Proof. We can see that the following equivalences hold:

1. 1. $z = \{x,y\} \iff (x\in z \land y\in z) \land \forall w\in z(w=x\lor w=y)$ .

2. 2. $\{x,y\}\in z \iff \exists w\in z (w = \{x,y\})$ .

3. 3. $z = \langle x,y\rangle \iff (\{x\}\in z\land \{x,y\}\in z)\land \forall w\in z (w = \{x\} \lor w = \{x,y\})$ .

4. 4. $z=\bigcup x \iff [\forall w\in z\exists y\in x (w\in y)] \land [\forall y\in x\forall w\in y (w\in z)]$ .

5. 5. $z=x\cup y \iff [\forall w\in x(w\in z)\land \forall w\in y(w\in z)] \land [\forall w\in z(z\in x \lor z\in y)]$ .

6. 6. $z=\varnothing \iff \forall w\in z (w\neq w)$ .

7. 7. $x \text {is transitive} \iff \forall y\in x \; \forall z\in y \; (z\in x)$ .

8. 8. $x \text {is an ordinal} \iff x \text {is transitive} \land \forall y\in x \; (y \text {is transitive})$ .

The last equivalence requires clarification: The classical definition of $\alpha $ being an ordinal is that $\alpha $ is transitive and $(\alpha ,\in )$ is well-ordered in the sense that $(\alpha ,\in )$ is linear and for every non-empty $y\subseteq \alpha $ , y has an $\in $ -minimal element. Unfortunately, the classical definition is $\Pi _1$ . However, we can prove that $\alpha $ is an ordinal if and only if $\alpha $ is a transitive set of transitive sets. The reader can see that the proof of Lemmas 6.3 and 6.4 of [Reference Jeon11] works over $\mathsf {KP}_1$ , and it shows that $(\alpha ,\in )$ is well-ordered if and only if $\alpha $ is a transitive set of transitive sets.

Proof. Suppose that $\langle x_0,x_1\rangle = \langle y_0,y_1\rangle $ , that is,

$$ \begin{align*} \{\{x_0\},\{x_0,x_1\}\} = \{\{y_0\},\{y_0,y_1\}\}. \end{align*} $$

Then we have $\{x_0\}=\{y_0\}\lor \{x_0\}=\{y_0,y_1\}$ . Similarly, we have $\{x_0,x_1\}=\{y_0\}\lor \{x_0,x_1\}=\{y_0,y_1\}$ . Based on this, let us divide the cases:

• • If $\{x_0\} = \{y_0,y_1\}$ , then $x_0=y_0=y_1$ . Thus we get $\{x_0,x_1\}=\{y_0\}$ , so $x_1=y_0$ . In sum, $x_0=x_1=y_0=y_1$ .

• • If $\{x_0,x_1\}=\{y_0\}$ , then by an argument similar to the previous one, we have $x_0=x_1=y_0=y_1$ .

• • Now consider the case $\{x_0\}=\{y_0\}$ and $\{x_0,x_1\}=\{y_0,y_1\}$ . Then we have $x_0=y_0$ , so $\{x_0,x_1\}=\{x_0,y_1\}$ . Hence $x_1\in \{x_0,y_1\}$ , so either $x_1=x_0$ or $x_1=y_1$ . But if $x_1=x_0$ , then from $y_1\in \{x_0,x_1\}$ we get $x_1=y_1$ .

Proof. Let us prove the first claim: Suppose that $x=\langle x_0,x_1\rangle $ . By taking $u=\{x_0,x_1\}$ , we have

$$ \begin{align*} \exists u\in x (x_0,x_1\in u\land x=\langle x_0,x_1\rangle). \end{align*} $$

Thus if $\varphi (x_0,x_1,v_0,\ldots ,v_{n-1})$ holds, then $\varphi '(x,v_0,\ldots ,v_{n-1})$ follows.

Conversely, suppose that $\varphi '(x,v_0,\ldots ,v_{n-1})$ holds and that for some $u\in x$ and some $y_0,y_1\in u$ :

$$ \begin{align*} x=\langle y_0,y_1\rangle\land \varphi(y_0,y_1,v_0,\ldots,v_{n-1}). \end{align*} $$

Now let us prove the second claim by induction on n. The left-to-right direction is clear: For given $x_0$ and $x_1$ , set $x=\langle x_0,x_1\rangle $ . For the other direction, suppose that for every x the following holds:

(2) $$ \begin{align} \exists v_0\forall v_1\ldots \mathsf{Q} v_{n-1} \varphi'(x,v_0,v_1,\ldots,v_{n-1}). \end{align} $$

By taking $x=\langle x_0, x_1\rangle $ from (2) and applying (1), we get the following:

$$ \begin{align*} \exists v_0\forall v_1\ldots \mathsf{Q} v_{n-1} \varphi(x_0,x_1,v_0,v_1,\ldots,v_{n-1}).\\[-34pt] \end{align*} $$

Proof of Lemma 2.8

1. 1. Suppose that $\varphi (x)$ is $\Sigma _n$ . We first claim that $\exists x\varphi (x)$ is equivalent to a $\Sigma _n$ -formula. Suppose that $\varphi (x)$ is of the form

   $$ \begin{align*} \exists v_0\forall v_1\ldots\mathsf{Q} v_{n-1} \varphi_0(x,v_0,\ldots,v_{n-1}) \end{align*} $$
   for some $\Delta _0$ -formula $\varphi _0$ . By Lemma 2.9, we have a $\Delta _0$ -formula $\varphi _0'$ satisfying the following:
   $$ \begin{align*} &\forall x\forall v_0\exists v_1\dots \overline{\mathsf{Q}} v_{n-1} \lnot\varphi(x,v_0,v_1,\ldots,v_{n-1}) \\ &\qquad\qquad\qquad\qquad\qquad\qquad\leftrightarrow \forall y\exists v_1\dots \overline{\mathsf{Q}} v_{n-1} \lnot\varphi_0'(y,v_1,\ldots,v_{n-1}), \end{align*} $$
   where $\overline {\mathsf {Q}}$ is the dual of the quantifier $\mathsf {Q}$ . By taking the negation, we have
   $$ \begin{align*} &\exists x\exists v_0\forall v_1\ldots \mathsf{Q} v_{n-1} \varphi(x,v_0,v_1,\ldots,v_{n-1}) \\ &\qquad\qquad\qquad\qquad\qquad\qquad\leftrightarrow \exists y\forall v_1\ldots \mathsf{Q} v_{n-1} \varphi_0'(y,v_1,\ldots,v_{n-1}). \end{align*} $$

   Now let us consider the bounded quantifier case. Formally, $\exists x\in a\varphi (x)$ is equivalent to $\exists x (x\in a\land \varphi (x))$ . Under the same assumption on $\varphi $ and $\varphi _0$ , $\exists x (x\in a\land \varphi (x))$ is equivalent to

   $$ \begin{align*} \exists x \exists v_0 \forall v_0\ldots\mathsf{Q}v_{n-1} [x\in a\land \varphi_0(x,v_0,\ldots,v_{n-1})]. \end{align*} $$

   The formula $x\in a\land \varphi _0(x,v_0,\ldots ,v_{n-1})$ is $\Delta _0$ , so by exactly the same argument we did for an unbounded $\exists $ , we can normalize $\exists x\in a\varphi (x)$ to a $\Sigma _n$ -formula. Normalizing $\forall x\varphi (x)$ or $\forall x\in a\varphi (x)$ when $\varphi $ is $\Pi _n$ is analogous, so we omit the proof.

2. 2. Suppose that $\varphi $ and $\psi $ are $\Sigma _n$ -formulas. Showing the normalizability of $\lnot \varphi $ is easy, so let us only consider the normalizability of $\varphi \land \psi $ . The case for $\varphi \lor \psi $ is analogous.

   Suppose that $\varphi $ and $\psi $ are $\Sigma _n$ -formulas. Then we have $\Delta _0$ -formulas $\varphi _0$ and $\psi _0$ such that

   $$ \begin{align*} \varphi(\vec{x}) \equiv \exists u_0 \forall u_1 \ldots \mathsf{Q} u_{n-1} \varphi_0(u_0,u_1,\ldots, u_{n-1},\vec{x}) \end{align*} $$
   and
   $$ \begin{align*} \psi(\vec{x}) \equiv \exists v_0 \forall v_1 \ldots \mathsf{Q} v_{n-1} \psi_0(v_0,v_1,\ldots, v_{n-1},\vec{x}). \end{align*} $$
   We may assume that all of $u_i$ and $v_i$ are different variables by substitution, so by basic predicate logic, the following holds:
   $$ \begin{align*} \varphi(\vec{x})\land\psi(\vec{x}) \iff \exists u_0\exists v_0 \forall u_1\forall v_1\ldots \mathsf{Q}u_{n-1}\mathsf{Q}v_{n-1}[\varphi_0\land \psi_0]. \end{align*} $$
   Now inductively reduce duplicated quantifiers from the inside to the outside by applying the second clause of Lemma 2.9.

   To illustrate what is happening, let us consider the case $n=2$ . In this case, we need to normalize the following formula:

   $$ \begin{align*} \exists u_0\exists v_0 \forall u_1\forall v_1[\varphi_0(u_0,u_1)\land \psi_0(v_0,v_1)]. \end{align*} $$
   By Clause 2 of Lemma 2.9, we can find some $\Delta _0$ -formula $\chi (u_0,v_0,w_1)$ such that for every $u_0$ and $v_0$ , we have
   $$ \begin{align*} \forall u_1\forall v_1[\varphi_0(u_0,u_1)\land \psi_0(v_0,v_1)] \leftrightarrow \forall w_1 \chi(u_0,v_0,w_1). \end{align*} $$
   Then applying the equivalent form of the second clause of Lemma 2.9, which is for repeating $\exists $ , we can find a $\Delta _0$ -formula $\chi '(w_0,w_1)$ , we have
   $$ \begin{align*} \exists u_0\exists u_1 \forall w_1 \chi(u_0,v_0,w_1) \leftrightarrow \exists w_0\forall w_1 \chi'(w_0,w_1). \end{align*} $$

3. 3. We will prove this claim by induction on n. $\Sigma _n$ -Collection states the following: For every $\Sigma _n$ -formula $\varphi (x,y,p)$ , we have for every a and p,

   $$ \begin{align*} \forall x\in a \exists y \varphi(x,y,p) \to \exists b \forall x\in a \exists y\in b \varphi(x,y,p). \end{align*} $$
   Clearly the consequent implies the antecedent, so the equivalence holds. Now assume that $\varphi $ by a $\Pi _{n-1}$ -formula. Then $\exists y\varphi (x,y,p)$ becomes a $\Sigma _n$ -formula.

   Thus, the claim that $\forall x\in a\psi (x)$ is equivalent to a $\Sigma _n$ -formula $\psi $ reduces to the claim that $\exists y\in b \chi (y)$ is equivalent to a $\Pi _{n-1}$ -formula $\chi $ . However, by the inductive assumption, $\Sigma _{n-1}$ -Collection shows that $\exists y\in b \chi (y)$ for a $\Pi _{n-1}$ -formula $\chi $ is equivalent to a $\Pi _{n-1}$ -formula.

   Reducing $\exists x\in a\varphi (x)$ for a $\Pi _n$ -formula $\varphi $ is analogous, so we omit it.

Proof. Suppose that F is defined by a $\Sigma _1$ -formula $\varphi (x,y)$ and $\psi $ is $\Sigma _1$ . $\psi (F(x))$ is equivalent to $\exists y \varphi (x,y)\land \psi (F(x))$ , which is normalizable to a $\Sigma _1$ -formula by Lemma 2.8. Similarly, if $\psi $ is $\Pi _1$ , then $\psi (F(x))$ is equivalent to $\forall y \varphi (x,y)\to \psi (y)$ , which is normalizable by a $\Pi _1$ -formula.

Proof. We will not prove the $\Sigma $ -Recursion theorem here, and the interested reader may consult [Reference Barwise3].

Now let us provide the complexity analysis for $z=F(x_0,\ldots ,x_m,y)$ when F is recursively defined from an $(m+2)$ -ary $\Sigma $ -definable function G. For notational convenience, let us consider the case $m=1$ .

Observe that $z=F(x,y)$ is equivalent to

$$ \begin{align*} &\exists f {\big[}f \text{ is a function of domain } \mathsf{TC}(\{y\}) \\ &\qquad\qquad\qquad\land \forall u\in\mathsf{TC}(\{y\})(f(x,u) = G(x,y,f\upharpoonright \mathsf{TC}(u) ) ){\big]} \land f(x,y)=z. \end{align*} $$

The reader can verify that the formula $y=\mathsf {TC}(x)$ is $\mathsf {KP}_1$ -provably $\Sigma _1$ (see [Reference Barwise3, Section I.6]), and the statement ‘f is a function of domain a’ is $\Delta _0$ . Also, if f is a function, then $g=f\upharpoonright a$ is also $\Delta _0$ since

$$ \begin{align*} g=f\upharpoonright a \iff \forall p\in g(\pi_0(p)\in a\land p\in f) \land \forall p\in f(\pi_0(p)\in a\to p\in g), \end{align*} $$

where $\pi _0(p)$ is a projection function satisfying $\pi _0(\langle u,v\rangle )=u$ , and the reader can see that both $\pi _0(p)=u$ and $\pi _0(p)\in a$ are $\Delta _0$ . Hence by applying Lemma 2.11, we have the following:

(3) $$ \begin{align} &\exists f {\Big[}\underbrace{f \text{ is a function of domain} \mathsf{TC}(\{y\})}_{\Delta_{1}} \nonumber\\ &\quad\quad\quad\land \forall \underbrace{u\in\mathsf{TC}(\{y\})}_{\Delta_1}(\underbrace {f(x,u) = G(x,y,f\upharpoonright \mathsf{TC}(u) )}_{\Delta_1} ) {\Big]} \land \underbrace{f(x,y)=z}_{\Delta_{0}}. \end{align} $$

Hence the above formula is $\Sigma _1$ . However, we can see that (3) is equivalent to

$$ \begin{align*}& \forall f {\big[}f\ \text{is a function of domain} \mathsf{TC}(\{y\}) \\ &\qquad\qquad\qquad\land \forall u\in\mathsf{TC}(\{y\})(f(x,u) = G(x,y,f\upharpoonright \mathsf{TC}(u) ) ){\big]} \to f(x,y)=z, \end{align*} $$

and the reader can verify the above formula is $\Pi _1$ .

Proof. Consider $R^\star (\xi )$ given by

$$ \begin{align*} \xi\ \text{is an ordinal}\land \exists x\in L_\xi A(x)\land \forall \eta<\xi \; \forall x\in L_\eta \lnot A(x). \end{align*} $$

Intuitively, $R^\star (\xi )$ says that $\xi $ is the least ordinal such that $L_\xi $ captures a witness of $\exists x A(x)$ .

We need to see that the above formula is $\mathsf {KP}_1$ -provably $\Sigma _1$ . At first glance, the second conjunct may appear $\Delta _0$ , but note that the bounded quantifier is bounded in $L_\xi $ , which is defined by $\Sigma $ recursion. That is, the second conjunct contains hidden quantifiers. We can precisely state the second conjunct as follows, which is $\mathsf {KP}_1$ -provably equivalent to $\Sigma _1$ :

$$ \begin{align*} \exists a \left[ a = L_\xi \land \left(\exists x\in a A(x)\right)\right]. \end{align*} $$

The third conjunct of $R^\star (\xi )$ also has the same problem. But the precise statement of the third conjunct is also $\mathsf {KP}_1$ -provably equivalent to a $\Sigma _1$ :

$$ \begin{align*} \forall \eta<\xi \exists a [a=L_\eta \land \forall x\in a \lnot A(x)]. \end{align*} $$

Hence if we let $R^{\boldsymbol {\gamma }}$ obtained from $R^\star $ by replacing the two conjuncts to their $\Sigma _1$ -equivalent form, respectively, then they satisfy the desired properties.

Proof. Suppose that $\beta <|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ , and without loss of generality, assume that $\beta $ is representable by ${\boldsymbol {\beta }}$ and $T\vdash \operatorname {\mathsf {Exists}}({\boldsymbol {\beta }})^{L_{\boldsymbol {\alpha }}}$ . Define a new representation $\boldsymbol {\beta +1}$ by

$$ \begin{align*} R^{\boldsymbol{\beta+1}}(\xi) \iff \forall\eta\in\xi [R^{\boldsymbol{\beta}}_*(\eta)\to \xi=\eta\cup\{\eta\}], \end{align*} $$

where $R^{\boldsymbol {\beta }}_*(x)$ is a $\Pi _1$ -formula given from $R^{\boldsymbol {\beta }}$ as presented in (4).

It is clear that $\mathsf {KP}_1$ proves that these two are equivalent, and an ordinal satisfying $R^{\boldsymbol {\beta +1}}(\xi )$ is unique if it exists. Furthermore, T proves $\exists \xi R^{\boldsymbol {\beta +1}}_\Sigma (\xi )$ , so $\beta +1<|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ . Thus $|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ is a limit of representable ordinals.

Proof. We prove $|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}\le |T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ first. Suppose that $\beta $ is an ordinal and $\beta <|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ . We claim that $\beta <|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ .

Since $\beta <|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ , Lemma 3.9 entails that there is a representable $\gamma \in (\beta ,|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}})$ such that T proves $\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ . Note that $\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ is a $\Sigma ^{\boldsymbol {\alpha }}_1$ formula, so by the definition of $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ , we have $L_{|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}}\models \exists \xi R^{\boldsymbol {\gamma }}(\xi )$ . That is, we have the following:

(A) $$\begin{align} L \models \exists \xi<|T|_{\Sigma^{\boldsymbol{\alpha}}_1}\ (R^{\boldsymbol{\gamma}}_\Sigma(\xi))^{L_{|T|_{\Sigma^{\boldsymbol{\alpha}}_1}}}. \end{align} $$

By upward absoluteness, (A) implies that there exists $\xi <|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ such that $ L_\alpha \models R^{\boldsymbol {\gamma }}(\xi )$ and so $\gamma <|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . Hence $\beta < \gamma < |T|_{\Sigma _1^{\boldsymbol {\alpha }}}$ .

Now let us prove $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}\le |T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ . Suppose that $\gamma < |T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . Then there is a $\Delta _0$ -formula $A(x)$ such that

$$ \begin{align*} T\vdash \exists x\in L_{\boldsymbol{\alpha}} A(x) \quad\text{but}\quad L_\gamma\nvDash \exists x A(x). \end{align*} $$

By Lemma 3.5, we can find a representation ${\boldsymbol {\delta }}$ such that $\mathsf {KP}_1 + (V=L)$ proves the following:

1. (C1) $\exists x A(x)\leftrightarrow \operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})$ .

2. (C2) $\exists x A(x)\to \exists x\in L_{\boldsymbol {\delta }} A(x)$ .

Then we have $T\vdash \operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})^{L_{\boldsymbol {\alpha }}}$ : Reasoning over T, we have $\exists x\in L_{\boldsymbol {\alpha }} A(x)$ by assumption. Since $L_{\boldsymbol {\alpha }}$ is a model of $\mathsf {KP}_1 + (V=L)$ , we have

$$ \begin{align*} L_{\boldsymbol{\alpha}} \vDash \exists x A(x)\leftrightarrow \operatorname{\mathsf{Exists}}({\boldsymbol{\delta}}), \end{align*} $$

so we get $\operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})^{L_{\boldsymbol {\alpha }}}$ . Reasoning externally, since T is $\Sigma _1^{\boldsymbol {\alpha }}$ -sound and $\operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})^{L_{\boldsymbol {\alpha }}}$ is $\Sigma _1^{\boldsymbol {\alpha }}$ , we have that ${\boldsymbol {\delta }}$ exists over the metatheory. Furthermore, we have ${\boldsymbol {\delta }} < |T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ .

Now let us prove $\gamma < {\boldsymbol {\delta }}$ : We know that $\operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})^{L_{\boldsymbol {\alpha }}}$ is true. Since $L_{\boldsymbol {\alpha }}$ is a model of $\mathsf {KP}_1 + (V=L)$ , we have

$$ \begin{align*} L_{\boldsymbol{\alpha}} \vDash \exists x A(x)\to \exists x\in L_{\boldsymbol{\delta}} A(x). \end{align*} $$

This implies that $\exists x\in L_{\boldsymbol {\delta }} A(x)$ is true. If $\gamma \ge {\boldsymbol {\delta }}$ , then $L_\gamma \vDash \exists x A(x)$ by upward absoluteness, a contradiction. Hence $\gamma < {\boldsymbol {\delta }}$ .

Proof. Clearly $T\equiv _{\Sigma ^{\boldsymbol {\alpha }}_1} U$ implies $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ since $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ is determined by the $\Sigma ^{\boldsymbol {\alpha }}_1$ -consequences of T. The remaining condition follows from [Reference Pohlers19, Theorem 1.2.5].

Proof. Let $A(x)$ be a $\Delta _0$ -formula and assume that $T\vdash \exists x\in L_\alpha A(x)$ . By Corollary 3.6, we can find a representation ${\boldsymbol {\gamma }}$ such that

$$ \begin{align*} \mathsf{KP}_1+\operatorname{\mathsf{Exists}}({\boldsymbol{\alpha}})\vdash \exists x\in L_{\boldsymbol{\alpha}}\ A(x)\leftrightarrow \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})^{L_{\boldsymbol{\alpha}}}. \end{align*} $$

Then we have ${\boldsymbol {\gamma }}<|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ , and $L_{|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}}\models \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ . By Lemma 3.9, $|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ is a limit of representable ordinals that U-provably exist, so we can find some representable ordinal ${\boldsymbol {\delta }}<|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ such that ${\boldsymbol {\gamma }}< {\boldsymbol {\delta }}$ , $U\vdash \operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})^{L_{\boldsymbol {\alpha }}}$ , and $L_{\boldsymbol {\delta }}\models \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ .Footnote ⁶

Now consider the sentence $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ given by

$$ \begin{align*} \operatorname{\mathsf{Comp}}({\boldsymbol{\gamma}},{\boldsymbol{\delta}})\equiv \forall\xi[R^{\boldsymbol{\delta}}(\xi)\to L_\xi \models \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})]. \end{align*} $$

Roughly, $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ says that if ${\boldsymbol {\delta }}$ exists then ${\boldsymbol {\gamma }}$ exists and ${\boldsymbol {\gamma }} < {\boldsymbol {\delta }}$ . So, in some sense, $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ says that ${\boldsymbol {\gamma }}<{\boldsymbol {\delta }}$ . Also, note that $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ is a true $\Pi ^{\boldsymbol {\alpha }}_1$ -sentence as long as $\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ and $\operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})$ are true and ${\boldsymbol {\gamma }}<{\boldsymbol {\delta }}$ .

Now we can see that U + $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ proves $L_{\boldsymbol {\delta }}\models \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ : Working inside U + $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ , we have $\operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})^{L_{\boldsymbol {\alpha }}}$ , which implies ${\boldsymbol {\delta }} < {\boldsymbol {\alpha }}$ . Then by $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ , we have $L_{\boldsymbol {\delta }}\vDash \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ . Now by upward absoluteness, we have $L_{\boldsymbol {\alpha }}\vDash \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ .

Returning back to reasoning externally, we have $U+\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})\vdash \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ . Hence $U+\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ also proves $\exists x A(x)$ . To rephrase the conclusion, we proved the following:

1. (*) For every $\Delta _0$ -formula $A(x)$ , if $T\vdash \exists x\in L_\alpha A(x)$ , then we can find two representable ordinals ${\boldsymbol {\gamma }}$ and ${\boldsymbol {\delta }}$ such that $U\vdash \operatorname {\mathsf {Exists}}({\boldsymbol {\delta }})$ and $U + \operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})\vdash \exists x\in L_{\boldsymbol {\alpha }} A(x)$ .

Define $T_{\Sigma ^{\boldsymbol {\alpha }}_1}$ as the set of $\Sigma ^{\boldsymbol {\alpha }}_1$ theorems of T and $U_{\Sigma ^{\boldsymbol {\alpha }}_1}$ as the set of $\Sigma ^{\boldsymbol {\alpha }}_1$ theorems of U. Now we define an extension $\widehat {U}$ of the theory U. Precisely, $\varphi \in \widehat {U}$ if and only if one of the following holds:

1. 1. $\varphi \in U$ .

2. 2. $\varphi $ has the form $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ , where ${\boldsymbol {\gamma }}$ and ${\boldsymbol {\delta }}$ are representations and

   $$ \begin{align*} \exists\psi\in T_{\Sigma^{\boldsymbol{\alpha}}_1} \left[ \mathsf{KP}_1+\operatorname{\mathsf{Exists}}({\boldsymbol{\alpha}}) \vdash \psi\leftrightarrow \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}}),\, U\vdash \operatorname{\mathsf{Exists}}({\boldsymbol{\delta}}),\, L_{\boldsymbol{\delta}}\models \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})\right]. \end{align*} $$

3. 3. $\varphi $ has the form $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ , where ${\boldsymbol {\gamma }}$ and ${\boldsymbol {\delta }}$ are representations and

   $$ \begin{align*} \exists\psi\in U_{\Sigma^{\boldsymbol{\alpha}}_1} \left[ \mathsf{KP}_1+\operatorname{\mathsf{Exists}}({\boldsymbol{\alpha}}) \vdash \psi\leftrightarrow \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}}),\, T\vdash \operatorname{\mathsf{Exists}}({\boldsymbol{\delta}}),\, L_{\boldsymbol{\delta}}\models \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})\right]. \end{align*} $$

We define $\widehat {T}$ analogously except that we replace clause (1) with $\varphi \in T$ . By ( $\star $ ), $\widehat {T}$ and $\widehat {U}$ prove all of the $\Sigma ^{\boldsymbol {\alpha }}_1$ -theorems of T and U.

Furthermore, $\widehat {T}$ and $\widehat {U}$ are also $\Pi ^{\boldsymbol {\alpha }}_1$ -definable since both of T and U are $\Pi ^{\boldsymbol {\alpha }}_1$ -definable and “ $L_{\boldsymbol {\delta }}\models \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ ” is expressible by the following $\Pi ^{\boldsymbol {\alpha }}_1$ formula:

$$ \begin{align*} \left[\forall \eta\ \left(R^{\boldsymbol{\delta}} (\eta)\to L_\eta\models \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})\right)\right]^{L_{\boldsymbol{\alpha}}}. \end{align*} $$

We extended T to $\widehat {T}$ by adding true $\Pi ^{\boldsymbol {\alpha }}_1$ -sentences; hence $T\equiv \widehat {T}$ by condition 2 from Definition 4.2. Similarly, $U\equiv \widehat {U}$ .

Now we claim that $\widehat {T}\equiv _{\Sigma ^{\boldsymbol {\alpha }}_1}\widehat {U}$ , which will finalize our proof since it implies $\widehat {T}\equiv \widehat {U}$ , so we have $T\equiv \widehat {T} \equiv \widehat {U} \equiv U$ . Indeed, it suffices to show that $\widehat {T}\subseteq _{\Sigma ^{\boldsymbol {\alpha }}_1} \widehat {U}$ since the converse containment is symmetric.

Suppose that $\widehat {T}\vdash \theta $ for some $\Sigma ^{\boldsymbol {\alpha }}_1$ -sentence $\theta $ . Then $T+\sigma \vdash \theta $ for some true $\Pi ^{\boldsymbol {\alpha }}_1$ -sentence $\sigma $ , which is a conjunction of sentences of the form $\operatorname {\mathsf {Comp}}({\boldsymbol {\gamma }},{\boldsymbol {\delta }})$ we added. Thus we have $T\vdash \sigma \to \theta $ , and $\sigma \to \theta $ is $\Sigma ^{\boldsymbol {\alpha }}_1$ . However, we defined $\widehat {U}$ so that it proves all $\Sigma ^{\boldsymbol {\alpha }}_1$ -consequences of T, so $\widehat {U}\vdash \sigma \to \theta $ . Since $\sigma \in \widehat {U}$ , we have $\widehat {U}\vdash \theta $ .

Proof. First, we need to see that every $\mathsf {Ord}_{\Sigma ^{\boldsymbol {\alpha }}_1}$ induced partition is $\boldsymbol {\alpha }$ -Kreiselian. Let $\equiv $ be $\mathsf {Ord}_{\Sigma ^{\boldsymbol {\alpha }}_1}$ induced.

1. 1. Suppose that $T\equiv _{\Sigma ^{\boldsymbol {\alpha }}_1}U$ . Then $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ , since the $\alpha $ -ordinal analysis partition is $\boldsymbol {\alpha }$ -Kreiselian. Then $T \equiv U$ since $\equiv $ coarsens the $\alpha $ -ordinal analysis partition.

2. 2. Let $\varphi $ be true $\Pi ^{\boldsymbol {\alpha }}_1$ . Then $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|T+\varphi |_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . Then $T \equiv T+\varphi $ since $\equiv $ coarsens the ${\boldsymbol {\alpha }}$ -ordinal analysis partition.

Next we must show that every $\boldsymbol {\alpha }$ -Kreiselian partition is $\mathsf {Ord}_{\Sigma ^{\boldsymbol {\alpha }}_1}$ induced. Let $\equiv $ be $\boldsymbol {\alpha }$ -Kreiselian. We define a partition $\sim $ on $\mathsf {Ord}_{\Sigma ^{\boldsymbol {\alpha }}_1}$ as follows: For $\beta ,\gamma \in \mathsf {Ord}_{\Sigma ^{\boldsymbol {\alpha }}_1}$ , $\beta \sim \gamma $ if and only if, for some $\Pi ^{\boldsymbol {\alpha }}_1$ -definable $\Sigma ^{\boldsymbol {\alpha }}_1$ -sound extensions V and W of $\mathsf {KP}_1 +\operatorname {\mathsf {Exists}}({\boldsymbol {\alpha }})$ , the following three claims hold:

1. 1. $|V|_{\Sigma ^{\boldsymbol {\alpha }}_1}=\beta .$

2. 2. $|W|_{\Sigma ^{\boldsymbol {\alpha }}_1}=\gamma .$

3. 3. $V\equiv W$ .

It is trivial to establish that $\sim $ is an equivalence relation on $\mathsf {Ord}_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . So we need only see that $\sim $ induces $\equiv $ . To this end, we define

$$ \begin{align*}T\equiv_\star U := |T|_{\Sigma^{\boldsymbol{\alpha}}_1}\sim |U|_{\Sigma^{\boldsymbol{\alpha}}_1}.\end{align*} $$

Since $\equiv _\star $ just is the $\sim $ -induced equivalence relation, it suffices to show that $T\equiv U$ if and only if $T\equiv _\star U$ .

Suppose that $T\equiv U$ . So by the definition of $\sim $ , we have $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}\sim |U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . But then $T\equiv _\star U$ .

Conversely, suppose that $T\equiv _\star U$ , i.e., that $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}\sim |U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . By definition of $\sim $ , we infer that for some $T'$ and $U'$ :

1. 1. $|T'|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ .

2. 2. $|U'|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}.$

3. 3. $T' \equiv U'$ .

Applying Theorem 4.5 to (1) yields $T\equiv T'$ . Likewise, applying Theorem 4.5 to (2) yields $U\equiv U'$ .

Combining these observations with (3) yields $T\equiv T' \equiv U' \equiv U$ , whence $T\equiv U$ .

Sketch of the proof. Formally, ‘M halts with an input x’ is equivalent to the existence of a computation d of p with input x, and we can see that d being a computation is $\Delta _1$ . If we take $\mathtt {T}(p,x,d)$ to be

$$ \begin{align*} \mathtt{T}(p,x,d):= d\ \text{is a computation of p with input x,} \end{align*} $$

and if $d\in L_\beta $ , then we have that the output of p under x is in $L_\beta $ since d contains the output of the computation.

Conversely, suppose that $\exists y \psi (x,y)$ is a $\Sigma _1$ -formula, where $\psi $ is a bounded formula. Now consider an SRM p trying to find y such that $\mathsf {Tr}_{\Delta _0}(\ulcorner \psi \urcorner ,x,y)$ holds via unbounded search. Then ‘p halts with input x’ is equivalent to $\exists y \psi (x,y)$ .

Sketch of the Proof.

In one direction, suppose that the graph of $F\colon L\to L$ is defined by a formula $\exists z \varphi _0(x,y,z,a)$ for some $\Delta _0$ formula $\varphi _0$ and a parameter a. Then consider the following SRM $p(x,a)$ : Enumerate all sets under the $<_L$ -order. If the set we enumerate is not an ordered pair, skip this set. If the set we enumerate is of the form $\langle y,z\rangle $ , compute $\mathrm {Tr}_{\Delta _0}(\ulcorner \varphi _0\urcorner , x,y,z,a)$ . If $\mathrm {Tr}_{\Delta _0}(\ulcorner \varphi _0\urcorner , x,y,z,a)=1$ , $p(x,a)$ returns y. Otherwise, proceed with the computation. Then by definition of p, $p(x,a)$ halts iff $x\in \operatorname {dom} F$ , and $p(x,a)=y$ iff $F(x)=y$ .

Conversely, suppose that F is a class function defined by an SRM $p(x,a)$ with parameter a. We can see that there is a $\Delta _1$ -definable function $\mathtt {U}$ , which reads off a computation d and extracts its output. Then $F(x)=y$ if and only if $\exists d\,\mathtt {T}(p,\langle x,a\rangle ,d)\land y=\mathtt {U}(d)$ , which is $\Sigma _1$ .

Proof. Pohlers [Reference Pohlers19, Lemma 1.2.3] proves $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}=|T|_{\Pi ^{\boldsymbol {\alpha }}_2}$ , so let us prove the following:

$|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}\le |T|_{{\boldsymbol {\alpha }}\text {-ht}}$ : To show this, it suffices to show $|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}} \le |T|_{{\boldsymbol {\alpha }}-ht}$ . Let ${\boldsymbol {\gamma }}$ be a representation such that $T\vdash \exists \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ . By Corollary 5.5, we can construct a following ${\boldsymbol {\alpha }}$ -SRM p: p tries to find an ordinal $\xi <\alpha $ such that $R^{\boldsymbol {\gamma }}(\xi )$ holds, and halts if there is such ordinal and returns that ordinal. We can decide $R^{\boldsymbol {\gamma }}(\xi )$ holds within time $<{\boldsymbol {\alpha }}$ by computing $R^{\boldsymbol {\gamma }}(\xi )$ and $\lnot R^{\boldsymbol {\gamma }}_*(\xi )$ simultaneously, where $R^{\boldsymbol {\gamma }}_*$ is the $\Pi _1$ statement for $R^{\boldsymbol {\gamma }}$ defined in (4). One of them must halt within time $<{\boldsymbol {\alpha }}$ , and we can decide the validity of $R^{\boldsymbol {\gamma }}(\xi )$ by using that computational result.

Since T proves $\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ , T proves p halts, and $\mathsf {height}(p)=\xi +1$ for the unique $\xi $ such that $L\models R^{\boldsymbol {\gamma }}(\xi )$ . That is, $\xi <|T|_{{\boldsymbol {\alpha }}\text {-ht}}$ . Then the desired inequality follows from the definition of $|T|_{{\boldsymbol {\alpha }}{\text {-}\mathsf {repr}}}$ .

$|T|_{{\boldsymbol {\alpha }}\text {-ht}}\le |T|_{{\boldsymbol {\alpha }}\text {-cl}}$ : ‘p halts’ for an ${\boldsymbol {\alpha }}$ -SRM p computing a constant function is a special case of ‘p halts for every input x.’ Thus, every T-provably-halting $\alpha $ -SRM that is computing a constant function outputs a value below $|T|_{{\boldsymbol {\alpha }}\text {-cl}}$ .

$|T|_{{\boldsymbol {\alpha }}\text {-cl}}\le |T|_{\Pi ^{\boldsymbol {\alpha }}_2}$ : By Corollary 5.5, ‘p halts with input x’ is equivalent to $\exists d\in L_{\boldsymbol {\alpha }} \; \mathtt {T}(p,d,x)$ . Thus the assertion ‘For every $x\in L_{\boldsymbol {\alpha }}$ , p halts with input x’ is $\Pi ^{\boldsymbol {\alpha }}_2$ . Thus if T proves p halts, then by the definition of $|T|_{\Pi ^{\boldsymbol {\alpha }}_2}$ , then $L_{|T|_{\Pi ^{\boldsymbol {\alpha }}_2}}\models \forall x \; \exists d \; \mathtt {T}(p,d,x)$ . By Corollary 5.5 again, we infer that the output of p belongs to $L_{|T|_{\Pi ^{\boldsymbol {\alpha }}_2}}$ . That is, $L_{|T|_{\Pi ^{\boldsymbol {\alpha }}_2}}$ is closed under computation by p.

Hence we get

$$ \begin{align*} |T|_{{\boldsymbol{\alpha}}-ht}\le |T|_{{\boldsymbol{\alpha}}-cl}\le |T|_{\Pi^{\boldsymbol{\alpha}}_2}=|T|_{\Sigma^{\boldsymbol{\alpha}}_1}\le|T|_{{\boldsymbol{\alpha}}-ht}.\\[-34pt] \end{align*} $$

Proof. Let us reason over $\mathsf {KP}_1 + \operatorname {\mathsf {Exists}}({\boldsymbol {\alpha }})$ and suppose that $\mathsf {RFN}_{\Sigma ^{\boldsymbol {\alpha }}_1}(T)$ holds. Now suppose that for a $\Sigma _1$ -sentence $\sigma $ , we have

$$ \begin{align*} T + \theta^{L_{\boldsymbol{\alpha}}} \vdash \sigma^{L_{\boldsymbol{\alpha}}}. \end{align*} $$

Then $T\vdash (\theta \to \sigma )^{L_{\boldsymbol {\alpha }}}$ , and $\theta \to \sigma $ is $\Sigma _1$ . Hence by $\mathsf {RFN}_{\Sigma ^{\boldsymbol {\alpha }}_1}(T)$ , we get $L_\alpha \vDash \theta \to \sigma $ . Since $\theta ^{L_\alpha }$ holds, we have $L_\alpha \vDash \sigma $ .

Proof. For the left-to-right, suppose that $T\subseteq ^{\Pi ^{\boldsymbol {\alpha }}_1}_{\Sigma ^{\boldsymbol {\alpha }}_1} U$ . Let $\gamma $ be a representable ordinal with a representation ${\boldsymbol {\gamma }}$ such that $\gamma <|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ . Then by Lemma 6.2, we have $T\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1}\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ . Since $T\subseteq ^{\Pi ^{\boldsymbol {\alpha }}_1}_{\Sigma ^{\boldsymbol {\alpha }}_1} U$ and $\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ is $\Sigma ^{\boldsymbol {\alpha }}_1$ , we have $U\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1}\operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ .

Hence we have a $\Pi _1$ -sentence $\psi $ such that $L_\alpha \vDash \psi $ and $U\vdash \psi ^{L_{\boldsymbol {\alpha }}}\to \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ . Since the statement $\psi \to \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ is $\Sigma _1$ , we have $L_{|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}}\models \psi \to \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ . However, since $\psi $ is $\Pi _1$ , $L_\alpha \models \psi $ implies $L_{|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}}\models \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})$ . This shows $\gamma <|U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ .

For the right-to-left direction, assume that $|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}\le |U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ and assume that $T\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1}\varphi ^{L_{\boldsymbol {\alpha }}}$ for a $\Sigma _1$ -formula $\varphi $ . Thus we can find a $\Pi _1$ -formula $\psi $ such that $L_\alpha \models \psi $ and $T\vdash (\psi \to \varphi )^{L_{\boldsymbol {\alpha }}}$ . We can see that $\psi \to \varphi $ is $\Sigma _1$ , so by Corollary 3.6, we can find a representation ${\boldsymbol {\gamma }}$ such that

$$ \begin{align*} \mathsf{KP}_1 \vdash \forall\beta\Big(\beta\ \text{is admissible}\to [(\psi\to\varphi)^{L_\beta} \leftrightarrow \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})^{L_\beta}]\Big). \end{align*} $$

Since ${\boldsymbol {\alpha }}$ is $\mathsf {KP}_1$ -provably admissible, we have

$$ \begin{align*} \mathsf{KP}_1 + \operatorname{\mathsf{Exists}}({\boldsymbol{\alpha}}) \vdash (\psi\to\varphi)^{L_{\boldsymbol{\alpha}}} \leftrightarrow \operatorname{\mathsf{Exists}}({\boldsymbol{\gamma}})^{L_{\boldsymbol{\alpha}}}. \end{align*} $$

Since T is an extension of $\mathsf {KP}_1 + \operatorname {\mathsf {Exists}}({\boldsymbol {\alpha }})$ , we have $T\vdash \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ . From this, we get ${\boldsymbol {\gamma }}<|T|_{\Sigma ^{\boldsymbol {\alpha }}_1}\le |U|_{\Sigma ^{\boldsymbol {\alpha }}_1}$ by Lemma 3.11. Thus by Lemma 6.2, $U\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1} \operatorname {\mathsf {Exists}}({\boldsymbol {\gamma }})^{L_{\boldsymbol {\alpha }}}$ , which implies $U\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1}(\psi \to \varphi )^{L_{\boldsymbol {\alpha }}}$ . Since $L_{\boldsymbol {\alpha }}\models \psi $ , we finally have $U\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1}\varphi ^{L_{\boldsymbol {\alpha }}}$ .

Proof. In this proof, we reason in $\mathsf {KP}_1 + \operatorname {\mathsf {Exists}}({\boldsymbol {\alpha }})$ with no other true $\Pi ^{\boldsymbol {\alpha }}_1$ sentences unless specified.

(1) $\leftrightarrow $ (2). $\mathsf {RFN}^{\Pi ^{\boldsymbol {\alpha }}_1}_{\Sigma ^{\boldsymbol {\alpha }}_1}(T)\to \mathsf {RFN}_{\Sigma ^{\boldsymbol {\alpha }}_1}(T)$ follows from the trivial fact that $T\vdash \varphi $ implies $T\vdash ^{\Pi ^{\boldsymbol {\alpha }}_1}\varphi $ .