mathsf {REF}^{<\omega }({\textsf {PA}})$ , i.e., for every $\Pi _1$ -formula $\theta (x)$ , the following theory is consistent:

$$\begin{align*}\mathsf{REF}^{<\omega}({\textsf{PA}}) + \forall x\bigl(\pi(x)\leftrightarrow \theta(x)\bigr).\end{align*}$$

For each $n\in \omega $ let $\sigma _n$ be the natural $\Sigma _1$ -definition of the following set of sentencesFootnote ¹² :

$$\begin{align*}{\textsf{PA}} + \left\{\pi(\underline{k}) \ \ | \ \ n\preceq k \right\}. \end{align*}$$

Now, condition (1) easily follows from the ( ${\textsf {PA}}$ -provable) transitivity of $\preceq $ . Condition (2) easily reduces to Condition (3), so let us now show the latter. Aiming at a contradiction assume $m\npreceq n$ and ${\textsf {CT}}_0\vdash {\textsf {Con}}_{\sigma _m}\rightarrow {\textsf {Con}}_{\sigma _n}$ . Let $\theta (x):= m\preceq x$ . By flexibility there exists model $\mathcal {M}$ such that

$$\begin{align*}\mathcal{M}\models \mathsf{REF}^{<\omega}({\textsf{PA}}) + \forall x\bigl(\pi(x)\leftrightarrow \theta(x)\bigr).\end{align*}$$

By the choice of $\mathcal {M}$ it follows that $\mathcal {M}\models \neg \pi (n)$ . As a consequence, by provable $\Sigma _1$ -completeness of ${\textsf {PA}}$ , $\mathcal {M}\models {\textsf {Pr}}_{{\textsf {PA}}}(\neg \pi (\underline {n})),$ and $\mathcal {M}\models \neg {\textsf {Con}}_{\sigma _{n}}$ . However, since $\mathcal {M}\models \mathsf {REF}({\textsf {PA}})$ , as viewed in $\mathcal {M}$ , ${\textsf {PA}}$ is consistent with $\Pi _1$ -truth (of $\mathcal {M}$ ). Consequently, since $\mathcal {M}\models \forall x\bigl ( m\preceq x \rightarrow \pi (x)\bigr )$ , it follows that $\mathcal {M}\models {\textsf {Con}}_{\sigma _{m}}$ . Hence $\mathcal {M}\models {\textsf {Con}}_{\sigma _n}$ as well, which contradicts our previous conclusions.

We are ready to construct the promised embedding. Fix the family $\{\sigma _n\}_{n\in \omega }$ as above and for each $m\in \omega $ choose $\delta _m \in \Delta ^*$ to be the natural elementary definition of the following set of sentences:

$$\begin{align*}{\textsf{PA}} + \left\{{\textsf{Con}}_{\sigma_m}(\underline{n}) \ \ | \ \ n\in\omega \right\}, \end{align*}$$

where ${\textsf {Con}}_{\sigma _m}(\underline {n})$ asserts that there is no proof of contradiction of $\sigma _m$ with Gödel code $\leq n$ . Since for every $m \in \omega $ , $\sigma _m$ is consistent, $\delta _m$ is really an axiomatization of ${\textsf {PA}}$ , hence $V_{\delta _m}\in \mathsf {Sch}_{{\textsf {PA}}}$ . We check that the map

$$ \begin{align*}m\mapsto V_{\delta_m}\end{align*} $$

is an embedding of $\langle \omega , \preceq \rangle $ into $\langle \mathsf {Sch}_{{\textsf {PA}}}, \leq _{{\textsf {CT}}^-}\rangle .$ Fix $m,n\in \omega $ and assume $m\preceq n$ . Then clearly ${\textsf {PA}}\vdash \forall x \bigl ({\textsf {Con}}_{\sigma _m}(x)\rightarrow {\textsf {Con}}_{\sigma _n}(x)\bigr )$ . Consequently, applying Proposition 35 to $\varphi (x):= {\textsf {Con}}_{\sigma _m}(x)$ and $\psi (x):= {\textsf {Con}}_{\sigma _n}(x)$ , we obtain

$$ \begin{align*}{\textsf{CT}}^-[{\textsf{PA}}]\vdash T[V_{\delta_m}]\rightarrow T[V_{\delta_n}].\end{align*} $$

Suppose now $m\npreceq n$ and aiming at a contradiction, assume that ${\textsf {CT}}^-[{\textsf {PA}}]\vdash T[V_{\delta _m}]\rightarrow T[V_{\delta _n}]$ . Then, by Corollary 37, ${\textsf {CT}}_0\vdash T[\delta _m]\rightarrow T[\delta _n]$ . However, since ${\textsf {CT}}_0 \vdash T[\delta _{{\textsf {PA}}}]$ , ${\textsf {CT}}_0\vdash T[\delta _i]\leftrightarrow {\textsf {Con}}_{\sigma _i}$ for every $i\in \omega $ . Hence ${\textsf {CT}}_0\vdash {\textsf {Con}}_{\sigma _m}\rightarrow {\textsf {Con}}_{\sigma _n}$ , which is impossible by our previous considerations, since $m\npreceq n$ .

Proof This follows since $\langle \mathsf {Sch}_{{\textsf {PA}}}, \leq _{{\textsf {CT}}^-}\rangle $ can be easily embedded into each of the above partial orders.

Proof This follows immediately from Theorem 6 that ${\textsf {CT}}_0\vdash \forall \varphi \bigl ({\textsf {Pr}}_{{\textsf {PA}}}(\varphi )\rightarrow T(\varphi )\bigr )$ .

Proof Let $\varphi '(x)$ formalize an elementary axiomatization of $\mathcal {T}$ (which exists by Craig’s trick). Define

$$\begin{align*}\varphi(x):= \exists \psi<x \bigl(\varphi'(\psi) \wedge x = \bigvee \psi\bigr).\end{align*}$$

That is to say that x satisfies $\varphi $ if it is a pleonastic disjunction of a formula from an elementary axiomatization of $\mathcal {T}$ . Observe first that

. Indeed, it is sufficient to show that for every sentence $\psi $ we have

Observe that, over ${\textsf {EA}}$ , $\varphi '(\psi )$ implies $\varphi (\bigvee \psi )$ , which in turn, over

implies $T(\bigvee \psi )$ . However, over pure ${\textsf {CT}}^-[{\textsf {EA}}]$ the last sentence implies $T(\psi )$ by compositional conditions, since $\bigvee \psi $ is a disjunction of length $\ulcorner \psi \urcorner $ and hence is standard.

We show conservativity: pick any model $\mathcal {M}\models \mathcal {T}$ . By Theorem 45 there is a $(\mathcal {N},T)\models {\textsf {CT}}^-[{\textsf {EA}}]$ such that $\mathcal {N}$ is an elementary extension of $\mathcal {M}$ and every disjunction of nonstandard length is made true by T. It follows that . Indeed, firstly observe that if $\mathcal {N}\models \varphi (a)$ then there exists $\psi $ such that $\mathcal {N}\models \varphi '(\psi ) \wedge a = \bigvee \psi $ . Now the argument splits into two cases:

1. 1. $\psi $ is a standard sentence. In this case $\bigvee \psi $ is standard and $\mathcal {N}\models \bigvee \psi $ , by elementarity. Consequently $(\mathcal {N}, T)\models T(\bigvee \psi )$ by compositional clauses; or

2. 2. $\psi $ is not a standard sentence. In this case $\bigvee \psi $ is a disjunction of nonstandard length, hence is made true in $(\mathcal {N}, T)$ .

Proof of Theorem 42

Fix $\mathcal {T}$ such that

$$\begin{align*}{\textsf{CT}}_0\vdash \mathcal{T}.\end{align*}$$

Let $\varrho $ be an arbitrary elementary axiomatization of $\mathcal {T}$ . Let $\sigma _{\varphi }[\theta ]$ denote the map from Definition 44. We now observe that the proof of the reflexive property of ${\textsf {PA}}$ is formalizable in I $\Sigma _1$ . This follows from two well-known facts: firstly, the cut-elimination theorem formalizes in I $\Sigma _1$ (thus provably in I $\Sigma _1$ , every provable sentence has a proof that has the subformula property) and secondly, I $\Sigma _1$ is enough for the formalization of the proof of existence of partial truth predicates in ${\textsf {PA}}$ . As a consequence, we obtain

$$\begin{align*}\text{I}\Sigma_1\vdash \forall \theta ~ {\textsf{Pr}}_{{\textsf{PA}}}(\neg ({\textsf{Pr}}_{\emptyset}(\ulcorner \theta \urcorner)\wedge \neg\theta)).\end{align*}$$

Hence $\delta (x)\in \Delta $ , where $\delta $ is defined as follows:

$$\begin{align*}\delta(x) := \delta_{{\textsf{PA}}}(x)\vee \exists \theta, \varphi<x\bigl( \varrho(\varphi) \wedge x = \sigma_{\bigvee \varphi}[\theta]\bigr).\end{align*}$$

Observe that $\delta $ naturally defines the following set of sentences:

$$\begin{align*}\mathsf{Q} \cup \left\{{\textsf{Ind}}(\varphi) \ \ | \ \ \varphi\in\mathcal{L}_{{\textsf{PA}}} \right\}\cup \left\{({\textsf{Pr}}_{\emptyset}(\ulcorner \theta \urcorner)\wedge\neg\theta) \rightarrow \bigvee \varphi \ \ | \ \ \theta\in\mathcal{L}_{{\textsf{PA}}}, \varphi \in \mathcal{T} \right\}.\end{align*}$$

We argue first that

is conservative over $\mathcal {T}$ . To see this, fix an arbitrary model $\mathcal {M}\models \mathcal {T}$ and let

be a model from Theorem 45. Then

since, reasoning by cases as in the proof of Proposition 47, for every $\varphi \in N$ such that $\mathcal {N}\models \varrho (\varphi )$ we have

$$\begin{align*}(\mathcal{N},T)\models T\left(\bigvee \varphi\right).\end{align*}$$

Now, we argue that

. Let $\varphi $ be an arbitrary $\varrho $ –axiom of $\mathcal {T}$ . We claim

To see why the last claim holds, reason in

. We have

$$\begin{align*}\forall \theta~ T(\sigma_{\bigvee\varphi}[\theta]).\end{align*}$$

By the axioms of ${\textsf {CT}}^-$ the above is equivalent to

(*) $$ \begin{align} \left(\exists \theta \bigl({\textsf{Pr}}_{\emptyset}(\ulcorner \theta \urcorner)\wedge \neg T(\theta)\bigr)\right)\rightarrow T\left(\bigvee\varphi\right). \end{align} $$

Now we reason by cases: either $\forall \theta ~ \bigl ({\textsf {Pr}}_{\emptyset }(\ulcorner \theta \urcorner )\rightarrow T(\theta )\bigr )$ or not. If the latter holds, we have $T(\bigvee \varphi )$ by Modus Ponens applied to (*). Hence $\varphi $ holds by compositional conditions, because $\bigvee \varphi $ is a disjunction of standard length and $\varphi $ is a standard sentence. If the former holds, we have ${\textsf {CT}}_0$ by Theorem 6 and $\varphi $ holds, because we assumed that ${\textsf {CT}}_0\vdash \mathcal {T}$ .

Proof Clearly $\Delta ^*$ is $\Pi _2$ -definable. Consider the map f that takes a scheme template $\tau $ as input and outputs the formula $\delta _{\tau }(x)$ that expresses “x is an instance of $\tau $ .” f is clearly recursive (indeed elementary) and satisfies

$$ \begin{align*} \tau \in \mathsf{Sch}_{{\textsf{PA}}} \text{ iff } \delta_{\tau} \in \Delta^*. \end{align*} $$

Therefore $\mathsf {Sch}_{{\textsf {PA}}}$ is many-one reducible to $\Delta ^*$ , which in light of the $\Pi _2$ -completeness of $\mathsf {Sch}_{{\textsf {PA}}}$ (established in Theorem 26), completes the verification of $\Pi _2$ -completeness of $\Delta ^*$ .

Proof Both sets are clearly r.e., because both definitions require just the provability of a particular sentence in I $\Sigma _1$ . To verify completeness we sketch the reduction f of the set of true $\Sigma _1$ -sentences to $\Delta $ (an analogous reduction works for $\Delta ^-$ ). Given a $\Sigma _1$ -sentence $\varphi $ we define a formula $f(\varphi ):= \delta _{{\textsf {PA}}}(x)\vee x = \varphi $ . It follows that

$$\begin{align*}\mathbb{N}\models \varphi \iff f(\varphi)\sim_{pt} \delta_{{\textsf{PA}}}.\end{align*}$$

The left-to-right direction follows by I $\Sigma _1$ -provable $\Sigma _1$ -completeness of ${\textsf {PA}}$ . The right-to-left direction follows by the soundness of I $\Sigma _1$ and ${\textsf {PA}}$ .

Proof of Theorem 50

Fix a $\Pi _2$ -sentence $\pi := \forall x \varphi (x)$ , where $\varphi (x)$ is $\Sigma _1$ . Let $\delta ^{\pi }$ be the formula in $\Delta $ that describes the union of (the canonical axiomatization of) ${\textsf {PA}}$ with the following set of sentences:

$$\begin{align*}\left\{{\textsf{Pr}}_{\emptyset}(\ulcorner \chi \urcorner)\wedge \neg \chi \rightarrow \bigvee \varphi(\underline{n}) \ \ | \ \ \chi\in\mathcal{L}_{{\textsf{PA}}}, n\in \omega \right\}.\end{align*}$$

The function $\pi \mapsto \delta ^{\pi }$ is clearly recursive, and $\delta ^{\pi } \in \Delta $ . Let $\theta (x):= \Pi _2\text{-}\text {REF}({\textsf {PA}})\vee \varphi (x)$ and observe that for every n, . Indeed, work in and assume $\neg \Pi _2\text{-}\text {REF}({\textsf {PA}})$ . Then clearly $\neg {\textsf {CT}}_0$ and consequently, as in the proof of Theorem 42 we get $T\left (\bigvee \varphi (\underline {n})\right )$ . Finally, the latter implies $\varphi (\underline {n})$ , since it is a standard sentence.

Let $\mathsf {True}^{\mathbb {N}}_{\Pi _2}$ be the set of $\Pi _2$ -statements that are true in $\mathbb {N}$ . We will prove

Assume first that $\pi \in \mathsf {True}^{\mathbb {N}}_{\Pi _2}$ and $\pi = \forall x ~\varphi (x)$ , for some $\varphi (x)\in \Sigma _1$ . In particular $\varphi (\underline {n})$ is a true $\Sigma _1$ sentence for every $n \in \omega $ , hence,

$$ \begin{align*} {\textsf{PA}}\vdash \varphi(\underline{n}) \text{ for every } n\in \omega. \end{align*} $$

As usual, fix any model $\mathcal {M}\models {\textsf {PA}}$ and take its elementary extension

in which every disjunction of nonstandard length is true. As previously, it follows that

.

Conversely, assume that is conservative over ${\textsf {PA}}$ . Then for every $n \in \omega $ , ${\textsf {PA}}\vdash \theta (\underline {n})$ . In particular, for every $n\in \omega $ , ${\textsf {PA}}+\neg \Pi _2\text{-}\text {REF}({\textsf {PA}})\vdash \varphi (\underline {n})$ . By the soundness of this theory we conclude that $\pi $ is true.

Observation 52. ${{\textsf {CT}}_0}/{{\textsf {PA}}}$ with the operations inherited from the Lindenbaum algebra of ${\textsf {PA}}$ is a lattice with a greatest but not a least element (obviously assuming the consistency of ${\textsf {CT}}_0$ ). The lack of the least element follows from the fact that the arithmetical consequences of ${\textsf {CT}}_0$ is a theory in $\mathcal {L}_{{\textsf {PA}}}$ which extends ${\textsf {PA}}$ . In particular this theory is not finitely axiomatizable by essential reflexivity of ${\textsf {PA}}$ . Moreover the greatest element has no immediate predecessors. This follows by the classical fact that the Lindenbaum Algebra of ${\textsf {PA}}$ is dense (see [Reference Shavrukov and Visser27]) for a proof.

Proof To each $[\varphi ]_{{\textsf {PA}}}$ we assign $\delta ^{\varphi } \in \Delta $ as in the proof of Theorem 42. $\varrho (x)$ is now simply $x = \underline {\ulcorner \varphi \urcorner }$ . Hence by compositional axioms, and the fact that $\varphi $ is standard we have

$$\begin{align*}{\textsf{CT}}^-[{\textsf{EA}}]\vdash \forall \theta \bigl(T(\sigma_{\bigvee \varphi}[\theta])\leftrightarrow T(\sigma_{\varphi}[\theta])\bigr).\end{align*}$$

Consequently, $\delta ^{\varphi }$ can be taken to axiomatize the (natural definition of the) following set of sentences:

$$\begin{align*}{\textsf{PA}} \cup \left\{({\textsf{Pr}}_{\emptyset}(\ulcorner \theta \urcorner)\wedge \neg\theta) \rightarrow \varphi \ \ | \ \ \theta\in\mathcal{L}_{{\textsf{PA}}} \right\}.\end{align*}$$

We claim that for an arbitrary $\varphi \in \mathcal {L}_{{\textsf {PA}}}$ , over , is equivalent to $\varphi $ . Working in assume first that $\varphi $ holds. Then for every $\theta $ we have

$$\begin{align*}T(\sigma_{\varphi}[\theta]),\end{align*}$$

since $T(\sigma _{\varphi }[\theta ])$ is equivalent to an implication with a true conclusion. Hence every sentence satisfying $\delta ^{\varphi }$ is true. For the converse implication, working over , assume . We argue by cases:

• • If ${\textsf {CT}}_0$ holds, then $\varphi $ holds, by assumption.

• • If ${\textsf {CT}}_0$ fails, then, as in the proof of Theorem 42, $\varphi $ holds.

We show that the mapping $\varphi \mapsto \delta ^{\varphi }$ is a lattice embedding. Firstly, we show that the mapping preserves the partial ordering. To this end, we prove that the following are equivalent for arbitrary arithmetical formulae $\varphi ,\psi $ that are provable in ${\textsf {CT}}_0$ :

• • ${\textsf {PA}}\vdash \varphi \rightarrow \psi .$

• • ${\textsf {CT}}^-[{\textsf {PA}}]\vdash T[\delta ^{\varphi }]\rightarrow T[\delta ^{\psi }].$

Indeed the top-to-bottom direction follows easily, since for an arbitrary $\varphi \in \mathcal {L}_{{\textsf {PA}}}$ , is equivalent to $\varphi $ and proves . The bottom-to-up direction uses the same observations plus additionally the conservativity of ${\textsf {CT}}^-[{\textsf {PA}}]$ over ${\textsf {PA}}$ . Finally, we show that the mapping preserves infima and suprema. It is enough to observe that for $\varphi $ and $\psi $ as above, is equivalent to and the same with $\wedge .$ This concludes the proof.

Proof The embedding $\iota $ is defined as in the proof of the previous proposition with the only exception that we do not add ${\textsf {PA}}$ to $\delta ^{\psi }$ . More concretely, if $[\varphi ]_{{\textsf {PA}}}\in {{\textsf {CT}}_0}/{{\textsf {PA}}}$ , then we put $\iota ([\varphi ]_{{\textsf {PA}}})$ to be the natural elementary definition of the following set of sentences:

$$\begin{align*}\left\{({\textsf{Pr}}_{\emptyset}(\ulcorner \theta \urcorner)\wedge \neg\theta)\rightarrow \varphi \ \ | \ \ \theta\in\mathcal{L}_{{\textsf{PA}}} \right\}.\end{align*}$$

Denote the canonical elementary definition of this set with $\delta ^{\varphi }$ . As in the proof of the previous proposition, we obtain that for every $[\varphi ]_{{\textsf {PA}}}\in {{\textsf {CT}}_0}/{{\textsf {PA}}}$ , provably in ${\textsf {CT}}^-[{\textsf {PA}}]$ , $\varphi $ is equivalent to $T[\delta ^{\varphi }]$ . Consequently, $\iota $ is a lattice embedding. Now we claim that $\iota $ is cofinal with the Tarski Boundary in the sense explained. Pick any $\alpha \in \mathcal {L}_{T}$ such that ${\textsf {CT}}^-[{\textsf {PA}}] + \alpha $ is non-conservative over ${\textsf {PA}}$ (but consistent). By definition, ${\textsf {CT}}^-[{\textsf {PA}}]+\alpha \vdash \varphi $ for some ${\textsf {PA}}$ - unprovable sentence $\varphi \in \mathcal {L}_{{\textsf {PA}}}$ . Then, since the Lindenbaum algebra of ${\textsf {PA}}$ is atomless there is a sentence $\psi \in \mathcal {L}_{{\textsf {PA}}}$ , which is logically strictly weaker than $\varphi $ . Then there is a sentence $\theta $ such that $[\theta ]_{{\textsf {PA}}}\in {{\textsf {CT}}_0}/{{\textsf {PA}}}$ and $\psi \vee \theta $ is unprovable in ${\textsf {PA}}$ . This holds, since it is known that over ${\textsf {PA}}$ , $\mathsf {REF}({\textsf {PA}})$ (which is a consequence of ${\textsf {CT}}_0$ ) does not follow from any finite, consistent, set of sentences. Hence $[\psi \vee \theta ]_{{\textsf {PA}}}\in {{\textsf {CT}}_0}/{{\textsf {PA}}}$ is not the greatest element. Consequently, $T[\iota (\psi \vee \theta )] = T[\delta ^{\psi \vee \theta }]$ is below the Tarski Boundary. However, since $\psi $ does not prove $\varphi $ (over ${\textsf {PA}}$ ), a fortiori $\psi \vee \theta $ does not prove $\varphi $ . Hence does not prove ${\textsf {CT}}^-[{\textsf {PA}}]+\alpha $ . Additionally, , since $\psi \vee \theta $ follows from $\alpha $ .

Proof We shall make use of a $\Pi _1$ -formula that is ${\textsf {PA}}$ -independent, i.e., for every binary sequence s of length $n\in \omega $ the following sentence is unprovable in ${\textsf {PA}}$ :

$$\begin{align*}\bigl(\pi(\underline{0})^{s(0)}\wedge \pi(\underline{1})^{s(1)} \wedge \ldots \wedge \pi(\underline{n-1})^{s(n-1)}\bigr),\end{align*}$$

where for any formula $\varphi $ , $\varphi ^{0} := \varphi ,$ and $\varphi ^{1}:= \lnot \varphi $ . We will use the construction of such a $\Pi _1$ -formula described in [Reference Lindström21, Theorem 9, Chapter 2]. Let $\pi (x)$ be such a formula. Assuming that each $\pi (\underline {k})$ is provable in ${\textsf {CT}}_0$ , $\{\pi (\underline {k})\}_{k\in \omega }$ is an infinite antichain in ${{\textsf {CT}}_0}/{{\textsf {PA}}}$ . By Proposition 53 this implies that $\{\delta ^{\pi (\underline {k})}\}_{k\in \omega }$ is an infinite antichain in $\Delta $ . These considerations show that it suffices to verify