Proof. Note that we can rephrase (i) as $s \sim _\alpha t \Rightarrow s \sim _{\alpha +1} t$ . We prove this by transfinite induction on $\alpha $ . For $\alpha =0$ this is clear because $s \simeq t$ implies $s x \simeq t x$ for every x. For a successor $\alpha +1$ , suppose that $s \sim _{\alpha +1} t$ , i.e., $\forall x \; s x \sim _\alpha t x$ . By the induction hypothesis, we then have $\forall x \forall y \; s x y \sim _\alpha t x y$ , hence $s x \sim _{\alpha +1} t x$ for every x. Finally, suppose that $s \sim _\alpha t$ for $\alpha $ a limit. Then $s \sim _\beta t$ for some $\beta <\alpha $ , so by the induction hypothesis $s x \sim _\beta t x$ for every x, hence $s x \sim _\alpha t x$ for every x by the definition of $\sim _\alpha $ .

For (ii), we again proceed by transfinite induction on $\alpha $ . For $\alpha =0$ this is trivial. The case where $\alpha $ is a successor follows from 1 and the induction hypothesis. Finally, if $\alpha $ is a limit and $s \sim _\beta t$ with $\beta \leqslant \alpha $ , then $s \sim _\alpha t$ by the definition of $\sim _\alpha $ .⊣

Proof. Observe that if t is a closed term and $t {\uparrow }$ , then $K t {\uparrow }$ as well, and therefore also $K t x {\uparrow }$ for every $x \in \mathcal {A}$ . Thus $K t x \simeq t$ for every closed term t and every $x \in \mathcal {A}$ , even when $t {\uparrow }$ .

Let s and t be closed terms. First suppose that $s \sim _\alpha t$ . Then

$$ \begin{align*} K s x \sim_0 s \sim_\alpha t \sim_0 K t x, \end{align*} $$

for every x, so $K s \sim _{\alpha + 1} K t$ .

Conversely, suppose that $K s \sim _{\alpha + 1} K t$ . Choose any $x \in \mathcal {A}$ . Then

$$ \begin{align*} s \sim_0 K s x \sim_\alpha K t x \sim_0 t, \end{align*} $$

so $s \sim _\alpha t$ .⊣

Proof. Since $\mathcal {A}$ is not extensional, the relations $\sim _0$ and $\sim _1$ differ on $\mathcal {A}$ . Suppose that $d,e\in \mathcal {A}$ are such that $d\neq e$ but $d\sim _1 e$ .

Inductively define elements $f_n$ and $g_n$ of $\mathcal {A}$ for $n \in \omega $ as follows, using the combinator K.

$$ \begin{align*} f_0 &= d, & g_0 &= e, \\ f_{n+1} &= K f_n, & g_{n+1} &= K g_n. \end{align*} $$

We have that $f_0 \not \sim _0 g_0$ and $f_0 \sim _1 g_0$ by the choice of $f_0$ and $g_0$ . By induction on n and Lemma 3.3, it follows that $f_n \not \sim _n g_n$ and $f_n \sim _{n+1} g_n$ for every n. Thus $\sim _{n+1}$ is a strictly weaker relation than $\sim _n$ for every n.

It also follows that $\sim _\omega $ strictly weaker than $\sim _n$ for every n. Suppose that there is a fixed n such that $f\sim _\omega g \Rightarrow f\sim _n g$ for every f and g. Then

$$ \begin{align*}f\sim_{n+1} g \Rightarrow f\sim_\omega g \Rightarrow f\sim_n g, \end{align*} $$

for every f and g, contradicting that $\sim _{n+1}$ is strictly weaker than $\sim _n$ .⊣

Proof. Let $\Omega $ denote the set of closed terms over $\mathcal {A}$ , and view each relation $\sim _\alpha $ as a subset of $\Omega \times \Omega $ . Theorem 3.2 tells us that these relations form an ascending sequence: $\alpha \leqslant \gamma \Rightarrow {\sim _\alpha } \subseteq {\sim _\gamma }$ for all ordinals $\alpha $ and $\gamma $ . Therefore there must be an $\alpha < |\Omega \times \Omega |^+$ for which ${\sim _\alpha } = {\sim _{\alpha +1}}$ because ${\sim _\alpha } \neq {\sim _{\alpha +1}}$ implies that there is an $(s, t) \in \Omega \times \Omega $ such that $s \sim _{\alpha +1} t$ but $(\forall \beta \leqslant \alpha )(s \not \sim _\beta t)$ , and there are only $|\Omega \times \Omega |$ many elements of $\Omega \times \Omega $ to choose among.

Now suppose that ${\sim _{\beta +1}} = {\sim _{\beta +2}}$ for some ordinal $\beta $ . We show that ${\sim _\beta } = {\sim _{\beta +1}}$ as well. It follows that the least $\alpha $ such that ${\sim _\alpha } = {\sim _{\alpha +1}}$ cannot be a successor. Consider closed terms s and t. We already know that $s \sim _\beta t \Rightarrow s \sim _{\beta +1} t$ . So suppose that $s \sim _{\beta +1} t$ . Then $K s \sim _{\beta +2} K t$ by Lemma 3.3, so $K s \sim _{\beta +1} K t$ by the assumption that ${\sim _{\beta +1}} = {\sim _{\beta +2}}$ . Thus $s \sim _\beta t$ , again by Lemma 3.3. Therefore $s \sim _{\beta +1} t \Rightarrow s \sim _\beta t$ , so ${\sim _\beta } = {\sim _{\beta +1}}$ .

Thus there is an $\alpha $ such that ${\sim _\alpha } = {\sim _{\alpha +1}}$ , and the least such $\alpha $ is either $0$ or a limit ordinal.⊣

Proof. By effective transfinite recursion, for every $\alpha < {\omega _1^{\textit {CK}}}$ we produce $f_\alpha , g_\alpha \in \mathcal {K}_1$ such that $f \not \sim _\alpha g$ and $f \sim _{\alpha + 1} g$ . Therefore ${\sim _\alpha } \neq {\sim _{\alpha + 1}}$ for all $\alpha < {\omega _1^{\textit {CK}}}$ .

The pca $\mathcal {K}_1$ is not extensional, so for the base case $\alpha = 0$ we may choose $f_0, g_0 \in \mathcal {K}_1$ such that $f_0 \not \sim _0 g_0$ and $f_0 \sim _1 g_0$ .

The successor case is similar to the proof of Theorem 3.4. Suppose that $\alpha = \beta + 1$ is a successor. By effective transfinite recursion, compute $f_\beta , g_\beta \in \mathcal {K}_1$ such that $f_\beta \not \sim _\beta g_\beta $ and $f_\beta \sim _\alpha g_\beta $ . Let $f_\alpha = K f_\beta $ and $g_\alpha = K g_\beta $ . Then $f_\alpha \not \sim _\alpha g_\alpha $ and $f_\alpha \sim _{\alpha + 1} g_\alpha $ by Lemma 3.3.

Suppose that $\alpha $ is a limit, and let $\beta _0 < \beta _1 < \cdots $ be a computable sequence of ordinals converging to $\alpha $ . By effective transfinite recursion, we can uniformly compute indices $f_{\beta _n}$ and $g_{\beta _n}$ such that $f_{\beta _n} \not \sim _{\beta _n} g_{\beta _n}$ and $f_{\beta _n} \sim _{\beta _n + 1} g_{\beta _n}$ for all $n \in \omega $ . Thus we can compute $f_\alpha $ and $g_\alpha $ so that $f_\alpha \cdot n = f_{\beta _n}$ and $g_\alpha \cdot n = g_{\beta _n}$ for all n. To see that $f_\alpha \not \sim _\alpha g_\alpha $ , consider a $\beta < \alpha $ , and let n be such that $\beta < \beta _n < \alpha $ . Then $f_\alpha \cdot n = f_{\beta _n} \not \sim _{\beta _n} g_{\beta _n} = g_\alpha \cdot n$ . Thus there is an n such that $f_\alpha \cdot n \not \sim _{\beta _n} g_\alpha \cdot n$ . Therefore $f_\alpha \not \sim _{\beta _n + 1} g_\alpha $ , so also $f_\alpha \not \sim _\beta g_\alpha $ . Thus $f_\alpha \not \sim _\beta g_\alpha $ for every $\beta < \alpha $ , so $f_\alpha \not \sim _\alpha g_\alpha $ . On the other hand, for every n, $f_\alpha \cdot n = f_{\beta _n} \sim _{\beta _n + 1} g_{\beta _n} = g_\alpha \cdot n$ , and therefore $f_\alpha \cdot n \sim _\alpha g_\alpha \cdot n$ . Thus $f_\alpha \sim _{\alpha + 1} g_\alpha $ . This completes the proof.⊣

Proof. We define $f \neq g$ so that for every x, $fx{\downarrow } = f$ and $gx{\downarrow } = g$ . This clearly implies that $f\not \sim _\alpha g$ for all $\alpha $ . First define f so that $fx{\downarrow } = f$ for all x using the recursion theorem. Second, we want to define $g\neq f$ so that $gx{\downarrow } = g$ for every x. Inspection of the proof of the recursion theorem shows that the fixed point can be chosen to be different from any given number by using a padding argument. This guarantees the existence of g.⊣

Proof. We prove that for closed terms s and t,

$$ \begin{align*} s \sim_{{\omega_1^{\textit{CK}}}+1} t \Longrightarrow s \sim_{\omega_1^{\textit{CK}}} t. \end{align*} $$

For the purpose of this proof, denote by $\Omega $ the set of closed terms over $\mathcal {K}_1$ , coded as elements of $\omega $ in some effective way. Now consider the operator $\Gamma \colon {\mathcal P}(\Omega \times \Omega ) \rightarrow {\mathcal P}(\Omega \times \Omega )$ defined by

$$ \begin{align*} \Gamma(X) = \big\{ (s,t) \in\Omega\times\Omega : \forall x \in \mathcal{K}_1 \; (sx,tx)\in X \big\}. \end{align*} $$

Define $\Gamma _\alpha $ for every ordinal $\alpha $ by

$$ \begin{align*} \Gamma_0 &= \big\{ (s,t) : s \simeq t \big\}, \\ \Gamma_{\alpha+1} &= \Gamma(\Gamma_\alpha), \\ \Gamma_\gamma &= \bigcup_{\alpha < \gamma} \Gamma_\alpha, \hspace{1cm} \text{for }\gamma \text{ a limit ordinal.} \end{align*} $$

Note that $s \sim _\alpha t$ if and only if $(s,t) \in \Gamma _\alpha $ and that $\mathrm {ord}(\mathcal {K}_1)$ is the least ordinal $\alpha $ such that $\Gamma _{\alpha +1} = \Gamma _\alpha $ . Also note that $\Gamma _\alpha \subseteq \Gamma _{\alpha +1}$ by Theorem 3.2.

The complexity of operators such as $\Gamma $ is measured by the complexity of the predicate “ $n\in \Gamma (X)$ .” Note that $\Gamma $ is a $\Pi ^0_1$ operator and that $\Gamma $ is monotone, i.e., $X\subseteq Y$ implies $\Gamma (X)\subseteq \Gamma (Y)$ . It follows that the closure ordinal of $\Gamma $ is at most ${\omega _1^{\textit {CK}}}$ because this holds for every $\Pi ^0_1$ operator on ${\mathcal P}(\omega )$ (Gandy [Reference Sacks21]), and also for every monotone $\Pi ^1_1$ operator (Spector [Reference Spector23], cf. [Reference Sacks21, Corollary III.8.6]).Footnote ³ Typically it is assumed that $\Gamma _0 = \emptyset $ , but these closure properties also hold when $\Gamma _0$ is an arithmetical set as it is here.

Since the operator $\Gamma $ above is both $\Pi ^0_1$ and monotone, the proof that ${\omega _1^{\textit {CK}}}$ is an upper bound can be somewhat simplified as follows. Suppose that $s \sim _{{\omega _1^{\textit {CK}}}+1} t$ . We have to prove that $s \sim _\alpha t$ for some computable $\alpha $ . Note that $s \sim _{{\omega _1^{\textit {CK}}}+1} t$ if

(2) $$ \begin{align} \forall x\in\omega \; \exists \alpha<{\omega_1^{\textit{CK}}} \; s x \sim_\alpha t x. \end{align} $$

This in itself is not enough to conclude that the $\alpha $ ’s have a bound smaller than ${\omega _1^{\textit {CK}}}$ because they could form a cofinal sequence in ${\omega _1^{\textit {CK}}}$ . The question is: How hard is it to find $\alpha $ for a given x?

Observe that the $\Gamma _\alpha $ are uniformly hyperarithmetical for $\alpha <{\omega _1^{\textit {CK}}}$ . Namely, by effective transfinite recursion we can define a computable function f such that for every $a\in \mathcal {O}$ , $f(a)$ is a $\Delta ^1_1$ index of $\Gamma _{|a|}$ (cf. [Reference Sacks21, Theorem II.1.5]). It follows that the statement

(3) $$ \begin{align} a\in\mathcal{O} \wedge (sx,tx) \in \Gamma_{|a|} \end{align} $$

is a $\Pi ^1_1$ predicate of x and a. Assuming (2), we have that for every x there is an a such that (3) holds. Since $\Pi ^1_1$ predicates can be uniformized by $\Pi ^1_1$ predicates (Kreisel [Reference Sacks21, Theorem II.2.3]), there is a partial $\Pi ^1_1$ function $p(x)$ such that

(4) $$ \begin{align} \forall x \big( p(x) \in\mathcal{O} \wedge (sx,tx) \in \Gamma_{|p(x)|} \big). \end{align} $$

Since p is in fact total, it is also $\Sigma ^1_1$ (cf. [Reference Sacks21, Proposition I.1.7]). Therefore the set $\{ p(x) : x\in \omega \}$ is a $\Sigma ^1_1$ subset of $\mathcal {O}$ , and it follows from Theorem 4.2 that there is a computable ordinal $\alpha $ such that $|p(x)| < \alpha $ for every x. It follows that $\forall x \; (sx,tx) \in \Gamma _\alpha $ and therefore that $(s,t) \in \Gamma _{\alpha +1}$ . Thus $s \sim _{\alpha +1} t$ .⊣

Proof. This follows from Theorems 5.1 and 5.4.⊣

Proof. Define a tree $T \subseteq \omega ^{<\omega }$ by $T = \{\sigma \in \omega ^{<\omega } : s \sigma \not \simeq t \sigma \}$ . We show that $s \approx t$ if and only if T is well-founded.

Suppose that $s \approx t$ . Observe that for any closed terms p and q, if $p \sim _\alpha q$ for some $\alpha> 0$ , then for every $x \in \mathcal {K}_1$ there is a $\beta < \alpha $ such that $p x \sim _\beta q x$ . Now consider any $f \colon \omega \rightarrow \omega $ . Define a sequence of ordinals $\alpha _0 \geqslant \alpha _1 \geqslant \cdots $ , where $s f(0) \cdots f(n-1) \sim _{\alpha _n} t f(0) \cdots f(n-1)$ for each n, as follows. First, let $\alpha _0$ be such that $s \sim _{\alpha _0} t$ . Then for each n, if $\alpha _n> 0$ , let $\alpha _{n+1} < \alpha _n$ be such that $s f(0) \cdots f(n-1)f(n) \sim _{\alpha _{n+1}} t f(0) \cdots f(n-1)f(n)$ . If $\alpha _n = 0$ , then let $\alpha _{n+1} = 0$ . The sequence $\alpha _0 \geqslant \alpha _1 \geqslant \cdots $ cannot be a strictly descending sequence of ordinals, so there must be an n such that $\alpha _n = 0$ . Then for this n, $s f(0) \cdots f(n-1) \simeq t f(0) \cdots f(n-1)$ , which means that f is not a path through T. Thus T is well-founded.

For the converse, suppose that T is well-founded. Recall the Kleene–Brouwer ordering of $\omega ^{<\omega }$ , where $\tau <_{\mathrm {KB}} \sigma $ if either $\tau $ is a proper extension of $\sigma $ or $\tau $ is to the left of $\sigma $ . That is, $\tau <_{\mathrm {KB}} \sigma $ if and only if

$$ \begin{align*} \tau \sqsupset \sigma \vee (\exists n < \min(|\sigma|, |\tau|))[\tau(n) < \sigma(n) \wedge (\forall i < n)(\sigma(i) = \tau(i))]. \end{align*} $$

The tree T is well-ordered by $<_{\mathrm {KB}}$ because we assume that T is well-founded, and a subtree of $\omega ^{<\omega }$ is well-founded if and only if it is well-ordered by $<_{\mathrm {KB}}$ . Let $\alpha $ be the ordinal isomorphic to $(T, <_{\mathrm {KB}})$ , and, for each $\sigma \in T$ , let $\alpha _\sigma $ be the ordinal corresponding to $\sigma $ . We show by transfinite induction that $s \sigma \sim _{\alpha _\sigma + 1} t \sigma $ for every $\sigma \in T$ .

Let $\sigma \in T$ and suppose inductively that $s \tau \sim _{\alpha _\tau + 1} t \tau $ for all $\tau \in T$ with $\tau <_{\mathrm {KB}} \sigma $ . Consider any x. If $\sigma ^\smallfrown x \notin T$ , then $s \cdot \sigma ^\smallfrown x \simeq t \cdot \sigma ^\smallfrown x$ , so $s \cdot \sigma ^\smallfrown x \sim _{\alpha _\sigma } t \cdot \sigma ^\smallfrown x$ . If $\sigma ^\smallfrown x \in T$ , then $\sigma ^\smallfrown x <_{\mathrm {KB}} \sigma $ , so $s \cdot \sigma ^\smallfrown x \sim _{(\alpha _{\sigma ^\smallfrown x} + 1)} t \cdot \sigma ^\smallfrown x$ . Thus $s \cdot \sigma ^\smallfrown x \sim _{\alpha _\sigma } t \cdot \sigma ^\smallfrown x$ because $\alpha _{\sigma ^\smallfrown x} < \alpha _\sigma $ . We have shown that $s \cdot \sigma ^\smallfrown x \sim _{\alpha _\sigma } t \cdot \sigma ^\smallfrown x$ for every x. Therefore $s \sigma \sim _{\alpha _\sigma + 1} t \sigma $ .

In the case of the empty string $\epsilon $ , we have that $s \epsilon = s$ and $t \epsilon = t$ and therefore that $s \sim _{\alpha _\epsilon + 1} t$ . We may also observe that $\epsilon $ is the $<_{\mathrm {KB}}$ -maximum element of T and therefore that $\alpha = \alpha _\epsilon + 1$ . So $s \sim _\alpha t$ . Thus $s \approx t$ .⊣

Proof. The relation $\approx $ is $\Pi ^1_1$ as discussed above. We show that $\approx $ is $\Pi ^1_1$ -hard.

A typical $\Pi ^1_1$ -complete set is the set of indices of partial computable functions computing well-founded subtrees of $\omega ^{<\omega }$ . Indeed, there is a uniformly computable sequence $(T_e)_{e \in \omega }$ of trees such that the set $\{e : T_e$ is well-founded $\}$ is $\Pi ^1_1$ -complete (see, for example, [Reference Cenzer and Remmel8, Theorem 15]).

Using the recursion theorem, fix an index $e_* \in \mathcal {K}_1$ such that $e_* \cdot x = e_*$ for every x.

Let $g(i, e, \sigma )$ be a total computable function such that for all i, e, $\sigma $ , and x,

$$ \begin{align*} \Phi_{g(i, e, \sigma)}(x) \simeq \Phi_i(e, \sigma^\smallfrown x). \end{align*} $$

By padding, we may define g so that $e_* \notin \operatorname {{\mathrm {ran}}}(g)$ . Using the recursion theorem, let i be such that

$$ \begin{align*} \Phi_i(e, \sigma) = \begin{cases} e_* & \text{if }\sigma \notin T_e,\\ g(i, e, \sigma) & \text{if }\sigma \in T_e. \end{cases} \end{align*} $$

Note that $\Phi _i$ is total because g is total. Also note that $\Phi _i(e, \sigma ) = e_*$ if and only if $\sigma \notin T_e$ because $e_* \notin \operatorname {{\mathrm {ran}}}(g)$ . Let f be the function computed by $\Phi _i$ .

Proof of Claim. If $\sigma \notin T_e$ , then $\sigma ^\smallfrown x \notin T_e$ as well, so $f(e, \sigma ) \cdot x = e_* \cdot x = e_* = f(e, \sigma ^\smallfrown x)$ . If $\sigma \in T_e$ , then

$$ \begin{align*} f(e,\sigma) \cdot x = \Phi_i(e, \sigma) \cdot x = g(i,e,\sigma) \cdot x = \Phi_{g(i,e,\sigma)}(x) = \Phi_i(e, \sigma^\smallfrown x) = f(e, \sigma^\smallfrown x). \end{align*} $$

⊣

We now show that, for every e, $f(e, \epsilon ) \approx e_*$ if and only if $T_e$ is well-founded. Notice that $e_* a_0 a_1 \cdots a_n = e_*$ for every sequence $a_0, a_1, \dots , a_n$ . Therefore, by Lemma 5.6, $f(e, \epsilon ) \approx e_*$ if and only if for every sequence $a_0, a_1, a_2, \dots $ , there is an n such that $f(e, \epsilon ) \cdot a_0a_1 \cdots a_n = e_*$ .

Suppose that $T_e$ is well-founded. Consider any sequence $a_0, a_1, a_2, \dots $ . This sequence is not a path through $T_e$ , so there is an n such that the string $\sigma = \langle a_0, \dots , a_n \rangle $ is not in $T_e$ . By the Claim and the fact that $\sigma \notin T_e$ , we have that $f(e, \epsilon ) \cdot a_0a_1 \cdots a_n = f(e, \sigma ) = e_*$ . Thus $f(e, \epsilon ) \approx e_*$ .

Conversely, suppose that $T_e$ is ill-founded, and let $a_0, a_1, a_2, \dots $ be a path through $T_e$ . For each n, let $\sigma _n = \langle a_0, \dots , a_n \rangle $ . By the Claim, for each n we have that $f(e, \epsilon ) \cdot a_0a_1 \cdots a_n = f(e, \sigma _n) \neq e_*$ , where the inequality is because $\sigma _n \in T_e$ . Thus $f(e, \epsilon ) \not \approx e_*$ .

We have shown that $f(e, \epsilon ) \approx e_*$ if and only if $T_e$ is well-founded. The map $e \mapsto (f(e, \epsilon ), e_*)$ is therefore a many-one reduction witnessing that $\{e : T_e$ is well-founded $\} \leqslant _{\mathrm {m}} {\approx }$ . Thus the $\approx $ relation is $\Pi ^1_1$ -hard and hence $\Pi ^1_1$ -complete.⊣

Proof. We first show that ${\sim _{\alpha + 1}} \leqslant _{\mathrm {m}} {\sim _{\alpha + 2}}$ . Given closed terms s and t, we can effectively produce indices $f, g \in \mathcal {K}_1$ where $f \sim _1 K s$ and $g \sim _1 K t$ . Then $f \sim _{\alpha + 2} g$ if and only if $s \sim _{\alpha + 1} t$ by Lemma 3.3 and because $\alpha + 1 \geqslant 1$ . Thus the map $(s, t) \mapsto (f, g)$ is a many-one reduction witnessing that ${\sim _{\alpha + 1}} \leqslant _{\mathrm {m}} {\sim _{\alpha + 2}}$ .

Now we show that ${\sim _{\alpha + 2}} \leqslant _{\mathrm {m}} {\sim _{\alpha + 1}}$ . For this, let $\langle \cdot , \cdot \rangle \colon \omega \times \omega \rightarrow \omega $ denote a computable bijection. Given closed terms s and t, we can effectively produce indices $f, g \in \mathcal {K}_1$ such that $f \cdot \langle a, b \rangle \simeq s a b$ and $g \cdot \langle a, b \rangle \simeq t a b$ for all a and b. The map $(s, t) \mapsto (f, g)$ is then a many-one reduction witnessing that ${\sim _{\alpha + 2}} \leqslant _{\mathrm {m}} {\sim _{\alpha + 1}}$ . To see this, first suppose that $s \sim _{\alpha + 2} t$ . Then $f \cdot \langle a, b \rangle \sim _0 s a b \sim _\alpha t a b \sim _0 g \cdot \langle a, b \rangle $ for all a and b. Thus $f \cdot \langle a, b \rangle \sim _\alpha g \cdot \langle a, b \rangle $ for all $\langle a, b \rangle $ , so $f \sim _{\alpha + 1} g$ . Conversely, suppose that $s \not \sim _{\alpha + 2} t$ . Then there are a and b such that $f \cdot \langle a, b \rangle \sim _0 s a b \not \sim _\alpha t a b \sim _0 g \cdot \langle a, b \rangle $ . Therefore there is an $\langle a, b \rangle $ such that $f \cdot \langle a, b \rangle \not \sim _\alpha g \cdot \langle a, b \rangle $ , so $f \not \sim _{\alpha + 1} g$ .⊣

Proof. Let $\forall n \exists m \; \psi (z, n, m)$ be a $\Pi ^0_2$ formula, where $\psi $ is $\Sigma ^0_0$ . We show that, given z and $\ell \geqslant 1$ , we can effectively produce indices f and g such that

• • $f \sim _1 g$ if $\forall n \exists m \; \psi (z, n, m)$ holds,

• • $f \not \sim _\ell g$ if $\forall n \exists m \; \psi (z, n, m)$ fails, and

• • $f \sim _{\ell + 1} g$ regardless of whether or not $\forall n \exists m \; \psi (z, n, m)$ holds.

Fix, say, $\ell = 1$ . Then the map $z \mapsto (f, g)$ witnesses that $\{z : \forall n \exists m \; \psi (z,n,m)\} \leqslant _{\mathrm {m}} {\sim _1}$ . It follows that $\sim _1$ is $\Pi ^0_2$ -hard. All that is required for this many-one reduction is that $f \sim _1 g$ if and only if $\forall n \exists m \; \psi (z, n, m)$ holds. The extra features are useful for the proof of Lemma 6.3.

Fix an index c such that $c x {\uparrow }$ for all x. Define indices $d_\ell $ for each $\ell \geqslant 0$ by $d_0 = c$ and $d_{\ell + 1} = K d_\ell $ . Then for every $\ell $ and every $x_0, \dots , x_\ell $ , we have that $d_\ell x_0 \cdots x_{\ell - 1}{\downarrow } = c$ and that $d_\ell x_0 \cdots x_{\ell - 1} x_\ell {\uparrow }$ .

Given z and $\ell \geqslant 1$ , let $g = d_\ell $ and compute an index f such that

$$ \begin{align*} \Phi_f(n) \simeq \begin{cases} d_{\ell-1} & \text{if }\exists m \; \psi(z, n, m),\\ {\uparrow} & \text{otherwise}. \end{cases} \end{align*} $$

Notice that $f x_0 \cdots x_\ell {\uparrow }$ and $g x_0 \cdots x_\ell {\uparrow }$ for all $x_0, \dots , x_\ell $ , so $f \sim _{\ell + 1} g$ . Suppose that $\forall n \exists m \; \psi (z, n, m)$ holds. Then $f n = d_{\ell -1} = g n$ for every n, so $f \sim _1 g$ . Conversely, suppose that $\exists m \; \psi (z, n, m)$ fails for some n. Let $x_1 = \cdots = x_{\ell -1} = 0$ . Then $f n {\uparrow }$ and so $f n x_1 \cdots x_{\ell -1}{\uparrow }$ , but $g n x_1 \cdots x_{\ell -1}{\downarrow } = c$ . Thus $f \not \sim _\ell g$ . This completes the proof.⊣

Proof. Let $\exists n \; \psi (z, n)$ be an arbitrary $\Sigma ^0_3$ formula, where $\psi $ is $\Pi ^0_2$ . We may assume that $\psi (z, n) \rightarrow \psi (z, n+1)$ for every z and n by replacing $\exists n \; \psi (z,n)$ by the equivalent $(\exists n)(\exists p < n)\psi (z, p)$ and pushing the bounded quantifier past the unbounded quantifiers.

We show that, given z, we can effectively produce indices $f, g \in \mathcal {K}_1$ such that

• • $f \sim _\omega g$ if and only if $\exists n \; \psi (z, n)$ holds, and

• • $f \sim _{\omega + 1} g$ regardless of whether or not $\exists n \; \psi (z, n)$ holds.

Given such a procedure, the map $z \mapsto (f, g)$ witnesses that $\{z : \exists n \; \psi (z, n)\} \leqslant _{\mathrm {m}} {\sim _\omega }$ . It follows that $\sim _\omega $ is $\Sigma ^0_3$ -hard. The additional $f \sim _{\omega + 1} g$ requirement is not needed for the many-one reduction, but it is helpful for the proof of Lemma 6.5.

By the proof of Lemma 6.2, given z, n, and $\ell \geqslant 1$ , we can effectively produce indices $a^{z, \ell }_n$ and $b^{z, \ell }_n$ such that

• • $a^{z, \ell }_n \sim _1 b^{z, \ell }_n$ if $\psi (z, n)$ holds,

• • $a^{z, \ell }_n \not \sim _\ell b^{z, \ell }_n$ if $\psi (z, n)$ fails, and

• • $a^{z, \ell }_n \sim _{\ell + 1} b^{z, \ell }_n$ regardless of whether or not $\psi (z, n)$ holds.

Now, given z, define f and g so that $f n = a^{z, n+1}_n$ and $g n = b^{z, n+1}_n$ for each n.

To see that $f \sim _{\omega + 1} g$ , observe that $f n = a^{z, n+1}_n \sim _{n+2} b^{z, n+1}_n = g n$ for every n. Thus $f n \sim _\omega g n$ for every n, so $f \sim _{\omega + 1} g$ .

Suppose that there is an $n_0$ such that $\psi (z, n_0)$ holds. We show that $f n \sim _{n_0 + 1} g n$ for every n. Therefore $f \sim _{n_0 + 2} g$ , so $f \sim _\omega g$ . Fix n. If $n < n_0$ , then $f n \sim _{n_0 + 1} g n$ because $f n \sim _{n+2} g n$ as above, and $n + 2 \leqslant n_0 + 1$ . Suppose instead that $n \geqslant n_0$ . Then $\psi (z, n)$ holds because $\psi (z, n_0)$ holds and $n \geqslant n_0$ . Thus $f n = a^{z, n+1}_n \sim _1 b^{z, n+1}_n = g n$ . That is, $f n \sim _1 g n$ , so also $f n \sim _{n_0 + 1} g n$ .

Finally, suppose that $\psi (z, n)$ fails for every n. Then $f n = a^{z, n+1}_n \not \sim _{n + 1} b^{z, n+1}_n = g n$ for every n. Thus also $f \not \sim _{n + 1} g$ for every n, so $f \not \sim _\omega g$ . This completes the proof.⊣

Proof. The proof is similar to that of Theorem 5.7. Let $g(i, e, \sigma )$ be a total computable function such that for all i, e, $\sigma $ , and x,

$$ \begin{align*} \Phi_{g(i, e, \sigma)}(x) \simeq \Phi_i(e, \sigma^\smallfrown x). \end{align*} $$

Using the recursion theorem, let i be such that

$$ \begin{align*} \Phi_i(e, \sigma) \simeq \begin{cases} \Phi_e(n,m) & \text{if }\sigma = \sigma_{n,m},\\ g(i, e, \sigma) & \text{if }\sigma \sqsubset \sigma_{n,m}\text{ for some }n\text{ and }m,\\ {\uparrow} & \text{otherwise.} \end{cases} \end{align*} $$

Let h be the function partially computed by $\Phi _i$ . Let e be an index for the sequence $(a_{n,m})_{n,m \in \omega }$ , so that $\Phi _e(n,m) = a_{n,m}$ for all n and m.

Proof of Claim. If $\sigma \sqsubset \sigma _{n,m}$ , then

$$ \begin{align*} h(e, \sigma) \cdot x \simeq \Phi_i(e, \sigma) \cdot x \simeq g(i,e,\sigma) \cdot x \simeq \Phi_{g(i,e,\sigma)}(x) \simeq \Phi_i(e, \sigma^\smallfrown x) \simeq h(e, \sigma^\smallfrown x). \end{align*} $$

⊣

Take $f = h(e, \epsilon ) = g(i,e,\epsilon )$ , which is defined because g is total. By the Claim, for each n and m we have that $f \sigma _{n,m} \simeq h(e, \epsilon ) \cdot \sigma _{n,m} \simeq h(e, \sigma _{n,m}) \simeq \Phi _e(n,m) = a_{n,m}$ . Thus $f \sigma _{n,m} = a_{n,m}$ for all n and m.

Suppose that $\tau $ is $\sqsubseteq $ -incomparable with $\sigma _{n,m}$ for all n and m. Let $\sigma \sqsubset \tau $ be the longest initial segment of $\tau $ that is $\sqsubseteq $ -comparable with $\sigma _{n_0, m_0}$ for some $n_0$ and $m_0$ . Then $\sigma \sqsubset \sigma _{n_0, m_0}$ , but $\sigma ^\smallfrown \tau (|\sigma |) = \tau {\upharpoonright } (|\sigma |+1)$ is $\sqsubseteq $ -incomparable with $\sigma _{n,m}$ for all n and m. By the Claim, we have that

$$ \begin{align*} f \cdot \tau {\upharpoonright} (|\sigma|+1) \simeq f \cdot \sigma^\smallfrown \tau(|\sigma|) \simeq h(e, \epsilon) \cdot \sigma^\smallfrown \tau(|\sigma|) \simeq h(e, \sigma^\smallfrown \tau(|\sigma|)). \end{align*} $$

However, $h(e, \sigma ^\smallfrown \tau (|\sigma |)){\uparrow }$ because $\sigma ^\smallfrown \tau (|\sigma |)$ is $\sqsubseteq $ -incomparable with $\sigma _{n,m}$ for all n and m. Therefore $(f \cdot \tau {\upharpoonright } (|\sigma |+1)){\uparrow }$ . Thus $f \tau {\uparrow }$ as well. Thus $f \tau {\uparrow }$ whenever $\tau $ is $\sqsubseteq $ -incomparable with $\sigma _{n,m}$ for all n and m.⊣

Proof. We induct on $k> 0$ to show the following. Given a $\Sigma ^0_{2k+1}$ formula $\exists n \forall m \; \psi (z, n, m)$ , where $\psi $ is $\Sigma ^0_{2k-1}$ , and a z, we can effectively produce indices $f, g \in \mathcal {K}_1$ such that

• • $f \sim _{\omega k} g$ if and only if $\exists n \forall m \; \psi (z, n, m)$ holds, and

• • $f \sim _{\omega k + 1} g$ regardless of whether or not $\exists n \forall m \; \psi (z, n, m)$ holds.

Given such a procedure, the map $z \mapsto (f, g)$ witnesses that $\{z : \exists n \forall m \; \psi (z,n,m)\} \leqslant _{\mathrm {m}} {\sim _{\omega k}}$ . It follows that $\sim _{\omega k}$ is $\Sigma ^0_{2k+1}$ -hard.

The base case $k = 1$ is by Lemma 6.3 and its proof. Thus suppose that $k> 1$ , and consider an arbitrary $\Sigma ^0_{2k+1}$ formula $\exists n \forall m \; \psi (z, n, m)$ , where $\psi (z, n, m)$ is $\Sigma ^0_{2k-1}$ . We may assume that $\forall m \; \psi (z, n, m) \rightarrow \forall m \; \psi (z, n+1, m)$ for every z and n by replacing $\exists n \forall m \; \psi (z,n,m)$ by the equivalent $(\exists n)(\exists p < n)(\forall m)\psi (z, p, m)$ and pushing the bounded quantifier past the unbounded quantifiers.

Inductively assume that, given z, n, and m, we can effectively produce indices $a^z_{n, m}$ and $b^z_{n, m}$ such that

• • $a^z_{n, m} \sim _{\omega (k-1)} b^z_{n, m}$ if and only if $\psi (z,n,m)$ holds, and

• • $a^z_{n, m} \sim _{\omega (k-1) + 1} b^z_{n, m}$ regardless of whether or not $\psi (z,n,m)$ holds.

Let $\sigma _{n,m} = {\langle n, m \rangle _2}^\smallfrown 0^n$ for each $n, m \in \omega $ as in Lemma 6.4. Now, given z, use Lemma 6.4 to effectively produce indices $f, g \in \mathcal {K}_1$ so that

• • $f \sigma _{n,m} = a^z_{n, m}$ and $g \sigma _{n,m} = b^z_{n, m}$ for all $n, m \in \omega $ , and

• • $f \tau {\uparrow }$ and $g \tau {\uparrow }$ if $\tau $ is $\sqsubseteq $ -incomparable with $\sigma _{n,m}$ for all $n, m \in \omega $ .

First, we show that $f \sim _{\omega k + 1} g$ . Consider any x, and decode x as $x = \langle n, m \rangle _2$ . Now consider any $\tau $ of length n. If $\tau = 0^n$ , then $x^\smallfrown \tau = \sigma _{n,m}$ , in which case

$$ \begin{align*} f \cdot x^\smallfrown\tau = f \sigma_{n,m} = a^z_{n, m} \sim_{\omega(k-1) + 1} b^z_{n, m} = g \sigma_{n,m} = g \cdot x^\smallfrown\tau. \end{align*} $$

If $\tau \neq 0^n$ , then $x^\smallfrown \tau $ is $\sqsubseteq $ -incomparable with $\sigma _{n^{\prime },m^{\prime }}$ for all $n^{\prime }$ and $m^{\prime }$ , so $(f \cdot x^\smallfrown \tau ){\uparrow }$ and $(g \cdot x^\smallfrown \tau ){\uparrow }$ , which means that $f \cdot x^\smallfrown \tau \simeq g \cdot x^\smallfrown \tau $ and hence that $f \cdot x^\smallfrown \tau \sim _{\omega (k-1) + 1} g \cdot x^\smallfrown \tau $ . We therefore have that $f \cdot x^\smallfrown \tau \sim _{\omega (k-1) + 1} g \cdot x^\smallfrown \tau $ for all $\tau $ of length n. This means that $f x \sim _{\omega (k-1) + n + 1} g x$ and therefore that $f x \sim _{\omega k} g x$ . So $f x \sim _{\omega k} g x$ for every x. Thus $f \sim _{\omega k + 1} g$ .

Now suppose that there is an $n_0$ such that $\forall m \; \psi (z, n_0, m)$ holds. In this case, we show that $f \tau \sim _{\omega (k-1)} g \tau $ for every $\tau $ of length $n_0 + 1$ . It then follows that $f \sim _{\omega (k-1) + n_0 + 1} g$ and hence that $f \sim _{\omega k} g$ . So consider a $\tau $ of length $n_0 + 1$ . If $\tau $ is $\sqsubseteq $ -incomparable with $\sigma _{n, m}$ for all n and m, then $f \tau {\uparrow }$ and $g \tau {\uparrow }$ , so $f \tau \simeq g \tau $ , so $f \tau \sim _{\omega (k-1)} g \tau $ . Suppose instead that $\sigma _{n, m} \sqsubset \tau $ for some n and m. Then $\tau = {\sigma _{n,m}}^\smallfrown \eta $ for some string $\eta $ of length $\ell> 0$ . So $f \tau = f \cdot {\sigma _{n,m}}^\smallfrown \eta \simeq a^z_{n, m} \cdot \eta $ , and $g \tau = g \cdot {\sigma _{n,m}}^\smallfrown \eta \simeq b^z_{n, m} \cdot \eta $ . We know that $a^z_{n, m} \sim _{\omega (k-1) + 1} b^z_{n, m}$ , so $a^z_{n, m} \cdot \eta \sim _{\omega (k-1)} b^z_{n, m} \cdot \eta $ because $|\eta |> 0$ . All together, we therefore have that $f \tau \sim _{\omega (k-1)} g \tau $ . Finally, suppose that $\tau \sqsubseteq \sigma _{n, m}$ for some n and m. It must then be that $n_0 + 1 = |\tau | \leqslant |\sigma _{n,m}| = n + 1$ , so $n_0 \leqslant n$ . Therefore $\forall m \; \psi (z, n, m)$ holds because $\forall m \; \psi (z, n_0, m)$ holds and $n_0 \leqslant n$ . So,

$$ \begin{align*} f \sigma_{n,m} = a^z_{n, m} \sim_{\omega (k-1)} b^z_{n, m} = g \sigma_{n,m}. \end{align*} $$

Thus there is an $\alpha < \omega (k-1)$ such that $f \sigma _{n,m} \sim _\alpha g \sigma _{n,m}$ . Now consider any $\eta $ of length $\ell = n - n_0$ . If $\eta = 0^\ell $ , then $\tau ^\smallfrown \eta = \sigma _{n,m}$ , so $f \cdot \tau ^\smallfrown \eta \sim _\alpha g \cdot \tau ^\smallfrown \eta $ . If $\eta \neq 0^\ell $ , then $\tau ^\smallfrown \eta $ is $\sqsubseteq $ -incomparable with $\sigma _{n^{\prime }, m^{\prime }}$ for all $n^{\prime }$ and $m^{\prime }$ , so $(f \cdot \tau ^\smallfrown \eta ) {\uparrow } \simeq (g \cdot \tau ^\smallfrown \eta ) {\uparrow }$ . Thus also $f \cdot \tau ^\smallfrown \eta \sim _\alpha g \cdot \tau ^\smallfrown \eta $ . So $f \cdot \tau ^\smallfrown \eta \sim _\alpha g \cdot \tau ^\smallfrown \eta $ for every $\eta $ of length $\ell $ . This means that $f \tau \sim _{\alpha + \ell } g \tau $ . We have that $\alpha + \ell < \omega (k-1)$ because $\alpha < \omega (k-1)$ and $\omega (k-1)$ is a limit. Thus $f \tau \sim _{\omega (k-1)} g \tau $ .

Finally, suppose that $\forall m \; \psi (z, n, m)$ fails for every n. We show that $f \not \sim _{\omega k} g$ . Given n, let m be such that $\psi (z, n, m)$ fails, and consider $\sigma _{n, m}$ . Then $f \sigma _{n,m} = a^z_{n, m} \not \sim _{\omega (k-1)} b^z_{n, m} = g \sigma _{n,m}$ . As $|\sigma _{n,m}| = n+1$ , this implies that $f \not \sim _{\omega (k-1) + n + 1} g$ . So $f \not \sim _{\omega (k-1) + n + 1} g$ for every n. Thus $f \not \sim _{\omega k} g$ .

Given z, we have effectively produced indices f and g such that

• • $f \sim _{\omega k} g$ if and only if $\exists n \forall m \; \psi (z,n,m)$ holds, and

• • $f \sim _{\omega k + 1} g$ regardless of whether or not $\exists n \forall m \; \psi (z,n,m)$ holds.

This completes the induction step and therefore completes the proof.⊣

Proof. For $k = 0$ , this is given by Lemma 6.2. Let $k> 0$ , and consider a $\Pi ^0_{2k+2}$ formula $\forall n \; \psi (z, n)$ , where $\psi (z,n)$ is $\Sigma ^0_{2k+1}$ . By Lemma 6.5 and its proof, given z and n, we can effectively produce indices $a^z_n, b^z_n \in \mathcal {K}_1$ such that $a^z_n \sim _{\omega k} b^z_n$ if and only if $\psi (z, n)$ holds. Thus given z, we can effectively produce indices $f, g \in \mathcal {K}_1$ where $f n = a^z_n$ and $g n = b^z_n$ for all $n \in \omega $ .

If $\forall n \; \psi (z, n)$ holds, then $f n = a^z_n \sim _{\omega k} b^z_n = g n$ for every n. Thus $f \sim _{\omega k + 1} g$ . On the other hand, if n is such that $\psi (z, n)$ fails, then $f n = a^z_n \not \sim _{\omega k} b^z_n = g n$ , so $f \not \sim _{\omega k + 1} g$ . Therefore the map $z \mapsto (f, g)$ witnesses that $\{z : \forall n \; \psi (z, n)\} \leqslant _{\mathrm {m}} {\sim _{\omega k + 1}}$ . Thus $\sim _{\omega k + 1}$ is $\Pi ^0_{2k + 2}$ -hard.⊣

Proof. It is straightforward to check that $\sim _{\omega k}$ is $\Sigma ^0_{2k + 1}$ -definable for every $k> 0$ and that $\sim _{\omega k + n}$ is $\Pi ^0_{2k + 2}$ -definable for every k and every $n> 0$ . See also Lemma 7.5 below. The relation $\sim _{\omega k}$ is $\Sigma ^0_{2k + 1}$ -hard for every $k> 0$ by Lemma 6.5. The relation $\sim _{\omega k + 1}$ is $\Pi ^0_{2k + 2}$ -hard for every k by Lemma 6.6. For $n> 0$ , we have that ${\sim _{\omega k + n}} \equiv _{\mathrm {m}} {\sim _{\omega k + 1}}$ by repeated applications of Lemma 6.1, so $\sim _{\omega k + n}$ is also $\Pi ^0_{2k + 2}$ -hard for every k and every $n> 0$ .

Proof. Clearly $G(0) = 0$ and $G(1) = 1$ . A straightforward argument shows that $G(\alpha )$ is a limit ordinal when $\alpha> 0$ is even and that $G(\alpha )$ is the successor of a limit ordinal when $\alpha> 1$ is odd.

To see that the range of G contains all limit ordinals, suppose for a contradiction that some limit ordinal is missing, and let $\gamma $ be the least limit ordinal not in the range of G. First suppose that there is a maximum limit ordinal $\delta < \gamma $ , in which case it must be that $\gamma = \delta + \omega $ . As $\delta < \gamma $ , there is an $\alpha $ such that $G(\alpha ) = \delta $ , and $\alpha $ must be even by the discussion above. Thus $G(\alpha + 2) = \delta + \omega = \gamma $ , which is a contradiction. Now suppose instead that there is no maximum limit ordinal $\delta < \gamma $ . Then $\gamma = \sup \{\delta < \gamma : \delta \text { is a limit}\}$ . For each limit ordinal $\delta < \gamma $ , let $\alpha _\delta $ be such that $G(\alpha _\delta ) = \delta $ , and let $\alpha = \sup \{\alpha _\delta : \delta < \gamma \text { is a limit}\}$ . Then $G(\alpha ) = \sup \{G(\alpha _\delta ) : \delta < \gamma \text { is a limit}\} = \gamma $ , which is again a contradiction. So the range of G contains all limit ordinals.

Finally, consider $\gamma + 1$ , where $\gamma $ is a limit ordinal. The range of G contains all limit ordinals, so there is an even ordinal $\alpha $ with $G(\alpha ) = \gamma $ . Then $G(\alpha + 1) = \gamma + 1$ . So the range of G contains all successors of limit ordinals as well.⊣

Proof. Proceed by transfinite induction on $\alpha $ . Clearly $F(G(0)) = 0$ . Now suppose that $F(G(\delta )) = \delta $ for all $\delta < \alpha $ .

We have two successor cases, depending on whether $\alpha $ is even or odd. First, suppose that $\alpha = \beta + 2d + 1$ , where $d < \omega $ and $\beta $ is either $0$ or a limit. Then

$$ \begin{align*} F(G(\alpha)) &= F(G(\beta + 2d + 1)) & &\\ &= F(G(\beta + 2d) + 1) & &\text{by the definition of }G\\ &= F(G(\beta + 2d)) + 1 & &\text{by the definition of }F, \text{as }G(\beta + 2d)\text{ is }0\text{ or a limit}\\ &=\beta + 2d + 1 & &\text{by the induction hypothesis}\\ &=\alpha. & & \end{align*} $$

Second, suppose that $\alpha = \beta + 2d + 2$ , where $d < \omega $ and $\beta $ is either $0$ or a limit. Then

$$ \begin{align*} F(G(\alpha)) &= F(G(\beta + 2d + 2)) & &\\[-1pt] &= F(G(\beta + 2d + 1) + \omega) & &\!\!\text{by the definition of } G\\[-1pt] &= \lim_{n < \omega}\Bigl[F(G(\beta + 2d + 1) + n) + 1\Bigr] & &\!\!\text{by the definition of }F\\[-1pt] &= \lim_{n < \omega}\Bigl[F(G(\beta + 2d) + 1 + n) + 1\Bigr] & &\!\!\text{by the definition of }G\\[-1pt] &= \lim_{n < \omega}\Bigl[F(G(\beta + 2d)) + 1 + 1\Bigr] & &\!\!\text{by the definition of }F, \text{ as }G(\beta + 2d)\\[-1pt]&&&\quad \!\!\text{ is }0\text{ or a limit}\\[-1pt] &= \lim_{n < \omega}\Bigl[\beta + 2d + 2 \Bigr] & &\!\!\text{by the induction hypothesis}\\[-1pt] &= \beta + 2d + 2 & &\\[-1pt] &= \alpha. & & \end{align*} $$

Finally, suppose that $\alpha $ is a limit. Then $G(\alpha )$ is a limit as well, and

$$ \begin{align*} F(G(\alpha)) &= \lim_{\beta < G(\alpha)}(F(\beta) + 1). \end{align*} $$

If $\beta < G(\alpha )$ , then $\beta < G(\delta )$ for some $\delta < \alpha $ because $G(\alpha ) = \lim _{\delta < \alpha }G(\delta )$ . Then $F(\beta ) + 1 \leqslant F(G(\delta )) + 1 = \delta + 1 < \alpha $ by the induction hypothesis and the fact that $\alpha $ is a limit. Thus $F(\beta ) + 1 < \alpha $ for all $\beta < G(\alpha )$ , so $F(G(\alpha )) \leqslant \alpha $ . On the other hand, if $\beta < \alpha $ , then $G(\beta ) < G(\alpha )$ , so $F(G(\alpha )) \geqslant F(G(\beta )) + 1 = \beta + 1> \beta $ by the induction hypothesis. Thus $F(G(\alpha )) \geqslant \alpha $ . So $F(G(\alpha )) = \alpha $ .⊣

Proof. By effective transfinite recursion on $\alpha $ , we uniformly produce a code for a formula $\varphi (u,v)$ with the following properties.

• • If $\alpha $ is an even, then $\varphi (u,v)$ is $\Sigma ^0_{1 + \alpha }$ .

• • If $\alpha $ is an odd, then $\varphi (u,v)$ is $\Pi ^0_{1 + \alpha }$ .

• • For all closed terms s and t, $s \sim _{G(\alpha )} t$ if and only if $\varphi (s,t)$ .

For the base case $\alpha = 1$ , we have that $1$ is odd, that $G(1) = 1$ , and that $\sim _1$ is defined by the fixed $\Pi ^0_2$ formula $\varphi (u, v) \equiv \forall x (u x \simeq v x)$ .

Suppose that $\alpha = \beta + 1$ is a successor. There are two cases, depending on whether $\beta $ even or odd. The coding of computable ordinals allows us to distinguish between the cases.

First, suppose that $\beta $ is even. In this case, $\alpha $ is odd and $G(\alpha ) = G(\beta + 1) = G(\beta ) + 1$ . By effective transfinite recursion, we can compute a (code for a) $\Sigma ^0_{1 + \beta }$ formula $\psi (u, v)$ such that $s \sim _{G(\beta )} t \Leftrightarrow \psi (s,t)$ for all closed terms s and t. Let $\varphi (u, v)$ be the $\Pi ^0_{1 + \beta + 1}$ formula $\varphi (u, v) \equiv \forall x \; \psi (u x, v x)$ . Then $s \sim _{G(\beta ) + 1} t \Leftrightarrow \varphi (s,t)$ for all closed terms s and t. Thus $\varphi (u, v)$ is the desired formula because $\alpha = \beta + 1$ is odd and $G(\alpha ) = G(\beta ) + 1$ .

Second, suppose that $\beta $ is odd. In this case, $\alpha $ is even and $G(\alpha ) = G(\beta + 1) = G(\beta ) + \omega $ . By effective transfinite recursion, we can compute a (code for a) $\Pi ^0_{1 + \beta }$ formula $\psi (u, v)$ such that $s \sim _{G(\beta )} t \Leftrightarrow \psi (s,t)$ for all closed terms s and t. From the code for $\psi $ , we can uniformly compute codes for $\Pi ^0_{1 + \beta }$ formulas $\psi _n(u, v)$ , where $\psi _n(u, v) \equiv \forall x_0, \dots , x_{n-1} \; \psi (u x_0 \cdots x_{n-1}, v x_0 \cdots x_{n-1})$ for each $n < \omega $ . Note that $s \sim _{G(\beta ) + n} t \Leftrightarrow \psi _n(s,t)$ for all closed terms s and t and all $n < \omega $ . We can then compute a code for the $\Sigma ^0_{1 + \beta + 1}$ formula $\varphi (u, v) \equiv \bigvee _{n \in \omega } \psi _n(u, v)$ . For closed terms s and t, $\varphi (s,t)$ holds if and only if $\psi _n(s,t)$ holds for some $n < \omega $ if and only if $s \sim _{G(\beta ) + n} t$ for some $n < \omega $ if and only if $s \sim _{G(\beta ) + \omega } t$ . Thus $\varphi (u, v)$ is the desired formula because $\alpha = \beta + 1$ is even and $G(\alpha ) = G(\beta ) + \omega $ .

Now suppose that $\alpha $ is a limit, and let $\beta _0 < \beta _1 < \cdots $ be a computable sequence of ordinals converging to $\alpha $ . We may assume that each $\beta _n$ is odd by adding to $\beta _n$ where necessary. Note that $\alpha $ is even and that $G(\alpha ) = \lim _{n < \omega }G(\beta _n)$ is a limit ordinal. By effective transfinite recursion, for each $n < \omega $ , we can uniformly compute (a code for) a $\Pi ^0_{1 + \beta _n}$ formula $\psi _n(u, v)$ such that $s \sim _{G(\beta _n)} t \Leftrightarrow \psi _n(s,t)$ for all closed terms s and t. We can thus compute a code for the $\Sigma ^0_{1 + \alpha }$ formula $\varphi (u, v) \equiv \bigvee _{n \in \omega } \psi _n(u, v)$ . For closed terms s and t, $\varphi (s,t)$ holds if and only if $\psi _n(s,t)$ holds for some $n < \omega $ if and only if $s \sim _{G(\beta _n)} t$ for some $n < \omega $ if and only if $s \sim _{G(\alpha )} t$ . Thus $\varphi (u, v)$ is the desired formula.⊣