Proof. Otherwise, $\delta _{\alpha } = \alpha ^+$ by Gostanian’s characterization of $\Sigma ^{1}_{1}$ -reflection, so $\alpha $ is $(\alpha ^++1)$ -stable. By Gostanian’s result mentioned in the introduction, if $\alpha $ is $(\alpha ^++1)$ -stable, then it is $\Sigma ^{1}_{1}$ -reflecting. ⊣

Proof. Let $\delta = \delta _{\sigma ^{1}_{1}}$ . Since $\delta <(\sigma ^{1}_{1})^+$ , it is not admissible. As we observed before, $\delta $ is a limit ordinal; thus, the failure of admissibility must be due to an instance of collection. Choose some $\Delta _{0}$ formula $\psi $ such that for some $\vec a \in L_{\delta }$ , $L_{\delta }\not \models \psi (\vec a)\text {-collection}.$ To see that $\sigma ^{1}_{1}$ is not $(\delta +1)$ -stable, consider the formula $\phi $ in the language of set theory asserting that there are sets A and B such that:

1. 1. A and B are transitive sets satisfying $V=L$ , A is admissible, $A \in B$ , and there is $\vec a \in B$ such that B does not satisfy $\psi (\vec a)\text {-collection}$ ;

2. 2. for each $({\mathsf {Ord}}\cap A)$ -recursive linear ordering $R \in B$ , either there is an infinite descending sequence b through R with $b\in A$ , or there is an ordinal $\beta \in B$ and an isomorphism $f \in B$ from R to $\beta $ ;

3. 3. for each $\beta \in B$ , there is an $({\mathsf {Ord}} \cap A)$ -recursive linear ordering $R \in B$ and an isomorphism $f \in B$ from R to $\beta $ .

Notice that $\phi $ is a $\Sigma _{1}$ formula, since the only unbounded quantifier is the one on B. Moreover, it does not hold in $L_{\sigma ^{1}_{1}}$ , for the sets A and B would need to be of the form $L_{\alpha }$ and $L_{\beta }$ , with $\alpha <\beta <\sigma ^{1}_{1}$ . Conditions (2) and (3) together imply that $\beta = \delta _{\alpha }$ , but Gostanian’s characterization of $\sigma ^{1}_{1}$ then implies $\beta = \alpha ^+$ , contradicting condition (1). Finally, it does hold in $L_{\delta +1}$ , as witnessed by $A = L_{\sigma ^{1}_{1}}$ and $B = L_{\delta }$ . To see that (2) holds, recall a theorem of Gostanian [Reference Gostanian8, Theorem 3.2] by which if $\alpha $ is $\Sigma ^{1}_{1}$ -reflecting, then every $\alpha $ -recursive linear ordering which is not a wellordering has an infinite descending sequence in $L_{\alpha }$ . Thus, every $\sigma ^{1}_{1}$ -recursive linear ordering R either has an infinite descending sequence in $L_{\sigma ^{1}_{1}}$ , or else is isomorphic to some ordinal $\beta <\delta $ . One can construct an isomorphism witnessing this by transfinite recursion: at stage $\gamma <\beta $ , one has defined $f\upharpoonright \gamma $ and sets $f(\gamma )$ equal to the R-least element not in the range of $f\upharpoonright \gamma $ . Since this process takes $\beta $ -many stages and $R \in L_{\sigma ^{1}_{1}+1}$ , such an isomorphism belongs to $L_{\sigma ^{1}_{1} + \delta }$ . Since $\delta $ is additively indecomposable, it belongs to $L_{\delta }$ . The proof that (3) holds is similar. ⊣

Proof. The proof of the proposition is as in the comment directly preceding its statement. The only additional observation needed is that the hypothesis implies that $\alpha $ is $\Sigma ^{1}_{1}$ -reflecting, by Lemma 9. ⊣

Proof. Let $\alpha $ be as in the statement. By Proposition 12, it is a limit of $\Sigma ^{1}_{1}$ -reflecting ordinals. This fact is first-order expressible over $\alpha $ , since it is recursively inaccessible (and thus $\Sigma ^{1}_{1}$ -correct). By Proposition 12 again, $\alpha $ is a limit of $\Sigma ^{1}_{1}$ -reflecting ordinals which are limits of $\Sigma _{1}^{1}$ -reflecting ordinals. ⊣

Proof. Write $\delta = \delta _{\alpha }$ . Since $\delta $ is a limit ordinal, it suffices to consider arbitrary $\gamma <\delta $ and show that

$$ \begin{align*}L_{\alpha}\prec_{1} L_{\gamma}.\end{align*} $$

Let $a \in L_{\alpha }$ and $\phi $ be a $\Sigma _{1}$ -formula such that $L_{\gamma }\models \phi (a)$ . Without loss of generality, assume that a is an ordinal. Let R be a $\alpha $ -recursive wellordering of length $\gamma $ . In particular, R is $\alpha $ -r.e., so there is a $\Sigma _{1}$ formula $\psi $ , such that for all $x,y \in L_{\alpha }$ ,

$$ \begin{align*}xRy \leftrightarrow L_{\alpha}\models\psi(x,y).\end{align*} $$

Let us assume for notational simplicity that $\psi $ is defined without parameters. Given an ordinal ${\alpha ^{\prime }}$ , let $R_{\alpha ^{\prime }}$ be the binary relation given by

$$ \begin{align*}xR_{\alpha^{\prime}} y \leftrightarrow L_{\alpha^{\prime}}\models\psi(x,y).\end{align*} $$

Since $\psi $ is $\Sigma _{1}$ , we have $R_{\alpha ^{\prime }}\subset R_{\tau }$ whenever ${\alpha ^{\prime }}\leq \tau \leq \alpha $ . In particular, $R_{\alpha ^{\prime }}$ is wellfounded for all $\alpha ^{\prime }<\alpha $ .

Because $L_{\gamma }\models \phi (a)$ , there is a subset A of $L_{\alpha }$ such that

1. 1. A codes a model $(M, E)$ of ${\mathsf {KP}} + V = L$ ;

2. 2. M has a largest admissible ordinal $\tau $ and $(\tau , E)$ is isomorphic to $(\alpha , \in )$ ;

3. 3. there is an ordinal $\beta $ of M and a function $f \in M$ which is an isomorphism between $R_{\tau }^{M}$ (i.e., $R_{\tau }$ computed within M) and $\beta $ , and $L_{\beta }^{M}\models \phi (a)$ .

The existence of such an A can be expressed by a set-theoretic $\Sigma ^{1}_{1}$ formula over $L_{\alpha }$ with parameter a (as well as any other parameters involved in the definition of R). In particular, reference to $\alpha $ can be made over $L_{\alpha }$ , since $\alpha = {\mathsf {Ord}}^{L_{\alpha }}$ . Thus, clause (2) can be expressed in a $\Sigma ^{1}_{1}$ way over $L_{\alpha }$ by asserting the existence of a bijection h with domain $\alpha $ and range (the set of codes of elements of) $\tau $ such that for all $\beta _{0},\beta _{1}<\alpha $ , $\beta _{0} \in \beta _{1}$ if, and only if, $h(\beta _{0})Eh(\beta _{1})$ . Note that quantification over $\alpha $ is first-order over $L_{\alpha }$ .

By $\Sigma ^{1}_{1}$ -reflection, there is some ${\alpha ^{\prime }}<\alpha $ and some $A_{\alpha ^{\prime }}\subset L_{\alpha ^{\prime }}$ such that $a \in L_{\alpha ^{\prime }}$ and

1. 4. $A_{\alpha ^{\prime }}$ codes a model $(N, F)$ of ${\mathsf {KP}} + V = L$ ;

2. 5. N has a largest admissible ordinal $\eta $ and $(\eta , F)$ is isomorphic to $({\alpha ^{\prime }},\in )$ ;

3. 6. there is an ordinal b of N and a function $g \in N$ which is an isomorphism between $R_{\eta }^{N}$ and b, and $L_{b}^{N}\models \phi (a)$ .

Here and for the rest of our lives, let us identify the wellfounded part of N with its transitive collapse. Condition (5) implies that $L_{\alpha ^{\prime }} \in N$ . By (6), there is an ordinal b of N and an isomorphism $g \in N$ from $R_{\alpha ^{\prime }}$ to b. Because $R_{\alpha ^{\prime }} \subset R$ , it is wellfounded, and so b really is an ordinal. Now, $L_{b}\models \phi (a)$ , and N has no admissible ordinals above ${\alpha ^{\prime }}$ , so $b<({\alpha ^{\prime }})^+<\alpha $ . Since $\phi $ is $\Sigma _{1}$ , we conclude that $L_{\alpha }\models \phi (a)$ , as was to be shown. ⊣

Proof. Let $\phi $ be the sentence that asserts the existence of some $A\subset L_{\alpha }$ coding a model $(M,E)$ of ${\mathsf {KP}} + V = L$ containing $\alpha $ and such that

$$ \begin{align*}M \models L_{\alpha}\prec_{1} L_{\delta_{\alpha}}.\end{align*} $$

Clearly, every $\Pi ^{1}_{1}$ -reflecting ordinal satisfies this sentence, as does every $\Sigma ^{1}_{1}$ -reflecting ordinal, by Proposition 15.

Suppose that $L_{\alpha } \models \phi $ , as witnessed by $(M,E)$ . Suppose moreover that $\alpha $ is not $\Sigma ^{1}_{1}$ -reflecting, so that $\delta _{\alpha } = \alpha ^+$ by Gostanian’s characterization. Since $\alpha \in M$ , a well-known theorem of F. Ville (see e.g., Barwise [Reference Barwise5] for a proof) implies that $L_{\alpha ^+}\subset M$ . Given an arbitrary $\beta <\alpha ^+$ , we then have $\beta \in M$ and $\beta < \delta _{\alpha }^{M}$ , for otherwise $\delta _{\alpha }^{M}<\alpha ^+$ , which is impossible, since any $\alpha $ -recursive wellordering of a subset of $\alpha $ of length $\delta _{\alpha }^{M}$ would belong to M. By choice of M,

$$ \begin{align*}M \models L_{\alpha}\prec_{1} L_{\delta_{\alpha}},\end{align*} $$

and so $M\models L_{\alpha }\prec _{1} L_{\beta }$ . However, being $\Sigma _{1}$ -elementary is absolute, so we really do have $L_{\alpha }\prec _{1} L_{\beta }$ and, since $\beta $ was arbitrary, we have $L_{\alpha } \prec _{1} L_{\alpha ^+}$ , so $\alpha $ is $\Pi ^{1}_{1}$ -reflecting.⊣

Proof. Let $\phi $ be the sentence from Proposition 16. Then, if $\psi $ is another $\Sigma ^{1}_{1}$ sentence, so is the conjunction $\phi \wedge \psi $ . ⊣

Proof. This is similar to the proof of Proposition 15. Again, it is easy to see that $\delta _{\alpha }$ is a limit. Let $\gamma = \delta _{\alpha }+1$ and $a \in L_{\alpha }$ be such that $L_{\gamma }\models \phi (a)$ , for some $\Sigma _{1}$ formula $\phi $ . Let $\psi $ be the $\Sigma ^{1}_{1}$ formula expressing that $\alpha $ is locally countable and there is a set $A\subset L_{\alpha }$ coding a model $(M,E)$ of ${\mathsf {KP}} + V = L$ with $\alpha \in M$ and such that

Then $L_{\alpha }\models \psi $ . By hypothesis, there is a $\Sigma ^{1}_{1}$ -reflecting $\tau <\alpha $ such that $L_{\tau }\models \psi $ . Thus, $\tau $ is locally countable and there is a model $(M,E)$ of ${\mathsf {KP}} + V = L$ with $\tau \in M$ and such that

By Ville’s theorem, $L_{\tau ^+}\subset M$ and, since $\tau $ is $\Sigma ^{1}_{1}$ -reflecting, $\delta _{\tau } < \tau ^+$ . Hence, M computes $\delta _{\tau }$ and $\delta _{\tau }+1$ correctly and so we really have $L_{\delta _{\tau }+1}\models \phi (a)$ . Since $\phi $ is $\Sigma _{1}$ , we conclude $L_{\alpha }\models \phi (a)$ , as desired. ⊣

Proof. If $\alpha $ is $\sigma \sigma s$ -reflecting and $L_{\alpha }$ satisfies a $\Sigma ^{1}_{1}$ sentence $\phi $ , then, by definition, there is a $\sigma s$ -reflecting $\tau <\alpha $ such that $L_{\tau } \models \phi $ . By $\sigma s$ -reflection, there is an s-reflecting $\eta <\tau $ such that $L_{\tau }\models \phi $ . Hence, $\alpha $ is $\sigma s$ -reflecting. The argument for $\pi \pi s$ -reflection is similar. ⊣

Proof. We prove the first claim; the second one is similar. An admissible $\alpha $ being $\pi s$ -reflecting means that for every $\Pi ^{1}_{1}$ sentence $\phi $ with a parameter $\gamma <\alpha $ , if $L_{\alpha }\models \phi (\gamma )$ , then there is $\beta $ with $\gamma <\beta <\alpha $ such that $\beta $ is s-reflecting and $L_{\beta }\models \phi (\gamma )$ . By choosing a suitably universal $\Pi ^{1}_{1}$ formula, we can reduce $\pi s$ -reflection to a single sentence: it suffices to show that for a fixed $\Pi ^{1}_{1}$ sentence $\phi $ and a fixed parameter $\gamma $ , the sentence

(1)

is uniformly equivalent to a $\Sigma ^{1}_{1}$ sentence, for then we can quantify over all $\gamma <\alpha $ and the resulting sentence is still $\Sigma ^{1}_{1}$ .

The key observation for this is the fact that $\Pi ^{1}_{1}$ (and thus also $\Sigma ^{1}_{1}$ ) sentences with parameters $a_{0},\ldots , a_{n}$ about a structure of the form $L_{\xi }$ are equivalent to first-order sentences with parameters $a_{0},\ldots , a_{n},\xi $ over any admissible $L_{\zeta }$ , with $\xi <\zeta $ , and the equivalence is uniform (see Aczel–Richter [Reference Aczel and Richter3, Theorem 6.2] for a proof of this fact). Hence, every recursively inaccessible ordinal is $\Sigma ^{1}_{1}$ -correct and a straightforward induction shows that whether an ordinal $\xi $ is t-reflecting is computed correctly by any $L_{\zeta }$ whenever $\xi <\zeta $ and $\zeta $ is recursively inaccessible, for any reflection pattern t. Since every $\Sigma ^{1}_{1}$ - or $\Pi ^{1}_{1}$ -reflecting ordinal is recursively inaccessible, (1) is equivalent to the conjunction of

1. 2. $\alpha $ is recursively inaccessible, and

2. 3. $\big (L_{\alpha }\models \phi (\gamma )\big ) \to \exists \beta \, \big (\gamma <\beta <\alpha \wedge L_{\alpha }\models $ “ $\beta $ is s-reflecting and $L_{\beta }\models \phi (\gamma )$ ” $\big )$ ,

and both of these conjuncts are easily seen to be $\Sigma ^{1}_{1}$ .⊣

Proof. Recall that if an ordinal $\alpha $ is s-reflecting, for any nontrivial reflection pattern s, then it is recursively inaccessible and, in fact, a limit of recursively inaccessible ordinals. (1) then follows from the fact that being $\sigma s$ -reflecting is expressible by a $\Pi ^{1}_{1}$ sentence $\psi $ . Thus, if $\alpha $ is $\sigma s \wedge \pi t$ -reflecting and satisfies some $\Pi ^{1}_{1}$ -sentence $\phi $ , then the conjunction $\phi \wedge \psi $ is also $\Pi ^{1}_{1}$ , and any ordinal satisfying it must be $\sigma s$ -reflecting.

(2) is similar. For (3), there are two cases: if $\alpha $ is $\pi $ -reflecting, then the result follows from (2). If $\alpha $ is not $\pi $ -reflecting, it is not $\alpha ^+$ -stable. Hence, there is a least $\gamma <\alpha ^+$ such that $\alpha $ is not $\gamma $ -stable, i.e., there is a $\Sigma _{1}$ -formula $\psi $ and some parameter $\beta <\alpha $ such that $L_{\gamma }\models \psi (\beta )$ , but $L_{\alpha }\not \models \psi (\beta )$ . The remainder of the proof is an adaptation of the proof of Corollary 17 presented in Simpson [Reference Simpson11]:

Let $\phi $ be the $\Sigma ^{1}_{1}$ statement expressing that there is a model $(M,E)$ of ${\mathsf {KP}} + V = L$ end-extending $L_{\alpha +1}$ such that for some $\gamma ^{\prime } \in M$ with $\gamma ^{\prime }<\alpha ^{+M}$ , $L_{\gamma ^{\prime }}^{M}\models \psi (\beta )$ and, moreover, if $\gamma ^{\prime }$ is least such, then $M\models $ “ $\alpha $ is $({<}\gamma ^{\prime })$ -stable.” Then $L_{\alpha }\models \phi $ . By choice of $\alpha $ , there is an s-reflecting ordinal $\tau <\alpha $ such that $L_{\tau }\models \phi $ . This means that there is a model $(M,E)$ of ${\mathsf {KP}} + V = L$ end-extending $L_{\tau +1}$ such that for some $\gamma ^{\prime }\in M$ with $\gamma ^{\prime } < \tau ^{+M}$ , $L^{M}_{\gamma ^{\prime }}\models \psi (\beta )$ and, for the least such $\gamma ^{\prime }$ , we have $M\models $ “ $\tau $ is $({<}\gamma ^{\prime })$ -stable.” Since $\psi $ is $\Sigma _{1}$ and $L_{\alpha }\not \models \psi (\beta )$ , $\gamma ^{\prime }$ must belong to the illfounded part of M. So $\tau $ is s-reflecting and, as in the proof of Proposition 16, $\tau $ is $\pi $ -reflecting. By taking conjunctions as before, one sees that every $\Sigma ^{1}_{1}$ sentence satisfied by $L_{\alpha }$ is satisfied by some $(s\wedge \pi )$ -reflecting $\tau <\alpha $ , as was to be shown. ⊣

Proof. The conclusion of the theorem follows from Lemma 25 if $\alpha $ is $\pi s$ -reflecting, so we may assume that it is not.

Since $\alpha $ is not $\pi s$ -reflecting, it is not $\alpha ^+$ -stable on s, so there is a least $\beta <\alpha ^+$ and a $\Sigma _{1}$ -formula $\exists x\,\phi (y,x)$ such that $L_{\beta }\models \exists x\,\phi (L_{\alpha },x)$ and whenever $\gamma <\alpha $ and $\gamma $ is s-reflecting, then $L_{\gamma ^+} \not \models \exists x\,\phi (L_{\gamma },x)$ . Let $\psi $ be the formula expressing that there is a model M of ${\mathsf {KP}} + V = L$ such that

1. 1. M contains $\alpha $ .

2. 2. $M\models $ “ $\exists x\, \phi (L_{\alpha },x)$ and, letting $\beta $ be least such that $M\models \phi (L_{\alpha },a)$ for some $a \in L_{\gamma }$ , $\alpha $ is ${<}\beta $ -stable on s.”

Since $\alpha $ is $\sigma s$ -reflecting, there is an s-reflecting $\tau <\alpha $ with $L_{\tau }\models \psi $ , as witnessed by some model N which end-extends $L_{\tau ^+}$ . Let $\gamma $ be N-least such that $N\models \exists x\in L_{\gamma }\, \phi (L_{\tau },x)$ . Then, we cannot have $\gamma < \tau ^+$ , for otherwise $\tau $ is an s-reflecting ordinal such that $L_{\tau ^+}\models \exists x\, \phi (L_{\tau },x)$ , contradicting the choice of $\phi $ . Thus, $\gamma $ belongs to the illfounded part of N and, in N, $\tau $ is ${<}\gamma $ -stable on s. Since $\tau $ is recursively inaccessible (this can be assumed also if $s = \varnothing $ ), N is correct about s-reflection below $\tau $ , so an argument as before shows that $\tau $ is $\tau ^+$ -stable on s and thus $\pi s$ -reflecting.⊣

Proof. Let $\theta (L_{\alpha })$ be a $\Sigma _{1}$ sentence with parameters in $L_{\alpha }$ , say, of the form $\exists x\,\theta _{0}(x,L_{\alpha })$ . Let $\eta <\delta _{\alpha }$ and $b \in L_{\eta }$ be such that $L_{\delta _{\alpha }}\models \theta _{0}(b,L_{\alpha })$ . Since $\eta <\delta _{\alpha }$ , there is an $\alpha $ -recursive wellorder R of length $\eta $ . Let $\psi $ be the sentence asserting the existence of a model M of ${\mathsf {KP}}$ such that

1. 1. M end-extends $L_{\alpha +1}$ and $M\models $ “ $\alpha ^+$ exists”;

2. 2. in M, R is isomorphic to an ordinal $\eta _{0}$ and there is $b_{0} \in L_{\eta _{0}}^{M}$ such that $L_{\eta _{0}}^{M} \models \theta _{0}(b_{0},L_{\alpha })$ .

Then $L_{\alpha }\models \psi $ . Moreover, $\psi $ is $\Sigma ^{1}_{1}$ so, by reflection, there is an s-reflecting $\tau <\alpha $ such that $L_{\tau }\models \psi $ , as witnessed by some model N which end-extends $L_{\tau ^+}$ . Now, in N, $L_{\tau ^+}^{N}\models \theta _{0}(b_{0}, L_{\tau })$ for some $b_{0} \in L^{N}_{\eta _{0}}$ , where $\eta _{0}$ is some N-ordinal isomorphic to $R_{\tau }$ . However, $R_{\tau }\subset R$ , since R is $\alpha $ -recursive, and $\theta _{0}$ is $\Sigma _{0}$ , so we really have $L_{\tau ^+}\models \theta (L_{\tau })$ , as desired. ⊣

Proof. Suppose $\alpha $ is $\pi $ -reflecting on $\sigma s$ -reflecting ordinals but not $\sigma $ -reflecting. Let $\psi $ be the statement expressing that whenever $(M,E)$ is an end-extension of $L_{\alpha +1}$ satisfying ${\mathsf {KPi}}$ ,Footnote ¹ then $M\models $ “ $\alpha $ is not $\sigma s$ -reflecting.” This sentence is $\Pi ^{1}_{1}$ and thus cannot be satisfied by $L_{\alpha }$ , for otherwise it would be reflected to a $\sigma s$ -reflecting ordinal. But clearly $L_{\beta }$ cannot satisfy $\psi $ if $\beta $ is $\sigma s$ -reflecting.

Thus, $L_{\alpha }\not \models \psi $ , so there is a model M of ${\mathsf {KPi}}$ end-extending $L_{\alpha +1}$ such that

For ordinals $\tau <\alpha $ , whether $\tau $ is t-reflecting is computed correctly by M, for any reflection pattern t. By Lemma 31 applied within M,

Let $\phi $ be a $\Pi ^{1}_{1}$ statement, and $a \in L_{\alpha }$ be a parameter such that $L_{\alpha } \models \phi (a)$ . By Barwise–Gandy–Moschovakis [Reference Barwise, Gandy and Moschovakis6], there is a $\Sigma _{1}$ formula $\phi ^{*}$ such that for all admissible $\beta $ with $a \in L_{\beta }$ , $L_{\beta }\models \phi (a)$ if, and only if, $L_{\beta ^+}\models \phi ^{*}(a,L_{\beta })$ ; thus, $L_{\alpha ^+} \models \phi ^{*}(a,L_{\alpha })$ . Let $b \in L_{\alpha ^+}$ be a witness for $\phi ^{*}$ and let $\gamma <\alpha ^+$ be large enough so that $b \in L_{\gamma }$ . Since $\alpha $ is not $\sigma ^{1}_{1}$ -reflecting (in the real world), $\delta _{\alpha } = \alpha ^+$ , and thus

$$ \begin{align*}\gamma < \delta_{\alpha}^{M}.\end{align*} $$

Since $\phi ^{*}$ is $\Sigma _{1}$ ,

so by the $\delta _{\alpha }$ -stability of $\alpha $ on s within M, there is an s-reflecting $\tau <\alpha $ such that

Since $\tau ^+ < \alpha $ , we really do have

$$ \begin{align*}L_{\tau^+}\models \phi^{*}(a,L_{\tau}),\end{align*} $$

and so $L_{\tau }\models \phi (a)$ . This completes the proof of the theorem. ⊣

Proof. Suppose $\alpha $ is $\pi \sigma (t\wedge \pi s)$ -reflecting but not $\pi s$ -reflecting. Let $\theta $ be a $\Sigma ^{1}_{1}$ sentence with parameters in $L_{\alpha }$ such that

$$ \begin{align*}L_{\alpha}\models\theta;\end{align*} $$

we need to find a $(t\wedge \pi s)$ -reflecting $\tau <\alpha $ such that

$$ \begin{align*}L_{\tau}\models\theta.\end{align*} $$

By Barwise–Gandy–Moschovakis [Reference Barwise, Gandy and Moschovakis6], there is a $\Pi _{1}$ formula $\theta ^{*}(a)$ such that for every admissible $\beta $ containing the parameters of $\theta $ , $L_{\beta }\models \theta $ if, and only if, $L_{\beta ^+}\models \theta ^{*}(L_{\beta }).$ In particular,

$$ \begin{align*}L_{\alpha^+}\models\theta^{*}(L_{\alpha}).\end{align*} $$

Since $\alpha $ is not $\pi s$ -reflecting, there is a least $\gamma <\alpha ^+$ such that $\alpha $ is not $\gamma $ -stable on s. Because $\theta ^{*}$ is $\Pi _{1}$ ,

$$ \begin{align*}L_{\gamma}\models\theta^{*}(L_{\alpha}).\end{align*} $$

Let $\phi $ be the sentence asserting the non-existence of a model M of ${\mathsf {KPi}} + V = L$ end-extending $L_{\alpha +1}$ in which $\alpha $ is $\sigma (t\wedge \pi s)$ -reflecting. This is a $\Pi ^{1}_{1}$ sentence and thus cannot be satisfied by any $\pi \sigma (t\wedge \pi s)$ -reflecting ordinal and, in particular, by $\alpha $ . Thus, there is a model M of ${\mathsf {KPi}} + V = L$ end-extending $L_{\alpha +1}$ and such that

Let $\chi $ be the sentence asserting the existence of a model N of ${\mathsf {KP}} + V = L$ such that

1. 1. N contains $\alpha $ ;

2. 2. in N, letting $\gamma $ be least such that $\alpha $ is not $\gamma $ -stable on s, we have

   

Since $\gamma <\alpha ^+$ and M must end-extend $L_{\alpha ^+}$ , $\gamma \in M$ and M is correct about $\gamma $ being the least ordinal at which $\alpha $ fails to be stable on s. Thus, we have

as witnessed, say, by $L^{M}_{\alpha ^+}$ . Within M, $\alpha $ is $\sigma (t\wedge \pi s)$ -reflecting and thus there is some $\tau <\alpha $ such that

and so we really do have

$$ \begin{align*}L_{\tau}\models \chi.\end{align*} $$

Moreover, M is correct about reflection below $\alpha $ , so we may assume that $\tau $ is $(t\wedge \pi s)$ -reflecting. By the definition of $\chi $ , there is a model N of ${\mathsf {KP}} + V = L$ such that

1. 1. N contains $\tau $ ;

2. 2. in N, letting $\gamma $ be least such that $\tau $ is not $\gamma $ -stable on s, we have

   

Since $\tau $ is $\pi s$ -reflecting, it is $\tau ^+$ -stable on s, and thus $\gamma $ cannot be a true ordinal smaller than $\tau ^+$ . By Ville’s Theorem, N must end-extend $L_{\tau ^+}$ . Because $\theta ^{*}$ is $\Pi _{1}$ , it follows that

$$ \begin{align*}L_{\tau^+} \models \theta^{*}(L_{\tau}),\end{align*} $$

and thus, that

$$ \begin{align*}L_{\tau}\models\theta,\end{align*} $$

as was to be shown. ⊣

Proof. Suppose $\alpha $ is $\sigma \pi \sigma \pi s$ -reflecting. For every $\Sigma ^{1}_{1}$ -sentence $\phi $ satisfied by $L_{\alpha }$ , one can find some $\pi \sigma \pi s$ -reflecting $\beta <\alpha $ such that $L_{\beta }\models \phi $ and $\beta $ is not $\sigma $ -reflecting. (To see this, use $\sigma \pi \sigma s$ -reflection to find some $\pi \sigma \pi s$ -reflecting ordinal $\beta _{0}$ such that $L_{\beta _{0}}\models \phi $ . If $\beta _{0}$ is not $\sigma $ -reflecting, then we are done. Otherwise, $\beta _{0}$ is $\sigma $ -reflecting and $\pi \sigma \pi s$ -reflecting, thus $\sigma \pi \sigma \pi s$ -reflecting by Lemma 25. Hence, there is some $\pi \sigma \pi s$ -reflecting $\beta _{1}<\beta _{0}$ such that $L_{\beta _{1}}\models \phi $ . Proceeding this way, one eventually finds some $\pi \sigma \pi s$ -reflecting $\beta _{n}$ such that $L_{\beta _{n}}\models \phi $ and $\beta _{n}$ is not $\sigma $ -reflecting. Then $\beta = \beta _{n}$ is as desired.) By Theorem 32, $\beta $ is $\pi \pi s$ -reflecting. By Lemma 23, $\beta $ is $\pi s$ -reflecting, as desired. ⊣

Proof. Recursively, we assign ordinals to reflection patterns without conjunction: we assign the ordinal $\omega ^{n}$ to the pattern

$$ \begin{align*}\sigma^{n}.\end{align*} $$

In particular, the ordinal $1$ is assigned to the empty pattern. If s and t are (possibly empty) patterns to which ordinals $o(s)$ and $o(t)$ have been assigned, we assign the ordinal

$$ \begin{align*}o(s) + o(t)\end{align*} $$

to the pattern

$$ \begin{align*}t \pi s.\end{align*} $$

We have to check that this assignment is well defined, in the sense that $s\equiv t$ if and only if $o(s) = o(t)$ . Note that if $0< n < m$ , then, on the one hand,

$$ \begin{align*}\alpha + \omega^{n} + \omega^{m} = \alpha + \omega^{m},\end{align*} $$

while, on the other,

$$ \begin{align*} \sigma^{m} \pi s &\to \sigma^{m-n}\sigma^{n} \pi s\\ &\to \sigma^{m-n}(\sigma^{n} \pi s \wedge \pi) & \text{by Lemma 25(3)}\\ &\to \sigma^{m-n}(\sigma^{n}\pi s \wedge \pi\sigma^{n} \pi s) & \text{by Lemma 25(1)}\\ &\to \sigma^{m-n}(\sigma^{n}(\pi s \wedge \pi\sigma^{n} \pi s) \wedge \pi\sigma^{n}\pi s) & \text{by Lemma 25(2)}\\ &\to \sigma^{m}(\pi s \wedge \pi\sigma^{n} \pi s) \\ &\to \sigma^{m} \pi s \wedge \sigma^{m}\pi\sigma^{n} \pi s\\ &\to \sigma^{m} \pi s. \end{align*} $$

Thus, every $\sigma ^{m} \pi s$ -reflecting ordinal is also $\sigma ^{m} \pi \sigma ^{n} \pi s$ -reflecting when $n < m$ ; the converse is also true, by Lemmas 23 and 40. By Lemma 23 and Theorem 29, we have

$$ \begin{align*}\sigma\pi^{n} \sigma \equiv \sigma\pi^{m}\sigma,\end{align*} $$

for any pair of nonzero numbers n and m. It follows that every linear pattern of reflection is equivalent to one of the form

(2) $$ \begin{align} \pi^{m} \sigma^{k_{0}}\pi\sigma^{k_{1}}\pi\cdots\pi\sigma^{k_{l}}, \end{align} $$

with $0 < k_{0} \leq k_{1} \leq \cdots \leq k_{l}$ , and moreover there is a unique choice of such $m,l, k_{0}, \dots , k_{l}$ . Observe that transforming a linear reflection pattern into one of the form (2) preserves the assigned ordinal. We conclude that if s and t are two linear patterns of reflection and $s \equiv t$ , then $o(s) = o(t)$ . Similarly, if $o(s) = o(t)$ , then s and t are each equivalent to the same pattern of the form (2) and hence to each other, thus proving the claim that the assignment of ordinals is well defined.

Now that we know that the assignment is well defined, we can prove that for all linear patterns s and t, we have $s < t$ if, and only if, $o(s) < o(t)$ . If $o(s) < o(t)$ , let $\gamma $ be so that $o(s) + \gamma = o(t)$ and let r be a pattern so that $o(r) = \gamma $ . Then, $o(r\pi s) = o(t)$ , so that $r \pi s\equiv t$ . It follows that every t-reflecting ordinal is a limit of s-reflecting ordinals.

Conversely, suppose that $s< t$ . We cannot have $o(s) = o(t)$ , as this would contradict well-definedness; we cannot have $o(t) < o(s)$ , as this would contradict the conclusion of the previous paragraph. Therefore, we must have $o(s) < o(t)$ .⊣

Proof. To prove the theorem, we shall prove that for every reflection pattern s there is some $n\in \mathbb {N}$ such that whenever $n\leq k$ , for every reflection pattern t, every $\sigma (\sigma ^{k} \wedge t)$ -reflecting ordinal is also $\sigma (\sigma ^{k}\wedge t \wedge s)$ -reflecting. If s is a reflection pattern, then for such an n we have $s < \sigma ^{n+1}$ (this follows by taking t to be trivial), so the theorem will follow.

This is done by induction on the construction of s. The case that s is of the form $\pi s^{\prime }$ is immediate from Theorem 29. Suppose s is of the form $s_{1} \wedge s_{2}$ . Use the induction hypothesis to find $n_{1},n_{2}\in \mathbb {N}$ such that whenever $n_{1}\leq k_{1}$ and $n_{2}\leq k_{2}$ , for every reflection pattern t,

$$ \begin{align*}\sigma(\sigma^{k_{1}}\wedge t)\equiv \sigma(\sigma^{k_{1}}\wedge t \wedge s_{1}),\end{align*} $$

and

$$ \begin{align*}\sigma(\sigma^{k_{2}}\wedge t)\equiv \sigma(\sigma^{k_{2}}\wedge t \wedge s_{2}).\end{align*} $$

Let $n = \max \{n_{1},n_{2}\}$ and $n\leq k$ . Then, we have

$$ \begin{align*}\sigma(\sigma^{k}\wedge t) \equiv \sigma(\sigma^{k}\wedge t \wedge s_{1})\equiv \sigma(\sigma^{k}\wedge t \wedge s_{1} \wedge s_{2}).\end{align*} $$

Finally, suppose that $s = \sigma s^{\prime }$ . Use the induction hypothesis to find $n\in \mathbb {N}$ such that whenever $n\leq k$ , for every reflection pattern t, every $\sigma (\sigma ^{k} \wedge t)$ -reflecting ordinal is also $\sigma (\sigma ^{k}\wedge t \wedge s^{\prime })$ -reflecting. Then,

$$ \begin{align*}\sigma(\sigma^{k+1} \kern1.4pt{\wedge}\kern1.4pt t) \equiv \sigma(\sigma\sigma^{k}\kern1.4pt{\wedge}\kern1.4pt t\kern-0.5pt) \equiv \sigma\big(\sigma(\sigma^{k} \kern1.4pt{\wedge}\kern1.4pt s^{\prime})\kern1.4pt{\wedge}\kern1.4pt t\kern-0.5pt\big) \equiv \sigma(\sigma^{k+1} \kern1.4pt{\wedge}\kern1.4pt \sigma s^{\prime} \kern1.4pt{\wedge}\kern1.4pt t\kern-0.5pt) \equiv \sigma(\sigma^{k+1}\kern1.4pt{\wedge}\kern1.4pt t\kern1.4pt{\wedge}\kern1.4pt s\kern-0.5pt),\end{align*} $$

as desired. ⊣

Proof. By Lemma 25,

$$ \begin{align*} \sigma\wedge\pi^{k+1} s \equiv \sigma\wedge \pi (\sigma\wedge\pi^{k} s) \equiv \sigma\wedge\pi (\sigma\pi s\wedge\pi^{k} s) \equiv \sigma\wedge\pi (\sigma\wedge\pi^{k}(s \wedge \sigma\pi s)). \end{align*} $$

We show by induction on n that

$$ \begin{align*}\sigma\wedge \pi c^{k}_{1} s \equiv \sigma\wedge \pi c^{k}_{1} c^{k}_{n} s.\end{align*} $$

Suppose that

$$ \begin{align*}\sigma\wedge \pi c^{k}_{1} s \equiv \sigma\wedge \pi c^{k}_{1} c^{k}_{n} s.\end{align*} $$

After some applications of Lemma 25, we have

$$ \begin{align*} \sigma\wedge \pi c^{k}_{1} c^{k}_{n} s &\equiv\sigma\wedge \pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s)\\ &\equiv\sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s)\wedge \pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s)\\ &\equiv\sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s)\wedge \pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s \wedge \sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s))\\ &\equiv\sigma \wedge \pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s \wedge \sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s))\\ &\equiv\sigma \wedge \pi (\sigma\wedge\pi^{k}(\sigma\pi c^{k}_{n} s \wedge \sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s))). \end{align*} $$

After some additional manipulations, we have

$$ \begin{align*} \sigma\pi (\sigma\wedge \pi^{k}\sigma\pi c^{k}_{n} s) &\equiv \sigma\pi (\sigma\wedge \pi^{k}\sigma\pi c^{k}_{n} s) \wedge \sigma\pi (\pi^{k}\sigma\pi c^{k}_{n} s)\\ &\equiv \sigma\pi (\sigma\wedge \pi^{k}\sigma\pi c^{k}_{n} s) \wedge \sigma\pi \sigma\pi c^{k}_{n} s\\ &\equiv \sigma\pi (\sigma\wedge \pi^{k}\sigma\pi c^{k}_{n} s) \wedge \sigma\pi c^{k}_{n} s\\ &\equiv \sigma\pi c^{k}_{n} s \wedge \sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s), \end{align*} $$

where the second equivalence follows from Lemma 23 and the third follows from Lemma 40. Putting this together with the previous computation,

$$ \begin{align*} \sigma\wedge \pi c^{k}_{1} c^{k}_{n} s &\equiv \sigma \wedge \pi (\sigma\wedge\pi^{k}(\sigma\pi c^{k}_{n} s \wedge \sigma\pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s)))\\ &\equiv \sigma\wedge \pi (\sigma\wedge\pi^{k}\sigma\pi (\sigma\wedge\pi^{k} \sigma\pi c^{k}_{n} s))\\ &\equiv \sigma\wedge \pi (\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n+1} s)\\ &\equiv \sigma\wedge \pi (c^{k}_{1} c^{k}_{n+1} s), \end{align*} $$

as desired. The chain of equivalences at the beginning of the proof shows that every $\sigma \wedge \pi ^{k+1}s$ -reflecting ordinal is also $\pi c^{k}_{1} s$ -reflecting (the converse is not true in general), and the equivalence just proved by induction shows that it is therefore $\pi c^{k}_{n} s$ -reflecting for any n, as desired. ⊣

Proof. By Lemma 46,

$$ \begin{align*}\sigma\wedge \pi^{k+1}s \to \pi c^{k}_{n+1}s.\end{align*} $$

By definition,

$$ \begin{align*}\pi c^{k}_{n+1}s \equiv \pi(\sigma\wedge\pi^{k}\sigma\pi c^{k}_{n} s);\end{align*} $$

in particular,

$$ \begin{align*}\sigma\wedge \pi^{k+1}s\to\pi^{k+1}\sigma\pi c^{k}_{n}s.\end{align*} $$

Applying Lemma 23 n times to see that

$$ \begin{align*}c^{k}_{n} s\to c^{l}_{n} s,\end{align*} $$

from which the result follows. ⊣

Proof. Observe that every reflection pattern of the form $\sigma \pi c^{l}_{m}$ , for $l,m\in \mathbb {N}$ , is of the form $t^{\prime } \sigma \pi $ . Thus, Lemma 40 implies that

$$ \begin{align*}\sigma\pi c^{l}_{m} \sigma \pi s \equiv \sigma\pi c^{l}_{m} s.\end{align*} $$

The result then follows from applying Corollary 47 repeatedly. ⊣

Proof. Let

$$ \begin{align*}s=\pi^{n}c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}.\end{align*} $$

Suppose first $n_{0} \neq 0$ . By Theorem 29,

$$ \begin{align*}\sigma\sigma \to \sigma(\sigma\wedge \pi^{\max\{n,k+1\}}).\end{align*} $$

By Lemmas 23 and 48,

$$ \begin{align*}\sigma(\sigma\wedge \pi^{\max\{n,k+1\}})\to \sigma(\sigma\wedge \pi^{n} \sigma\pi c^{0}_{n_{0}-1}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}),\end{align*} $$

as desired. Suppose now that $n_{0} = 0$ and let l be least such that $n_{l} \neq 0$ (if there is no such l, the lemma follows from Theorem 29). Thus,

$$ \begin{align*}s=\pi^{n}c^{l}_{n_{l}}\dots c^{k}_{n_{k}} = \pi^{n}(\sigma\wedge \pi^{l}\sigma\pi c^{l}_{n_{l}-1}\dots c^{k}_{n_{k}}).\end{align*} $$

As before,

$$ \begin{align*}\sigma\sigma\to\sigma(\sigma\wedge \pi^{l} \sigma\pi c^{l}_{n_{l}-1}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}).\end{align*} $$

By Theorem 29,

$$ \begin{align*}\sigma(\sigma\wedge \pi^{l} \sigma\pi c^{l}_{n_{l}-1}c^{1}_{n_{1}}\dots c^{k}_{n_{k}})\to \sigma\pi^{n}(\sigma\wedge \pi^{l} \sigma\pi c^{l}_{n_{l}-1}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}).\end{align*} $$

as desired. ⊣

Proof. Let

$$ \begin{align*}\pi t = \pi^{n}c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}},\end{align*} $$

and

$$ \begin{align*}\pi s = \pi^{m}c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c^{l}_{m_{l}},\end{align*} $$

where n and m are nonzero. Without loss of generality, we assume that $n_{k}$ and $m_{l}$ are also nonzero. It follows that $k \leq l$ . Suppose that $o(\pi t) < o(\pi s)$ . It will be convenient, for illustrative purposes, to consider the case that $k < l$ first. If so, it suffices to show that every $\pi (\sigma \wedge \pi ^{l})$ -reflecting ordinal which is not $\sigma $ -reflecting is $(\pi t)$ -reflecting, for then the result follows from Theorem 32 and contraction. By Lemma 48,

$$ \begin{align*} \pi(\sigma\wedge\pi^{l}) \to \pi(\sigma\wedge \pi^{l} \sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}). \end{align*} $$

By Lemma 25,

$$ \begin{align*}\pi(\sigma\wedge \pi^{l} \sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}})\to \pi(\sigma \pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}),\end{align*} $$

and, by Theorem 29,

$$ \begin{align*}\pi(\sigma \pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}})\to \pi(\sigma \pi^{n} c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}),\end{align*} $$

so that, if a $\pi (\sigma \wedge \pi ^{l})$ -reflecting ordinal is not $\sigma $ -reflecting, then it is

$$ \begin{align*}\pi(\pi^{n} c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}})\text{-reflecting,}\end{align*} $$

by Theorem 32, i.e., $\pi t$ -reflecting.

The case $k = l$ is not very different: let $i\leq k$ be greatest such that $n_{i} < m_{i}$ and notice that

$$ \begin{align*}c_{m_{i}}^{i} = c_{m_{i}-n_{i}}^{i}c_{n_{i}}^{i}.\end{align*} $$

Thus,

$$ \begin{align*}\pi t = \pi^{n}c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}t^{\prime},\end{align*} $$

where $t^{\prime } = c^{i}_{n_{i}}\dots c^{k}_{n_{k}}$ ; and

$$ \begin{align*}\pi s = \pi^{m}c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c_{m_{i}-n_{i}}^{i}t^{\prime}.\end{align*} $$

It suffices to show that every $\pi (\sigma \wedge \pi ^{i}\sigma \pi t^{\prime })$ -reflecting ordinal which is not $\sigma $ -reflecting is $\pi t$ -reflecting, for then the result follows from Theorem 32 and contraction. Lemma 48 (with $\sigma \pi t^{\prime }$ being the s in the statement) shows that

$$ \begin{align*} \pi(\sigma\wedge\pi^{i}\sigma\pi t^{\prime})\to \pi(\sigma\wedge \pi^{i}\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}\sigma\pi t^{\prime}). \end{align*} $$

By Lemma 25,

$$ \begin{align*} \pi(\sigma\wedge \pi^{i}\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}\sigma\pi t^{\prime})\to \pi(\sigma\pi^{i}\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}\sigma\pi t^{\prime}). \end{align*} $$

By contraction,

$$ \begin{align*} \pi(\sigma\pi^{i}\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}\sigma\pi t^{\prime}) &\to \pi(\sigma\pi\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}\sigma\pi t^{\prime})\\ &\to \pi(\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}\sigma\pi t^{\prime})\\ &\to \pi(\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}t^{\prime}). \end{align*} $$

By Theorem 29,

$$ \begin{align*} \pi(\sigma\pi c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}t^{\prime})\to \pi(\sigma\pi^{n} c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}t^{\prime}), \end{align*} $$

so that if a $\pi (\sigma \wedge \pi ^{i}\sigma \pi t^{\prime })$ -reflecting ordinal is not $\sigma $ -reflecting, then it is

$$ \begin{align*} \pi(\pi^{n} c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{i-1}_{n_{i-1}}t^{\prime})\text{-reflecting,} \end{align*} $$

by Theorem 32, as desired. ⊣

Proof. We restrict to the case $n = 1$ and let $s_{0} = t$ and $s_{1} = s$ , since the general case is similar. We prove $\sigma s\equiv \sigma (s\wedge t)$ , since clearly every $\sigma (s\wedge t)$ -reflecting ordinal is $(\sigma s\wedge \sigma t)$ -reflecting, and every $(\sigma s\wedge \sigma t)$ -reflecting ordinal is $\sigma s$ -reflecting. Suppose $\alpha $ is $\sigma s$ -reflecting and let $\phi $ be a $\Sigma ^{1}_{1}$ -sentence satisfied by $L_{\alpha }$ . We must find some $(s\wedge t)$ -reflecting $\beta <\alpha $ such that $L_{\beta }\models \phi $ . By Lemma 51, it suffices that $\beta $ be s-reflecting and not $\sigma $ -reflecting. $\beta $ is found as in the proof of Lemma 40:

By $\sigma s$ -reflection, there is some s-reflecting ordinal $\beta _{0}$ such that $L_{\beta _{0}}\models \phi $ . If $\beta _{0}$ is not $\sigma $ -reflecting, then we are done. Otherwise, $\beta _{0}$ is $\sigma $ -reflecting and s-reflecting, thus (because s is of the form $\pi v$ ) $\sigma s$ -reflecting by Lemma 25. Hence, there is some s-reflecting $\beta _{1}<\beta _{0}$ such that $L_{\beta _{1}}\models \phi $ . Proceeding this way, one eventually finds some s-reflecting $\beta _{n}$ such that $L_{\beta _{n}}\models \phi $ and $\beta _{n}$ is not $\sigma $ -reflecting. Then $\beta = \beta _{n}$ is as desired. ⊣

Proof. Put $t = \pi ^{k}c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c^{l}_{m_{l}}$ . The lemma is immediate unless $k \neq 0$ and there is some least $i \leq l$ such that $m_{i} \neq 0$ . Thus,

$$ \begin{align*} t &= \pi^{k}c^{i}_{m_{i}}c^{i+1}_{m_{i+1}}\dots c^{l}_{m_{l}}\\ &= \pi^{k}(\sigma \wedge \pi^{i} \sigma\pi c^{i}_{m_{i}-1}c^{i+1}_{m_{i+1}}\dots c^{l}_{m_{l}}). \end{align*} $$

Let

$$ \begin{align*}s = c^{k+i}_{m_{k+i}}c^{k+ i+1}_{m_{k+i+1}}\dots c^{l}_{m_{l}},\end{align*} $$

so that

$$ \begin{align*}t = \pi^{k}(\sigma \wedge \pi^{i} \sigma\pi c^{i}_{m_{i}-1}c^{i+1}_{m_{i+1}}\dots c^{k+i-1}_{m_{k + i-1}}s).\end{align*} $$

If all the displayed $m_{j}$ are zero, then the result follows easily; otherwise, by Lemma 25 and contraction,

$$ \begin{align*}t \equiv \pi^{k}(\sigma \wedge \pi^{i} \sigma\pi c^{i}_{m_{i}-1}c^{i+1}_{m_{i+1}}\dots c^{k + i-1}_{m_{k + i-1}}\sigma\pi s).\end{align*} $$

By Lemma 25,

$$ \begin{align*} \sigma\wedge t &\equiv \sigma \wedge \pi^{k}(\sigma \wedge \pi^{i} \sigma\pi c^{i}_{m_{i}-1}c^{i+1}_{m_{i+1}}\dots c^{k+i-1}_{m_{k+i-1}}\sigma\pi s)\\ &\equiv \sigma \wedge \pi^{k}(\pi^{i} \sigma\pi c^{i}_{m_{i}-1}c^{i+1}_{m_{i+1}}\dots c^{k+i-1}_{m_{k+i-1}}\sigma\pi s)\\ &\equiv \sigma \wedge \pi^{k+i}\sigma\pi c^{i}_{m_{i}-1}c^{i+1}_{m_{i+1}}\dots c^{k+i-1}_{m_{k+i-1}}\sigma\pi s). \end{align*} $$

By Lemma 48 on the one hand and Lemma 23 and contraction on the other,

(3) $$ \begin{align} \sigma \wedge \pi^{k+i}\sigma\pi s \equiv \sigma \wedge t, \end{align} $$

but the reflection pattern on the left-hand side is readily seen to be equivalent to a normal one. ⊣

Proof. We shall prove the following claims:

1. 1. If s is a normal reflection pattern, then so too is $\pi s$ and, if s is of the form $\pi s^{\prime }$ , $\sigma s$ is also equivalent to a normal reflection pattern.

2. 2. If s and t are normal reflection patterns, and at least one of them is not of the form $\pi v$ , then $s\wedge t$ is equivalent to a normal reflection pattern.

3. 3. If s is a quasi-normal reflection pattern which is not normal, then $\sigma s$ and $\sigma \wedge s$ are equivalent to normal reflection patterns, provided the string $\sigma \sigma $ does not occur in them.

Let us derive the conclusion of the theorem from (1) (3) by induction on the construction of s. Suppose s is quasi-normal. Then so too is $\pi s$ by definition. Suppose s is quasi-normal and the string $\sigma \sigma $ does not occur in $\sigma s$ . If s is not normal, then $\sigma s$ is normal by (3). Otherwise, if s is normal, then, since $\sigma \sigma $ does not occur in s, s must be of the form $\pi s^{\prime }$ , and thus $\sigma s$ is normal by (1). Lastly, suppose that s and t are quasi-normal reflection patterns. If none of them is normal, then $s\wedge t$ is quasi-normal by definition. Otherwise, say t is normal and s is not. If t is of the form $\pi t^{\prime }$ , then $s\wedge t$ is quasi-normal by definition. Otherwise, $t \equiv \sigma \wedge t$ , so that

$$ \begin{align*}s\wedge t \equiv \sigma\wedge s\wedge t.\end{align*} $$

$\sigma \wedge s$ is normal by (3) and thus $s\wedge t$ is normal by (2). Finally, if both s and t are normal, then either they are both of the form $\pi u$ , so that $s\wedge t$ is quasi-normal by definition, or $s\wedge t\equiv \sigma \wedge s\wedge t$ is normal by (2).

We begin with the proof of (1). Clearly, if s is equivalent to a normal reflection pattern, then so too is $\pi s$ . We only need to consider the string $\sigma s$ in the case that s is of the form

$$ \begin{align*}\pi^{m}c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c^{l}_{m_{l}},\end{align*} $$

where m is nonzero. Then, it is easy to see that

$$ \begin{align*} \sigma\pi^{m}c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c^{l}_{m_{l}} &\equiv \sigma\pi c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c^{l}_{m_{l}}\\ &\equiv c^{0}_{m_{0}+1}c^{1}_{m_{1}}\dots c^{l}_{m_{l}}. \end{align*} $$

This proves (1).

To prove (2), let

$$ \begin{align*}s = \pi^{m}c^{0}_{m_{0}}c^{1}_{m_{1}}\dots c^{l}_{m_{l}}\end{align*} $$

be as above, and let

$$ \begin{align*}t = \pi^{n}c^{0}_{n_{0}}c^{1}_{n_{1}}\dots c^{k}_{n_{k}}.\end{align*} $$

We need to show that $s\wedge t$ is equivalent to a normal reflection pattern. By the case hypothesis, at least one of m and n is zero, so that

$$ \begin{align*}s\wedge t \equiv \sigma \wedge s \wedge t.\end{align*} $$

Write $n = n_{-1}$ and $m = m_{-1}$ and let i and j be least such that $n_{i}$ and $m_{j}$ are nonzero, respectively. There are four cases to consider. The first one is that in which both i and j are equal to $0$ . Then, there are reflection patterns u and v, both normal, such that

$$ \begin{align*} t = \sigma\pi u \end{align*} $$

and

$$ \begin{align*} s = \sigma \pi v. \end{align*} $$

The conclusion then follows from Lemma 52.

We now consider the case that both i and j are nonzero. The remaining cases, in which one of them is $0$ and the other one is not, are treated similarly, and we leave them to the reader. We need to show that the pattern

$$ \begin{align*}s\wedge t \equiv \sigma \wedge s \wedge t \equiv (\sigma \wedge s) \wedge (\sigma \wedge t)\end{align*} $$

is normal. It was shown as part of Lemma 53 (see Equation (3)) that if s is normal, then $\sigma \wedge s$ has the form

$$ \begin{align*}\sigma\wedge \pi^{p}\sigma\pi c^{p}_{m_{p}-1}\dots c^{l}_{m_{l}},\end{align*} $$

for some suitably chosen $p\in \mathbb {N}$ . Similarly, $\sigma \wedge t$ has the form

$$ \begin{align*}\sigma\wedge \pi^{q}\sigma\pi c^{q}_{n_{q}-1}\dots c^{k}_{n_{k}},\end{align*} $$

for some suitably chosen $q\in \mathbb {N}$ . Put $u = c^{p}_{m_{p}-1}\dots c^{l}_{m_{l}}$ and $v = c^{q}_{n_{q}-1}\dots c^{k}_{n_{k}}$ , so

$$ \begin{align*}\sigma\wedge s \equiv \sigma\wedge \pi^{p}\sigma\pi u\end{align*} $$

and

$$ \begin{align*}\sigma\wedge t \equiv \sigma\wedge \pi^{q}\sigma \pi v.\end{align*} $$

By Lemma 52 one of u and v, say w, has the property that

$$ \begin{align*}\sigma \pi w \equiv \sigma(\pi u \wedge \pi v).\end{align*} $$

Let $r = \max \{p,q\}$ .⊣

Claim 56. $\sigma \wedge \pi ^{r} \sigma \pi w \equiv s\wedge t$ .

Proof. We have to show that

$$ \begin{align*}\sigma \wedge \pi^{r} \sigma\pi w \equiv \sigma\wedge \pi^{p}\sigma\pi u \wedge \pi^{q}\sigma\pi v.\end{align*} $$

By direct computation, using the usual tools:

$$ \begin{align*} \sigma \wedge \pi^{r} \sigma\pi w &\equiv \sigma \pi^{r} \sigma\pi w \wedge \pi^{r} \sigma\pi w\\ &\equiv \sigma \pi \sigma\pi w \wedge \pi^{r} \sigma\pi w\\ &\equiv \sigma\pi w \wedge \pi^{r} \sigma\pi w\\ &\equiv \sigma(\pi u \wedge \pi v) \wedge \pi^{r} \sigma\pi w\\ &\equiv \sigma \wedge \pi^{r} (\sigma\pi w\wedge \sigma(\pi u \wedge \pi v))\\ &\equiv \sigma \wedge \pi^{r} \sigma\pi w\wedge \pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v. \end{align*} $$

Since w is either u or v, we have

$$ \begin{align*} \sigma \wedge \pi^{r} \sigma\pi w\wedge \pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v &\equiv \sigma \wedge \pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v. \end{align*} $$

Now, on the one hand, $r = \max \{p,q\}$ and this implies that

$$ \begin{align*}\sigma \wedge \pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v\to \sigma \wedge \pi^{p} \sigma \pi u \wedge \pi^{q}\sigma \pi v.\end{align*} $$

On the other hand,

$$ \begin{align*} \sigma \wedge \pi^{p} \sigma \pi u \wedge \pi^{q}\sigma \pi v &\to \sigma (\pi^{p} \sigma \pi u \wedge \pi^{q}\sigma \pi v)\\ &\to \sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v), \end{align*} $$

where the last implication follows by Theorem 29. Since r is one of p and q, we have

$$ \begin{align*}\sigma \wedge \pi^{p} \sigma \pi u \wedge \pi^{q}\sigma \pi v\to (\sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v)\wedge \pi^{r}).\end{align*} $$

With one last computation, we obtain:

$$ \begin{align*} \sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v) \wedge \pi^{r} &\to \sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v) \wedge \pi^{r} \sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v)\\ &\to \sigma \wedge \pi^{r} \sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v)\\ &\to \sigma \wedge \pi^{r} \sigma \pi^{r} \sigma \pi u \wedge \pi^{r} \sigma \pi^{r}\sigma \pi v\\ &\to \sigma \wedge \pi^{r} \sigma \pi u \wedge \pi^{r} \sigma \pi v\\ &\to \sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v) \wedge \pi^{r}, \end{align*} $$

so that

$$ \begin{align*}\sigma (\pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v) \wedge \pi^{r}\equiv \sigma \wedge \pi^{r} \sigma \pi u \wedge \pi^{r} \sigma \pi v \equiv \sigma \wedge \pi^{p} \sigma \pi u \wedge \pi^{q}\sigma \pi v.\end{align*} $$

We have shown that

$$ \begin{align*} \sigma\wedge \pi^{r} \sigma \pi w &\equiv \sigma \wedge \pi^{r} \sigma \pi u \wedge \pi^{r}\sigma \pi v\\ &\equiv \sigma \wedge \pi^{p} \sigma \pi u \wedge \pi^{q}\sigma \pi v, \end{align*} $$

as had been claimed. ⊣

Proof. This is immediate from the definition. ⊣

Proof. This is proved by induction on the construction of s. If s is a conjunction of two normal reflection patterns, both of the form $\pi u$ , then the lemma follows from Lemma 51. If s is of the form $s_{0}\wedge s_{1}$ , where, say, $s_{1}$ is not normal, then $o(s) = \max \{o(s_{0}),o(s_{1})\}$ . Both $o(s_{0})$ and $o(s_{1})$ are successor ordinals, so the normal reflection patterns $t_{0}$ and $t_{1}$ such that $o(t_{0}) = o(s_{0})$ and $o(t_{1}) = o(s_{1})$ are both of the form $\pi t^{\prime }$ . Assume without loss of generality that $o(s_{0}) < o(s_{1})$ . By Lemma 51, every $t_{1}$ -reflecting ordinal is either $\sigma $ -reflecting or $t_{0}$ -reflecting. By the induction hypothesis applied to each of $s_{0}$ and $s_{1}$ , every $t_{1}$ -reflecting ordinal is either $\sigma $ -reflecting or both $s_{0}$ -reflecting and $s_{1}$ -reflecting.

Finally, if s is of the form $\pi s^{\prime }$ , for some $s^{\prime }$ which is quasi-normal but not normal, then t is of the form $\pi t^{\prime }$ , for some $t^{\prime }$ . By induction hypothesis, every $t^{\prime }$ -reflecting ordinal is either $\sigma $ -reflecting or $s^{\prime }$ -reflecting. Suppose $\alpha $ is t-reflecting but not $\sigma $ -reflecting. There are two cases: if $\alpha $ is $\pi $ -reflecting on ordinals which are $t^{\prime }$ -reflecting but not $\sigma $ -reflecting, then $\alpha $ is $\pi $ -reflecting on ordinals which are $s^{\prime }$ -reflecting, by induction hypothesis, so $\alpha $ is $\pi s^{\prime }$ -reflecting.

The other case is a bit subtler: suppose $\alpha $ is $\pi (\sigma \wedge t^{\prime })$ -reflecting. $s^{\prime }$ is quasi-normal but not normal, so $o(s^{\prime })$ is a successor ordinal, and thus $t^{\prime }$ is of the form $\pi t^{\prime \prime }$ . It follows that $\alpha $ is $\pi (\sigma t^{\prime } \wedge t^{\prime })$ -reflecting and in particular $\pi \sigma t^{\prime }$ -reflecting. We claim that $\alpha $ is $\pi $ -reflecting on ordinals which are $\sigma $ -reflecting on ordinals which are $t^{\prime }$ -reflecting but not $\sigma $ -reflecting. To see this, let $\phi $ be any $\Pi ^{1}_{1}$ sentence satisfied by $L_{\alpha }$ and find a $\sigma t^{\prime }$ -reflecting ordinal $\beta <\alpha $ such that $L_{\beta }$ satisfies $\phi $ . Now, let $\psi $ be a $\Sigma ^{1}_{1}$ sentence satisfied by $L_{\beta }$ . We must find some $\gamma <\beta $ which is $t^{\prime }$ -reflecting and not $\sigma $ -reflecting and such that $L_{\gamma }\models \psi $ . By $\sigma t^{\prime }$ -reflection, there is some $\gamma _{0}<\beta $ which is $t^{\prime }$ -reflecting and such that $L_{\gamma _{0}}\models \psi $ . If $\gamma _{0}$ is not $\sigma $ -reflecting, then we are done. Otherwise, $\gamma _{0}$ is $\sigma $ -reflecting and $t^{\prime }$ -reflecting, thus $\sigma t^{\prime }$ -reflecting. Hence, there is $\gamma _{1}<\gamma _{0}$ which is $t^{\prime }$ -reflecting and such that $L_{\gamma _{1}}\models \psi $ . Since there cannot be an infinite descending sequence of ordinals, this procedure eventually produces a $\gamma $ as desired. Thus, $\alpha $ is $\pi $ -reflecting on ordinals which are $\sigma $ -reflecting on ordinals which are $s^{\prime }$ -reflecting (by induction hypothesis), i.e., $\alpha $ is $\pi \sigma s^{\prime }$ -reflecting. Since $\alpha $ is not $\sigma $ -reflecting (by assumption), it is $\pi s^{\prime }$ -reflecting, by Theorem 32, as desired. ⊣

Proof. Let us assume that s is not normal, for otherwise the result is trivial. That the least t-reflecting ordinal is s-reflecting follows from Lemma 59 and the observation that the least t-reflecting ordinal is not $\sigma $ -reflecting, by Lemmas 25 and 58. That the least s-reflecting ordinal is t-reflecting is immediate from the definition of $o(s)$ . ⊣

Proof. If a reflection pattern contains the string $\sigma \sigma $ , then naturally, it cannot be strictly smaller than $\sigma \sigma $ . If it does not contain the string $\sigma \sigma $ , then it is equivalent to a quasi-normal reflection pattern by Theorem 55. By Corollary 60, the least ordinal which reflects according to a quasi-normal reflection pattern is the least ordinal which reflects according to some normal reflection pattern, and these ordinals are all smaller than the least $\sigma \sigma $ -reflecting ordinal, by Lemma 50. Thus, the rank of $\sigma \sigma $ in the order of reflection is the order-type of the set of normal reflection patterns. An easy induction using Lemma 48 shows that, for normal reflection patterns u and v, $u < v$ if, and only if, $o(u) < o(v)$ , so the result follows. ⊣

Question 62. What is the length of the order of reflection?

Proof. Since

$$ \begin{align*}L_{\alpha}\prec_{1} L_{\alpha^+},\end{align*} $$

it follows that $\alpha $ is $\pi $ -reflecting. By Gostanian’s theorem [Reference Gostanian8] mentioned in the introduction, $\alpha $ is $\sigma $ -reflecting. Inductively, suppose $\alpha $ is s-reflecting and let $\psi $ be a $\Sigma ^{1}_{1}$ sentence such that

$$ \begin{align*}L_{\alpha}\models \psi.\end{align*} $$

Choose a $\Pi _{1}$ sentence $\psi ^{*}$ such that for all admissible $\beta $ containing all relevant parameters,

$$ \begin{align*}L_{\beta}\models\psi,\end{align*} $$

if, and only if,

$$ \begin{align*}L_{\beta^+}\models\psi^{*}(L_{\beta}),\end{align*} $$

so that, in particular,

$$ \begin{align*}L_{\alpha^+}\models\psi^{*}(L_{\alpha}).\end{align*} $$

Then, from the point of view of $L_{\alpha ^++1}$ , there are admissible sets $L_{\alpha }$ and $L_{\alpha ^+}$ such that

1. 1. $\alpha $ is s-reflecting (this is a first-order statement about $L_{\alpha ^+}$ ),

2. 2. $L_{\alpha ^+}\models \psi ^{*}(L_{\alpha })$ .