Proof. Suppose g is single-valued, has codomain $\mathbb {N}^{\mathbb {N}}$ , and $g \leq _{\mathrm {W}} f$ . Define a single-valued problem h with codomain $\textbf {2}$ as follows: Given $n,m \in \mathbb {N}$ and a g-instance x, produce $1$ if $g(x)(n) \geq m$ , otherwise produce $0$ . It is easy to see that $g \leq _{\mathrm {W}} \widehat {h}$ and $h \leq _{\mathrm {W}} f$ . The latter implies $h \leq _{\mathrm {W}} \mathrm {Det}_{ \mathbf {2} }(f)$ and so $g \leq _{\mathrm {W}} \widehat {h} \leq _{\mathrm {W}} \widehat {\mathrm {Det}_{ \mathbf {2} }(f)}$ .

Proof. Given a non-well quasi-order $(Q,\preceq _Q)$ where $Q \subseteq \mathbb {N}$ , compute the set $S := \{a \in Q {}\,:\,{} (\forall b <_{\mathbb {N}} a)(a \not \preceq _Q b \text { or }b \not \preceq _Q a)\}$ . The restriction $(S,\preceq _Q)$ is a non-well partial order because it is isomorphic to the partial order of $\preceq _Q$ -equivalence classes.

Proof. For $(1) \Rightarrow (2)$ , let g be upwards closed with codomain $\mathbb {N}$ and assume $f \leq _{\mathrm {W}} g$ via $\Phi $ and $\Psi $ . Given $x \in \operatorname {dom}(f)$ , run the computations $(\Psi (x,m))_{m \in \mathbb {N}}$ in parallel. Once some $\Psi (x,m)$ halts, we output its result and cancel $\Psi (x,n)$ for all $n < m$ . This produces a sequence of numbers. The fact that g is upwards closed guarantees that cofinitely many elements of this sequence are elements of $f(x)$ .

For the converse direction, for every $x\in \operatorname {dom}(f)$ , let $p_x\in \mathbb {N}^{\mathbb {N}}$ be as in the hypothesis. Define $M_x:=\max \{m {}\,:\,{} p_x(m)\notin f(x)\}$ and let $g(x):=\{n {}\,:\,{} n>M_x\}$ . Clearly g is upwards closed. The fact that $f\leq _{\mathrm {W}} g$ follows from the fact that $x\mapsto p_x$ is computable.

Proof. The fact that every partial order that admits a tree decomposition does not have an infinite descending sequence follows from the fact that if ( $v_n)_{n \in \mathbb {N}}$ is an infinite descending sequence in P, then since every interval $(\lceil v_n \rceil ]$ is finite, up to removing duplicates, the sequence $(\lceil v_n\rceil )_{n\in \mathbb {N}}$ would be an infinite descending sequence in T.

If $(w_n)_{n \in \mathbb {N}}$ is an infinite antichain in T then, by definition of tree decomposition, $(\iota (w_n))_{n \in \mathbb {N}}$ is an infinite antichain in P. Conversely, if $(v_n)_{n \in \mathbb {N}}$ is an infinite antichain in P, then for every n, for all but finitely many m, $\lceil v_n \rceil $ is $\sqsubseteq $ -incomparable with $\lceil v_m \rceil $ . In particular, we can obtain an infinite antichain in T by choosing a subsequence $(v_{n_i})_{i \in \mathbb {N}}$ such that, for every $i\neq j$ , $\lceil v_{n_i} \rceil $ and $\lceil v_{n_j} \rceil $ are $\sqsubseteq $ -incomparable.

Proof. Fix a computable “guessing” procedure g that receives as input a partial order (admitting a tree decomposition) and outputs an infinite sequence of elements in that partial order. We shall build a partial order P together with a tree decomposition in stages such that, infinitely often, g outputs an element of P that does not extend to an infinite bad sequence.

Start with $T_0 = \{\varepsilon \}$ and $P_0$ having a single element $v_\varepsilon $ , with $\iota _0(\varepsilon ) = v_\varepsilon $ . In stage s, we have built a finite tree decomposition and wish to extend it to some . The tree $T_{s+1}$ will always be obtained by giving each leaf in $T_s$ a single child, and then adding two children to exactly one of the new leaves. To decide which leaf gets two children, say a finite extension Q of $P_s$ is suitable for if for every $v \in Q \setminus P_s$ , there is exactly one leaf $w \in T_s$ such that $\iota _s(w) <_Q v$ . Pick the left-most leaf $\sigma $ of $T_s$ with the following property:

To see that such $\sigma $ must exist, consider extending $P_s$ by adding an “infinite comb” (i.e., a copy of $\{0^n1^i {}\,:\,{} n \in \mathbb {N}, i \in \{0,1\}\}$ ) above the $\iota _s$ -image of a single leaf in $T_s$ . The resulting partial order $\overline {Q}$ is non-wqo, admits a tree decomposition (obtained by extending $T_s$ and $\iota _s$ in the obvious way), and its finite approximations (extending $P_s$ ) are suitable for $\iota _s$ . Hence, by hypothesis, g eventually guesses some element, which must be comparable with $\iota _s(\sigma )$ for some leaf $\sigma \in T_s$ (because all elements of $\overline {Q}$ are).

Having identified $\sigma $ , we fix any corresponding suitable extension Q of $P_s$ . In order to extend $\iota _s$ , we further extend Q to $Q'$ by adding a new maximal element $v_w$ to Q for each leaf $w \in T_s$ as follows: $v_w$ lies above all $v \in Q\setminus P_s$ such that $\iota _s(w) <_P v$ , and is incomparable with all other elements (including the other new maximal elements $v_{w'}$ ). To extend $T_s$ , we add a new leaf to $T_s$ for each leaf $\tau $ , obtaining a tree $T'$ . We extend $\iota _s$ to yield a tree decomposition in the obvious way.

Finally, we add two children to in $T'$ , i.e., define . We also add two children $v_1$ , $v_2$ to in $Q'$ to obtain $P_{s+1}$ , and extend $\iota '$ to $\iota _{s+1}$ by setting . This concludes stage s.

Figure 1 Schematic representation of the construction used in the proof of Lemma 3.5. The dashed box contains the partial construction up to stage s. The gray boxes contain intervals in the partial order P. For simplicity, each interval up to stage s only contains one point (i.e., the partial order $P_s$ is isomorphic to a tree), but this need not be the case in general. The black nodes are those in the range of the (partial) tree decomposition $\iota _s$ . The square gray node is the node guessed by g at stage $s+1$ (which identifies $\iota (\sigma )$ and ).

It is clear from the construction that $\iota : T \to P$ is a tree decomposition. Let us discuss the shape of the tree T. In stage s, we introduced a bifurcation above a leaf $\sigma _s$ of $T_s$ . These are the only bifurcations in T. Observe that, whenever $s' < s$ , $\sigma _s$ is either above or to the right of $\sigma _{s'}$ , because every suitable extension of $P_s$ is also a suitable extension of $P_{s'}$ and, at stage $s'$ , the chosen leaf was the left-most. This implies that the sequence $(\sigma _s)_s$ converges to a path $p\in [T]$ , i.e., for every n there is a stage s such that for every $s'>s$ , $p[n]\sqsubseteq \sigma _{s'}$ . This also implies that p is the unique non-isolated path (if $\tau \not \sqsubseteq p$ then there are only finitely many bifurcations above $\tau $ ). Observe also that a vertex w in T is extendible to an infinite antichain in T if and only if it does not belong to p.

We may now apply Lemma 3.4 to analyze P. First, since T is not wqo, neither is P. Second, we claim that if $v <_P \iota (\sigma )$ for some vertex $\sigma $ on p, then v is not extendible to an infinite bad sequence. To prove this, suppose v is extendible. Then so is $\iota (\sigma )$ . The proof of Lemma 3.4 implies that $\lceil \iota (\sigma ) \rceil = \sigma $ is extendible to an infinite antichain in T. So $\sigma $ cannot lie on p, proving our claim.

To complete the proof, observe that our construction of $\iota $ ensures that for each s, $g(P)$ eventually outputs a guess which is below . Whenever lies along p (which holds for infinitely many s), this guess is wrong by the above claim (Figure 1).

Proof of Theorem 3.1.

Suppose towards a contradiction that ${}^1\mathsf {BS} \leq _{\mathrm {W}} \boldsymbol {\Pi }^1_1\mathsf {-Bound}$ . Since the problem of finding an element in a non-wqo which extends to an infinite bad sequence is first-order, it is Weihrauch reducible to $\boldsymbol {\Pi }^1_1\mathsf {-Bound}$ as well. Now, $\boldsymbol {\Pi }^1_1\mathsf {-Bound}$ is upwards closed, so there is a computable guessing procedure for this problem (Lemma 3.2). However such a procedure cannot exist, even for partial orders which admit a tree decomposition (Lemma 3.5).

Proof. To prove the first part of the theorem, fix $\alpha \trianglelefteq \beta $ and let $A\in \mathbb {N}^{\mathbb {N}}$ be an infinite $\le _P$ -bad sequence extending $\alpha $ . Let also $i<|{\alpha }|$ be a witness for $\alpha \trianglelefteq \beta $ . For every $j>i$ and every $k<|{\beta }|$ , $\beta (k)\not \le _P A(j)$ (otherwise $A(i) = \alpha (i) \le _P\beta (k)\le _P A(j)$ would contradict the fact that A is a $\le _P$ -bad sequence), which implies that $\beta $ is extendible.

Assume now that $\alpha $ is non-extendible and let $F \in \mathbb {N}^{\mathbb {N}}$ be a $\le _P$ -bad sequence. We show that there is $i<|{\alpha }|$ and infinitely many k such that $\alpha (i) <_P F(k)$ . This is enough to conclude the proof, as we could take B as any subsequence of F with $\alpha (i) <_P B(k)$ for every k (i.e., $\alpha \trianglelefteq B$ ).

Assume that, for every $i<|{\alpha }|$ there is $k_i$ such that for every $k\ge k_i$ , $\alpha (i)\not \le _P F(k)$ (since P is a partial order, there can be at most one k such that $\alpha (i)=F(k)$ ). Since $\alpha $ is finite, we can take $k:=\max _{i<|{\alpha }|} k_i$ and consider the sequence which extends $\alpha $ by the tail $F(k+1), F(k+2), \dots $ of F. We have now reached a contradiction as this is an infinite $\le _P$ -bad sequence extending $\alpha $ .

Proof. The implication from (1) to (2) holds because $\widehat {\mathsf {ACC}_{\mathbb {N}}} \times \mathsf {lim} \equiv _{\mathrm {W}} \mathsf {lim}$ . The implications from (2) to (3) and from (3) to (4) hold because $\mathsf {lim} \leq _{\mathrm {W}} \mathsf {DS} \leq _{\mathrm {W}} \mathsf {BS}$ [Reference Goh, Pauly and Valenti18, Theorem 4.16]. For the implication from (4) to (1), we consider a name x for an input to f together with witnesses $\Phi ,\Psi $ for the reduction to $\mathsf {BS}$ . We show that, from them, we can uniformly compute an input q to $\widehat {\mathsf {ACC}_{\mathbb {N}}}$ together with an enumeration of the set W of all finite sequences $\sigma $ in the (non-well) partial order P built by $\Phi $ on $(q,x)$ such that $\sigma $ does not extend to an infinite bad sequence. We can then use $\mathsf {lim}$ to obtain the characteristic function of W. Having access to this lets us find an infinite bad sequence in P greedily by avoiding sequences in W. From such a bad sequence (and $(q,x)$ ), $\Psi $ then computes a solution to f for x.

It remains to construct $q = (q_\sigma )_{{\sigma \in \mathbb {N}}}$ and W to achieve the above. At the beginning, W is empty, and we extend each $q_\sigma $ in a way that removes no solution from its $\mathsf {ACC}_{\mathbb {N}}$ -instance. As we do so, for each $\sigma \notin W$ (in parallel), we monitor whether the following condition has occurred:

Once the above occurs for $\sigma $ (if ever), we remove m as a valid solution to $q_\sigma $ by enumerating it. This ensures that $\tau $ , and hence $\sigma $ by Lemma 4.2, cannot extend to an infinite bad sequence in P. We shall then enumerate $\sigma $ into W. This completes our action for $\sigma $ , after which we return to monitoring the above condition for sequences not in W.

It is clear that each $q_\sigma $ is an $\mathsf {ACC}_{\mathbb {N}}$ -instance (with solution set $\mathbb {N}$ if the condition is never triggered, otherwise with solution set $\mathbb {N}\setminus \{m\}$ ). Hence $P:=\Phi (q,x)$ is a non-well partial order. As argued above, no $\sigma \in W$ extends to an infinite bad sequence in P. Conversely, suppose $\sigma $ is a bad sequence in P which does not extend to an infinite bad sequence. Since P is non-wqo, by Lemma 4.2 there is an infinite bad sequence r such that $\sigma \trianglelefteq ^P r$ . Then $\Psi $ has to produce all $\mathsf {ACC}_{\mathbb {N}}$ answers upon receiving $(q,x)$ and r, including an answer to $q_\sigma $ . By continuity, this answer is determined by finite prefixes only. In particular, after having constructed a sufficiently long prefix of q, some finite prefix $\tau $ of r will trigger the condition for $\sigma $ (unless something else triggered it previously), which ensures that $\sigma $ gets placed into W. This shows that W contains exactly the non-extendible finite bad sequences in P, thereby concluding the proof.

Proof. Since $\mathsf {ACC}_{\mathbb {N}} \leq _{\mathrm {W}} f \leq _{\mathrm {W}} \mathsf {BS}$ and f is parallelizable, we have $\widehat {\mathsf {ACC}_{\mathbb {N}}} \times f \leq _{\mathrm {W}} f \leq _{\mathrm {W}} \mathsf {BS}$ . By the previous theorem, $f \leq _{\mathrm {W}} \mathsf {lim}$ .

Proof. Recall that $\mathsf {KL}$ is parallelizable, $\mathsf {ACC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {KL}$ , yet $\mathsf {KL} \nleq _{\mathrm {W}} \mathsf {lim}$ , hence the claim follows from Corollary 4.4.

Proof. As proved in [Reference Kihara, Marcone and Pauly17, Propositions 6.12 and 6.13], $\mathsf {wList}_{2^{\mathbb {N}},\leq \omega }$ is parallelizable and $\mathsf {ACC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {lim} \leq _{\mathrm {W}} \mathsf {wList}_{2^{\mathbb {N}},\leq \omega }$ . On the other hand, $\mathsf {wList}_{2^{\mathbb {N}},\leq \omega } \nleq _{\mathrm {W}} \mathsf {lim}$ [Reference Kihara, Marcone and Pauly17, Corollary 6.16], hence the claim follows from Corollary 4.4.

Proof. If ${}^1f$ is continuous, the conclusion of the first statement is satisfied. Otherwise, $\mathsf {ACC}_{\mathbb {N}} \leq _{\mathrm {W}}^* f$ by [Reference Pauly, Soldà, Patey, Pimentel, Galeotti and Manea25, Theorem 1]. The relativization of Theorem 4.3 then implies $\widehat {f} \leq _{\mathrm {W}}^* \mathsf {lim}$ . We conclude ${}^1f \leq _{\mathrm {W}}^* {}^1\mathsf {lim} \equiv _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}$ . The second statement then follows from $\widehat {\mathsf {C}_{\mathbb {N}}} \equiv _{\mathrm {W}} \mathsf {lim} \leq _{\mathrm {W}} \mathsf {DS}$ .

Proof. Let f be a problem with codomain $\mathbb {N}$ and assume $f\leq _{\mathrm {W}} \mathsf {BS}$ via $\Phi ,\Psi $ . Fix $x\in \operatorname {dom}(f)$ . Let $(P,\leq _P)$ denote the non-well partial order defined by $\Phi (x)$ . We say that a finite $\leq _P$ -bad sequence $\beta $ is sufficiently long if $\Psi (x, \beta )$ returns a natural number in at most $|{\beta }|$ steps.

Consider the non-empty $\Sigma ^{1,x}_1$ set of sufficiently long finite extendible bad sequences. To show that $f\leq _{\mathrm {W}} \boldsymbol {\Sigma }^1_1\mathsf {-DUCC}$ , it is enough to notice that Lemma 4.2 implies that the aforementioned set is dense and upwards-closed with respect to the quasi-order $\trianglelefteq ^P$ (Definition 4.1).

Proof. Clearly, we only need to show that $\boldsymbol {\Sigma }^1_1\mathsf {-DUCC} \leq _{\mathrm {W}} \boldsymbol {\Sigma }^1_1\mathsf {-DUCC}(2^{<\mathbb {N}}, \cdot )$ . Suppose we are given as input a countable quasi-order $(Q,\preceq _Q)$ and a $\boldsymbol {\Sigma }^1_1$ -code for a set $A \subseteq Q$ which is dense and upwards-closed. We shall define a labelling which is computable uniformly in $(Q,\preceq _Q)$ , such that $\lambda ^{-1}(A)$ is dense and upwards-closed. This suffices to prove the claimed reduction, as the preimage of A via $\lambda $ is uniformly $\Sigma ^1_1$ relative to the input and, given $\sigma \in \lambda ^{-1}(A)$ , we may use the input $(Q,\preceq _Q)$ to compute $\lambda (\sigma )\in A$ .

The labelling $\lambda : 2^{<\mathbb {N}} \to Q$ is defined via an auxiliary function (useful to describe $\lambda $ in a recursive manner). First, for every $i \in \mathbb {N}$ , $\overline {\lambda }(x, 0^i):=\overline {\lambda }(x, 0^i1):= x$ . To evaluate $\overline {\lambda }$ on other strings, we fix a standard Q-computable enumeration $(x_n)_{{n}}$ of Q and define, recursively:

We then define $\lambda (\sigma ):=\overline {\lambda }(x_0,\sigma )$ . It is clear that $\lambda $ is total and it is uniformly computable in $(Q,\preceq _Q)$ . Rephrasing the definition, we can observe that, for every x and $\sigma $ , . Besides, . In particular,

Intuitively, $\lambda $ guides the construction of an ascending $\preceq _Q$ -sequence from $x_0$ . Indeed, we will show that $\lambda $ is weakly monotone, and therefore each $\sigma $ induces a finite ascending $\preceq _Q$ -sequence from $x_0$ to $\lambda (\sigma )$ . At the same time, the labelling is designed to be highly redundant, which we use to prove that $\lambda ^{-1}(A)$ is dense.

To show that $\lambda ^{-1}(A)$ is upwards-closed, we claim that for every $\sigma ,\tau \in 2^{<\mathbb {N}}$ , . The claim is proved by first using structural induction on $\sigma $ to show that for all $x \in Q$ , we have $x \preceq _Q \overline {\lambda }(x,\sigma )$ . Using this fact, one can perform another structural induction on $\sigma $ to show that for all x and $\tau $ , we have . Taking $x = x_0$ yields . Since A is upwards-closed in Q, it follows that $\lambda ^{-1}(A)$ is upwards-closed in the prefix ordering.

To prove that $\lambda ^{-1}(A)$ is dense, fix $\rho \in 2^{<\mathbb {N}}$ and assume $\lambda (\rho )\notin A$ . Since A is dense, there is some n such that $\lambda (\rho )\preceq _Q x_n \in A$ . Observe that $\overline {\lambda }(\lambda (\rho ), 0^n11) = \overline {\lambda }(x_n, \varepsilon ) = x_n$ , so it suffices to find some $\rho ' \sqsupseteq \rho $ such that $\lambda (\rho ') = \overline {\lambda }(\lambda (\rho ), 0^n11)$ . The construction of $\rho '$ depends on the form of $\rho $ (all the following claims can be proven by structural induction on $\rho $ ):

• • if $\rho $ ends with $0$ , then , so ;

• • if $\rho $ ends with an odd number of $1$ s, then , so ;

• • otherwise, $\rho $ is either $\varepsilon $ or ends with a positive even number of $1$ s. Then , so .

In all cases, we have found an extension of $\rho $ which maps to $x_n$ , as desired.

Proof. We have $\mathsf {RT}^{{1}}_{{k}} \equiv _{\mathrm {W}} \operatorname {Fin}_{k}\!\left (\mathsf {DS}\right ) \leq _{\mathrm {W}} \operatorname {Fin}_{k}\!\left (\mathsf {BS}\right ) \leq _{\mathrm {W}} \operatorname {Fin}_{k}\!\left (\boldsymbol {\Sigma }^1_1\mathsf {-DUCC}\right )$ by [Reference Goh, Pauly and Valenti18, Theorem 4.31] and Proposition 5.1. It remains to show that $\operatorname {Fin}_{k}\!\left (\boldsymbol {\Sigma }^1_1\mathsf {-DUCC}\right ) \leq _{\mathrm {W}} \mathsf {RT}^{{1}}_{{k}}$ . By Lemma 5.2, it is enough to show that $\operatorname {Fin}_{k}\!\left (\boldsymbol {\Sigma }^1_1\mathsf {-DUCC}(2^{<\mathbb {N}}, \cdot )\right ) \leq _{\mathrm {W}} \mathsf {RT}^{{1}}_{{k}}$ .

Let f be a problem with codomain k and assume $f\leq _{\mathrm {W}}\boldsymbol {\Sigma }^1_1\mathsf {-DUCC}(2^{<\mathbb {N}}, \cdot )$ via $\Phi ,\Psi $ . Observe that every $x\in \operatorname {dom}(f)$ induces a coloring as follows: run $\Psi (x,\sigma )$ in parallel on every $\sigma \in 2^{<\mathbb {N}}$ . Whenever we see that $\Psi (x,\sigma )$ returns a number less than k, we define $c(\tau ):=\Psi (x,\sigma )$ for every $\tau \sqsubseteq \sigma $ such that $c(\tau )$ is not defined yet. By density of the set coded by $\Phi (x)$ , c is total.

By the Chubb–Hirst–McNicholl tree theorem [Reference Chubb, Hirst and McNicholl8], there is some $\rho \in 2^{<\mathbb {N}}$ and some color $i<k$ such that i appears densely above $\rho $ . We claim that such i would be an f-solution to x. To prove this, fix (by density) some $\rho ' \sqsupseteq \rho $ which lies in the set coded by $\Phi (x)$ . Then fix some $\tau \sqsupseteq \rho '$ with color i. Finally, fix some $\sigma \sqsupseteq \tau $ such that $c(\tau )$ was defined to be $\Psi (x,\sigma )$ . Now $\sigma $ lies in the set coded by $\Phi (x)$ (by upwards-closure), so $\Psi (x,\sigma )$ is an f-solution to x. Since $\Psi (x,\sigma ) = c(\tau ) = i$ , we conclude that i is an f-solution to x.

This implies that f reduces to the problem “given a k-coloring of $2^{<\mathbb {N}}$ , find i such that, for some $\sigma $ , i appears densely above $\sigma $ ”. We claim that the latter can be solved using $\mathsf {RT}^{{1}}_{{k}}$ . This follows from the fact that, as shown in [Reference Pauly, Pradic and Soldà24, Corollary 42], $\mathsf {RT}^{{1}}_{{k}}$ can solve the problem $(\mathsf {c}(\eta )^1_{<\infty })_k$ defined as “given a k-coloring , produce $i<k$ such that, for some interval I with rational endpoints, $\alpha ^{-1}(i)$ is dense in I”.

We introduce the order $\prec $ on $2^{<\mathbb {N}}$ defined by setting for all $\sigma , \tau , \rho \in 2^{<\mathbb {N}}$ . This is a computable dense linear order with no endpoints, therefore there is a computable order isomorphism $\varphi $ between $(2^{<\mathbb {N}}, \prec )$ and ${(\mathbb {Q},<)}$ .

Every open interval in $(\mathbb {Q},<)$ contains the $\varphi $ -image of an upper cone in $(2^{<\mathbb {N}}, \prec )$ and vice versa. Specifically, the interval $(\varphi (\sigma ), \varphi (\tau ))$ contains the image of the cone above if is not a prefix of $\tau $ , and the image of the cone above if is a prefix of $\tau $ . Thus, by translating the coloring along $\varphi $ to $\mathbb {Q}$ , we find that a color which appears densely in an interval in $\mathbb {Q}$ also appears densely in a cone in $2^{<\mathbb {N}}$ . This means we can use $(\mathsf {c}(\eta )^1_{<\infty })_k$ to find a color appearing densely in a cone.

Proof. By algebraic properties of $\operatorname {Det}\!\left (\cdot \right )$ , $\operatorname {Fin}_{k}\!\left (\cdot \right )$ , and the parallelization, we have

$$\begin{align*}\operatorname{Det}\!\left(f\right) \leq_{\mathrm{W}} \widehat{\mathrm{Det}_{ \mathbf{2} }(f)} \leq_{\mathrm{W}} \widehat{\operatorname{Fin}_{2}\!\left(f\right)} \leq_{\mathrm{W}} \widehat{\mathsf{RT}^{{1}}_{{2}}}. \end{align*}$$

Since $\operatorname {Det}\!\left (f\right )$ is deterministic, it remains to show that $\operatorname {Det}\!\left (\widehat {\mathsf {RT}^{{1}}_{{2}}}\right ) \leq _{\mathrm {W}} \mathsf {lim}$ . This is known, but for completeness we give a proof using results from the survey of Brattka, Gherardi, and Pauly [Reference Brattka, Gherardi, Pauly, Brattka and Hertling3]:

$$ \begin{align*} \widehat{\mathsf{RT}^{{1}}_{{2}}} &\equiv_{\mathrm{sW}} \widehat{\mathsf{C}_{2}'} & [3, Theorem 11.8.11, Proposition 11.6.10] \\ &\equiv_{\mathrm{sW}} \left(\widehat{\mathsf{C}_{2}}\right)' & [3, Proposition 11.6.12] \\ &\equiv_{\mathrm{sW}} \mathsf{C}_{{2^{\mathbb{N}}}}' & [3, Theorem 11.7.23] \\ &\equiv_{\mathrm{W}} \mathsf{C}_{{2^{\mathbb{N}}}} * \mathsf{lim} & [3, Proposition 11.7.6, Proposition 11.6.14]. \end{align*} $$

By choice elimination (see [Reference Brattka, Gherardi, Pauly, Brattka and Hertling3, Theorem 11.7.25] or [Reference Goh, Pauly and Valenti18, Theorem 3.9]), $\operatorname {Det}\!\left (\mathsf {C}_{{2^{\mathbb {N}}}} * \mathsf {lim}\right ) \leq _{\mathrm {W}} \mathsf {lim}$ . Therefore $\operatorname {Det}\!\left (\widehat {\mathsf {RT}^{{1}}_{{2}}}\right ) \leq _{\mathrm {W}} \mathsf {lim}$ .

Proof. Since $\mathbb {N}$ computably embeds in $\mathbb {N}^{\mathbb {N}}$ , for every problem f we have $\mathrm {Det}_{ \mathbf {\mathbb {N}} }(f)\leq _{\mathrm {W}} \operatorname {Det}\!\left (f\right )$ . In particular, by Corollary 5.5, $\mathrm {Det}_{ \mathbf {\mathbb {N}} }(\mathsf {BS}) \leq _{\mathrm {W}}\operatorname {Det}\!\left (\mathsf {BS}\right ) \equiv _{\mathrm {W}} \mathsf {lim}$ . Since ${}^1\mathsf {lim}\equiv _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}$ ([Reference Brattka, Gherardi and Marcone2, Proposition 13.10], see also [Reference Soldà and Valenti28, Theorem 7.2]), this implies $\mathrm {Det}_{ \mathbf {\mathbb {N}} }(\mathsf {BS})\leq _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}$ . The converse reduction follows from the fact that $\mathsf {C}_{\mathbb {N}}\equiv _{\mathrm {W}} \mathrm {Det}_{ \mathbf {\mathbb {N}} }(\mathsf {DS})$ [Reference Goh, Pauly and Valenti18, Proposition 4.14].

Proof. The reductions $\mathsf {C}_{\mathbb {N}} \leq _{\mathrm {W}} \operatorname {Det} (\boldsymbol {\Pi }^1_1\mathsf {-Bound}) \leq _{\mathrm {W}} \operatorname {Det} ({}^1\mathsf {BS} )\leq _{\mathrm {W}} \operatorname {Det} (\boldsymbol {\Sigma }^1_1 \mathsf {-DUCC} )$ are straightforward: the first one follows from $\mathsf {C}_{\mathbb {N}}\equiv _{\mathrm {W}} \mathsf {UC}_{\mathbb {N}}$ and $\mathsf {C}_{\mathbb {N}}\leq _{\mathrm {W}} \boldsymbol {\Pi }^1_1\mathsf {-Bound}$ , the second one follows from $\boldsymbol {\Pi }^1_1\mathsf {-Bound}\leq _{\mathrm {W}} {}^1\mathsf {BS}$ , and the last one follows from Proposition 5.1. It remains to show that $\operatorname {Det} (\boldsymbol {\Sigma }^1_1\mathsf {-DUCC} )\leq _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}$ . In light of Lemma 5.2, it suffices to show that if $f:\subseteq \mathbb {N}^{\mathbb {N}} \to \mathbb {N}^{\mathbb {N}}$ is such that $f\leq _{\mathrm {W}} \boldsymbol {\Sigma }^1_1\mathsf {-DUCC}(2^{<\mathbb {N}}, \cdot )$ via $\Phi ,\Psi $ then $f\leq _{\mathrm {W}}\mathsf {C}_{\mathbb {N}}$ . Given $x\in \operatorname {dom}(f)$ , we can compute the set

$$ \begin{align*} X:=\{ \sigma \in 2^{<\mathbb{N}} {}\,:\,{} & (\forall \tau_0,\tau_1\sqsupseteq \sigma)(\forall n\in\mathbb{N}) \\ & ((\Psi(x,\tau_0)(n)\downarrow \land \Psi(x,\tau_1)(n)\downarrow)\rightarrow \Psi(x,\tau_0)(n)=\Psi(x,\tau_1)(n)) \}. \end{align*} $$

Observe that this is a $\Pi ^{0,x}_1$ subset of $2^{<\mathbb {N}}$ and it is non-empty: indeed, letting $A\subseteq 2^{<\mathbb {N}}$ be the set described by $\Phi (x)$ , the fact that f is deterministic implies that $A\subseteq X$ . We can therefore use $\mathsf {C}_{\mathbb {N}}$ to compute some $\sigma \in X$ . Observe that, by the density of A, there is $\tau \sqsupseteq \sigma $ such that $\tau \in A$ . In particular, $\Psi (x,\sigma )=\Psi (x,\tau )\in f(x)$ .

Proof. Notice $\operatorname {dom}(g^{\mathsf {G}}) = \operatorname {dom}(g)$ since g is computably true. It is clear that $g \leq _{\mathrm {W}} g^{\mathsf {G}}$ . Next, suppose h has codomain $\mathbb {N}$ and $g \leq _{\mathrm {W}} h$ via $\Gamma $ and $\Delta $ . To reduce $g^{\mathsf {G}}$ to h, observe that if $p \in \operatorname {dom}(g^{\mathsf {G}})$ , then $\Gamma (p) \in \operatorname {dom}(h)$ , and if $n \in h(\Gamma (p))$ , then $\Delta (p,n) \in g(p)$ . Using an index for $\Delta $ , we can define a functional $\Psi $ such that $\Phi _{\Psi (p,n)}(p) = \Delta (p,n)$ for every $p \in \mathbb {N}^{\mathbb {N}}$ and $n \in \mathbb {N}$ . Then if $n \in h(\Gamma (p))$ , we have $\Psi (p,n) \in g^{\mathsf {G}}(p)$ as desired. (Indeed $\Psi $ does not even need access to p, so $g^{\mathsf {G}} \leq _{\mathrm {sW}} h$ .)

Proof. Since g is single-valued and $g \leq _{\mathrm {W}} g^{\mathsf {G}}$ , we have $g \leq _{\mathrm {W}} \operatorname {Det}\!\left (g^{\mathsf {G}}\right ) <_{\mathrm {W}} g^{\mathsf {G}}$ . So $\operatorname {Det}\!\left (g^{\mathsf {G}}\right )$ cannot be first-order, by Theorem 6.1. We conclude that ${}^1\operatorname {Det}\!\left (g^{\mathsf {G}}\right ) <_{\mathrm {W}} \operatorname {Det}\!\left (g^{\mathsf {G}}\right )$ .

Proof. Suppose $h : \subseteq \mathbb {N}^{\mathbb {N}} \to \mathbb {N}^{\mathbb {N}}$ satisfies $f \equiv _{\mathrm {W}} h$ . Fix $\Phi $ and $\Psi $ witnessing that $f \leq _{\mathrm {W}} h$ . We define $r : \subseteq \mathbb {N}^{\mathbb {N}} \to \mathbb {N}^{\mathbb {N}}$ by $r(x) = \Psi (x,h(\Phi (x)))$ . By construction, r is a realizer of f, thus it in particular holds that $f \leq _{\mathrm {W}} r$ . Also by construction, $r \leq _{\mathrm {W}} h$ . Since $h \equiv _{\mathrm {W}} f$ , it follows that $r \equiv _{\mathrm {W}} f$ .

Proof. Suppose towards a contradiction that $U \supseteq A$ is c.e. and disjoint from $\mathrm {Tot}\setminus A$ . Then we can compute halting as follows. Given e, compute an index $f(e)$ such that $\phi _{f(e)}$ copies $\phi _{e_0}$ until $\phi _e(e)$ halts (if ever), after which $\phi _{f(e)}$ flips each output of $\phi _{e_0}$ . Notice if $\phi _e(e)$ halts, then $f(e) \in \mathrm {Tot}\setminus A$ , else $f(e) \in A \subseteq U$ . So we can compute whether $\phi _e(e)$ halts by waiting for it to halt, or for $f(e)$ to appear in U.

Proof. We shall construct a sequence of computable reals $(g_n)_{n \in \mathbb {N}}$ such that for every pair of Turing functionals $\Gamma $ and $\Delta $ (which form a potential Weihrauch reduction from some realizer of $g^{\mathsf {G}}$ to $g^{\mathsf {G}}$ itself), one of the following holds:

1. (1) some $\Gamma (n)$ is not a name for a natural number (i.e., an instance of g as defined below);

2. (2) there are n, e, $e' \in \mathbb {N}$ such that $\Phi _e(\Gamma (n)) = g_{\Gamma (n)} = \Phi _{e'}(\Gamma (n))$ but $\Delta (n,e) \neq \Delta (n,e');$

3. (3) there are n, $e \in \mathbb {N}$ such that $\Phi _e(\Gamma (n)) = g_{\Gamma (n)}$ but $\Phi _{\Delta (n,e)}(n) \neq g_n$ .

If we then define by $g(n) := g_n$ , we see that g is single-valued, computably true, and $g^{\mathsf {G}}$ is not Weihrauch equivalent to any of its realizers (hence not deterministic, by Proposition 6.3).

We shall construct $(g_n)_{n \in \mathbb {N}}$ in stages. At each stage, we handle a single pair $\Gamma $ and $\Delta $ by defining at most two new computable reals $g_n$ . We assume that for every $n \in \mathbb {N}$ , $\Gamma (n)$ names a natural number (otherwise no action is needed). Consider the following cases.

Case 1. For some $n \in \mathbb {N}$ such that $\Gamma (n) \neq n$ , $g_n$ has not been defined. By defining $g_{\Gamma (n)}$ if necessary, we may fix some e such that $\Phi _e(\Gamma (n)) = g_{\Gamma (n)}$ . Since $\Gamma (n) \neq n$ , we may then define $g_n$ to be a computable real which differs from $\Phi _{\Delta (n,e)}(n)$ (regardless of whether the latter is total), ensuring (3).

Case 2. Fix some $n \in \mathbb {N}$ such that $g_n$ is not defined. Since Case 1 fails, $\Gamma (n) = n$ .

Case 2a. There are $e,e' \in \mathbb {N}$ such that $\Phi _e(\Gamma (n)) = \Phi _{e'}(\Gamma (n))$ and $\Delta (n,e) \neq \Delta (n,e')$ (this includes the case where either or both sides diverge). We define $g_{\Gamma (n)} = \Phi _e(\Gamma (n))$ , ensuring (2).

Case 2b. If there is some $e \in \mathbb {N}$ such that $\Phi _e(\Gamma (n))$ is total and not equal to $\Phi _{\Delta (n,e)}(n)$ , we define $g_n = \Phi _e(\Gamma (n))$ . This ensures (3).

To complete the current stage of the construction, we shall show that the above cases encompass all possibilities. Suppose otherwise. Since Case 2b fails, for all $e \in \mathbb {N}$ , whenever $\Phi _e(\Gamma (n))$ is total, so is $\Phi _{\Delta (n,e)}(n)$ and they must be equal. Since Case 2a fails, whenever $\Phi _e(\Gamma (n)) = \Phi _{e'}(\Gamma (n))$ , we have $\Delta (n,e) = \Delta (n,e')$ . We deduce that whenever $\Phi _e(\Gamma (n))$ and $\Phi _{e'}(\Gamma (n))$ are total, $\Delta (n,e) = \Delta (n,e')$ if and only if $\Phi _e(\Gamma (n)) = \Phi _{e'}(\Gamma (n))$ . This contradicts Lemma 6.4: Define $F: \subseteq \mathbb {N} \to \mathbb {N}$ by $F(d) = \Delta (n,e)$ , where $\Phi _e$ upon input $\Gamma (n)$ merely simulates $\phi _d$ .

This completes the current stage of the construction. Observe we have ensured that one of (1)–(3) hold for $\Gamma $ and $\Delta $ .

Proof. By Proposition 6.5, fix a problem g which is computably true and single-valued, such that $g^{\mathsf {G}}$ is not deterministic. By Proposition 6.2, ${}^1\operatorname {Det}\!\left (g^{\mathsf {G}}\right ) <_{\mathrm {W}} \operatorname {Det}\!\left (g^{\mathsf {G}}\right )$ . Let $f = g^{\mathsf {G}}$ . Then ${}^1\operatorname {Det}\!\left (f\right ) \equiv _{\mathrm {W}} {}^1\operatorname {Det}\!\left (g^{\mathsf {G}}\right ) <_{\mathrm {W}} \operatorname {Det}\!\left (g^{\mathsf {G}}\right ) \equiv _{\mathrm {W}} \operatorname {Det}\!\left ({}^1f\right )$ , where the last equivalence holds because f is first-order and $\operatorname {Det}\!\left (\cdot \right )$ is degree-theoretic.

Proof. Given some $p \in \mathbf {X}^{\mathbb {N}}$ , we observe that $\{n \in \mathbf {X} {}\,:\,{} n \neq \lim _{i \to \infty } p_i\}$ is a $\Pi ^{0,p}_2$ -set omitting at most one element. That shows that $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbf {X}}$ can solve the problem of finding a non-limit point.

Conversely, let A be a $\boldsymbol {\Pi }^0_2$ -subset of $\mathbf {X}$ with $|\mathbf {X}\setminus A|\le 1$ . We can A-compute such that, for each $n\in \mathbf {X}$ , $q(n,\cdot )$ contains infinitely many $1$ s iff $n \in A$ . We define a sequence $p \in \mathbf {X}^{\mathbb {N}}$ as follows: for each $n\in \mathbf {X}$ and $k\in \mathbb {N}$ , let $M_{n,k}:= |\{ j \le k {}\,:\,{} q(n,j)=0 \}|$ .

For the sake of readability, we distinguish the cases where $\mathbf {X}$ is finite or not. If $\mathbf {X}=k$ , we define

$$\begin{align*}p_i:= \min \{ n < k {}\,:\,{} (\forall m < k)(M_{n,i}\ge M_{m,i})\}. \end{align*}$$

If $\mathbf {X}=\mathbb {N}$ , at each stage i, for each $n\le i$ we only see $j(n)$ -many elements of the sequence $(q(n,j))_{{j\in \mathbb {N}}}$ , where $j(n)$ is least such that $\langle n,j(n) \rangle>i$ . In this case, we define

$$\begin{align*}p_i:= \min \{ n \le i {}\,:\,{} (\forall m\le i)(M_{n,j(n)}\ge M_{m,j(m)})\}. \end{align*}$$

Observe that, in both cases, if $A = \mathbf {X} \setminus \{m\}$ for some m, then $\lim _{i\to \infty } p_i = m$ , hence any $n\neq \lim _{i\to \infty } p_i$ is a valid solution for $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbf {X}}(A)$ .

Proof. As $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbb {N}}$ is a first-order problem, $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {lim}$ would already imply $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}$ . The characterization in Proposition 7.2 shows that $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbb {N}}$ is a closed fractal. By the absorption theorem for closed fractals [Reference Le Roux and Pauly20, Theorem 2.4], this means that $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbb {N}} \leq _{\mathrm {W}} \mathsf {C}_{\mathbb {N}}$ in turn would imply that $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{\mathbb {N}}$ is computable, which is readily seen to be false.

Proof. Let $g:\subseteq \mathbb {N}^{\mathbb {N}} \rightrightarrows k$ be such that $g\leq _{\mathrm {W}}\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+2}^\diamond $ and let $\Phi $ be the computable functional witnessing the winning strategy for Player 2 for the game $G= G(\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+2}\to g)$ . For each input $x\in \operatorname {dom}(g)$ , we can build the c.e. tree T of all possible runs of the game G where Player 1 starts by playing x. More precisely, for $\sigma \in (k+2)^{<\mathbb {N}}$ and $i<k+2$ , we enumerate if $\sigma \in T$ and Player 2 does not declare victory on the $(|{\sigma }|+1)$ -th move. Moreover, if Player 2 declares victory on the $(|{\sigma }|+1)$ -th move, we enumerate in T and commit to never extend this branch further. Intuitively, if Player 2 declares victory then we reached a leaf of the tree, and the leaf has a special label that contains the last Player 2’s move, i.e., the solution to $g(x)$ obtained using $\Phi $ when the oracle answers are .

Observe that, each non-leaf $\sigma \in T$ has exactly $k+2$ -many children, and at least $k+1$ of them correspond to valid Player 1 moves (i.e., possible $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+2}$ -answers). Observe also that the tree T need not be well-founded and not all the leaves are necessarily labelled as such. Indeed, the hypothesis that Player 2 has a winning strategy only guarantees that the subtree $S\subseteq T$ corresponding to valid runs of the game is well-founded (and it is computable to tell if a string in S is a label).

By unbounded search, we can compute a well-founded tree $S\subseteq T$ such that, $\varepsilon \in S$ , each non-leaf $\sigma \in S$ has exactly $(k+1)$ -many children, and each leaf of S is of the form for some $\tau \in (k+2)^{<\mathbb {N}}$ and $\Phi (x, \tau )<k$ . To compute a solution for $g(x)$ we proceed recursively on the (finite) rank of S: each leaf of S is simply labelled with $\Phi (x, \tau )$ . To compute the label of a non-leaf $\sigma $ , observe that, by the pigeonhole principle, we can choose $i<k$ such that at least two of the $k+1$ children of $\sigma $ are labelled with i. We can then label $\sigma $ with i and move to the next stage.

We claim that the label of $\varepsilon $ is a valid solution for $g(x)$ . Indeed, assume $\tau \in S$ corresponds to a valid run of the game, namely for each $i<|{\tau }|$ , $\tau (i)$ is a valid $(n+2)$ -th move for Player 1 when Player 2 plays according to $\Phi $ and the first $(n+1)$ moves of Player 1 are . By induction, any such $\tau $ receives a label $i<k$ such that $i\in g(x)$ . The statement follows from the fact that $\varepsilon $ always represents a valid run of the game.

Proof. This can be proved observing that, for every f, $\operatorname {Fin}_{k}\!\left (f\right ) \leq _{\mathrm {W}} {}^1f$ , and therefore $\operatorname {Fin}_{k}\!\left (f\right ) \equiv _{\mathrm {W}} \operatorname {Fin}_{k}\!\left ({}^1f\right )$ , as $\operatorname {Fin}_{k}\!\left (f\right )$ is the strongest problem with codomain k that is reducible to f. This implies that

$$\begin{align*}\operatorname{Fin}_{k}\!\left(\widehat{\boldsymbol{\Gamma}\mathsf{-ACC}_{k+2}}\right) \equiv_{\mathrm{W}} \operatorname{Fin}_{k}\!\left({}^1\left(\widehat{\boldsymbol{\Gamma}\mathsf{-ACC}_{k+2}} \right) \right) \leq_{\mathrm{W}} \operatorname{Fin}_{k}\!\left(\boldsymbol{\Gamma}\mathsf{-ACC}_{k+2}^\diamond \right)\leq_{\mathrm{W}} \mathrm{id}, \end{align*}$$

where the second inequality follows from the fact, for every first-order f, ${}^1\widehat {f}\leq _{\mathrm {W}} f^\diamond $ (see [Reference Soldà and Valenti28, Theorem 5.7 and Proposition 6.3]). The last inequality is the statement of Proposition 7.5.

Proof. Proposition 2.1 and the definitions of deterministic/k-finitary part imply that, for every f and every $k\ge 2$ ,

$$\begin{align*}\operatorname{Det}\!\left(f\right) \leq_{\mathrm{W}} \widehat{\mathrm{Det}_{ \mathbf{k} }(f)} \leq_{\mathrm{W}} \widehat{\operatorname{Fin}_{k}\!\left(f\right)}. \end{align*}$$

Letting $f= \widehat {\boldsymbol {\Gamma }\mathsf {-ACC}_{k+2}}$ , the claim follows from Corollary 7.6.

Proof. We use the equivalent formulation of $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}$ introduced in Proposition 7.2. For every $A\in \operatorname {dom}(\mathsf {ACC}_k)$ , we uniformly compute a sequence $p\in \mathbb {N}^{\mathbb {N}}$ as follows:

$$\begin{align*}p(s):=\begin{cases} i & \text{if } i \text{ is enumerated outside of }A \text{ by stage }s \\ k & \text{otherwise.} \end{cases} \end{align*}$$

Let $n<k+1$ be such that $n\neq \lim _{s\to \infty } p(s)$ . If $n=k$ then $A\neq k$ , therefore we can computably find (by unbounded search) the unique $m \in k\setminus A$ . Conversely, if $n\neq k$ then it follows by the definition of p that $n\in A$ .

Proof. We shall use the characterization of $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k}$ introduced in Proposition 7.2, i.e., we can assume that a name for $A\in \operatorname {dom}(\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k})$ is a sequence $p\in k^{\mathbb {N}}$ such that if $\lim _n p(n)$ exists then it is the unique $i<k$ not in A.

Let us first show that $F\leq _{\mathrm {W}} \boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}$ . Let $\langle p, q \rangle $ be a name for $(A,n)\in \operatorname {dom}(F)$ . We define a sequence $r\in (k+1)^{\mathbb {N}}$ as follows: as long as $q(i)=0$ , let $r(i):=k$ . If $q(i)\neq 0$ for some i, then, for every $j\ge i$ , $r(j):=p(j)$ .

Clearly, r is a (name for a) valid input for $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}$ . Observe that if q is not constantly $0$ then $|A|<k$ , hence the sequence p has a limit $l<k$ . In particular, r always has a limit. We claim that, given $m\neq \lim _{i\to \infty } r(i)$ , we can uniformly compute a valid solution for $F(A,n)$ . Indeed, if $m<k$ , then $m\in A$ . This follows from the fact that if q is constantly $0$ then $|A|=k$ . Conversely, if q is not constantly $0$ then $m\neq \lim _i p(i)$ , and therefore $m\in A$ . On the other hand, if $m=k$ then q is not constantly $0$ , so, by hypothesis, if $q(i)$ is the first non-zero element of q then $q(i)-1\in A$ . This shows that $F\leq _{\mathrm {W}} \boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}$ .

Assume now that $f:\subseteq \mathbb {N}^{\mathbb {N}} \rightrightarrows k$ is such that $f\leq _{\mathrm {W}} \boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}$ via $\Phi ,\Psi $ . For the sake of readability, we can assume that $\Psi $ has codomain $\mathbb {N}$ . To reduce f to F, suppose we are given some $x \in \operatorname {dom}(f)$ . In parallel, we can simulate the computations $\Psi (x,i)$ for every $i<k+1$ . We can uniformly compute two sequences $p, q\in k^{\mathbb {N}}$ as follows: we wait until we either find $i<j<k+1$ such that $\Psi (x,i)=\Psi (x,j)=n$ or we see that $|\{ \Psi (x,i) {}\,:\,{} i<k+1\}|=k$ (by the pigeonhole principle, at least one of two cases must happen). If we first find $i<j<k+1$ such that ${\Psi (x,i)=\Psi (x,j)=n}$ , we define $p,q$ so that $0^s (n+1)\sqsubseteq q$ for some s and p is any convergent sequence with $\lim p \neq n$ . On the other hand, if we first see that $|\{ \Psi (x,i) {}\,:\,{} i<k+1\}|=k$ , we proceed as follows: we keep searching for $i<j$ as above. If they are never found, we let $q:=0^\omega $ and p be any non-convergent sequence. Conversely, if they are found, we define q so that $0^s (n+1)\sqsubseteq q$ for some s. Moreover, we let p be a convergent sequence with $\lim p\neq n$ such that if $\lim _t \Psi (x,\Phi (x)(t))$ exists and is different from n then $\lim p = \lim _t \Psi (x,\Phi (x)(t))$ . This can be done by computing $\Psi (x,\Phi (x)(t))$ and replacing each occurrence of n with some $m\neq n$ .

Observe that $\langle p,q \rangle $ is a valid name for an input $(A,n_\bot )$ of F. Indeed, q is constantly $0$ iff p is not convergent, i.e., iff p is a name for the input A of $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k}$ with $|A|=k$ . Moreover, if q is not constantly $0$ then it is a name for some $n\in \mathbb {N}$ with $n\neq \lim p$ . Given $m\in F(A,n_\bot )$ , we can uniformly compute a solution for $f(x)$ as follows: we wait until we see that q is the name for some $n\neq \bot $ or that $|\{ \Psi (x,i) {}\,:\,{} i<k+1\}|=k$ . If the former is observed first, then $n\in f(x)$ , as there are $i<j$ with $\Psi (x,i)=\Psi (x,j)=n$ , and either i or j is a valid solution for $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}(\Phi (x))$ . Conversely, if we first see that $|\{ \Psi (x,i) {}\,:\,{} i<k+1\}|=k$ , then any $m\in F(A,n_\bot )$ is a valid solution for $f(x)$ . Indeed, if $|f(x)|<k$ then the (unique) wrong solution is $\lim \Psi (x,\Phi (x))$ (as $\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+1}(\Phi (x))$ has at most one wrong solution). By construction, $m\neq \lim p = \lim \Psi (x,\Phi (x))$ , and therefore $m\in f(x)$ . Observe that, in this case, q is not the constantly $0$ sequence (if q is constantly $0$ then every $i<k+1$ is valid solution for $f(x)$ ).

Proof. For the first part of the statement, fix $k\ge 1$ . By Proposition 8.2 and Corollary 7.6,

$$\begin{align*}\operatorname{Fin}_{k}\!\left(\mathsf{BS}|_{\operatorname{Tree}}\right)\leq_{\mathrm{W}} \operatorname{Fin}_{k}\!\left(\widehat{\boldsymbol{\Pi}^0_2\mathsf{-ACC}_{k+2}}\right) \equiv_{\mathrm{W}} \mathrm{id}. \end{align*}$$

Similarly, the second part of the statement then follows from $\operatorname {Det}\!\left (\widehat {\boldsymbol {\Pi }^0_2\mathsf {-ACC}_{k+2}}\right ) \equiv _{\mathrm {W}} \mathrm {id}$ (Corollary 7.7).

Proof. We only need to show that ${}^1\mathsf {BS}|_{\operatorname {Tree}} \leq _{\mathrm {W}} \operatorname {ExtVer}$ . An instance of ${}^1\mathsf {BS}|_{\operatorname {Tree}}$ is a tree T of infinite width together with a notion of “sufficiently large finite antichain” (more formally, the latter is a computable functional which, when given T and an infinite T-antichain, must halt on some initial segment thereof). A solution is a sufficiently large finite antichain which is extendible to an infinite one. We shall compute a tree $T' \subseteq \mathbb {N}^{<\mathbb {N}}$ of infinite width such that each vertex $v \in T'$ (other than the root) is labelled with a sufficiently large finite antichain $A_v$ in T, and if v extends to an infinite $T'$ -antichain, then $A_v$ extends to an infinite T-antichain. For clarity, we shall use Greek letters ( $\sigma $ , $\tau $ ) to denote the vertices of T, and Roman letters (v, w) to denote the vertices of $T'$ .

The tree $T'$ is defined recursively as follows. Start by putting the root in $T'$ . The children of the root will each be labelled with a sufficiently large finite antichain $A_i$ in T. Specifically, the i-th child of the root is defined recursively as follows: First search for $A_0$ and $A_1$ in T, disjoint and each sufficiently large, such that $A_0 \cup A_1$ forms an antichain. Since T has infinite width, this search is successful and we label the first child with $A_0$ and the second child with $A_1$ . Then search for $A_2$ sufficiently large and disjoint from $A_0 \cup A_1$ such that $A_0 \cup A_1 \cup A_2$ forms an antichain. If found, we label the third child with $A_2$ . Proceed to search for $A_3$ , and so on. Note that $(A_i)_{{i}}$ may not be infinite. The reason why we start by searching for two antichains $A_0$ and $A_1$ , and then continue by searching one extra antichain at a time (first $A_2$ , then $A_3$ , and so on) is to guarantee that if a node has a child, then it has at least two incomparable children. More generally, if $v \in T'$ is not the root, then we define the children of v by searching as above among the sufficiently large antichains B such that $B \trianglerighteq ^T A_v$ (i.e., there is some $\sigma \in A_v$ such that $\sigma \sqsubseteq \tau $ for all $\tau \in B$ ). Note that for certain v, this search may turn up empty. This completes the construction of $T'$ , excepting a final modification to ensure that $T'$ is computable (uniformly in T): Each child of a vertex in $T'$ shall encode the stage at which its corresponding T-antichain was found.

We claim that $T'$ has infinite width. If any vertex has infinitely many children, we are done. Otherwise, we shall prove by induction that at each level of $T'$ (beyond the root), there is some vertex v and some $\sigma \in A_v$ such that $[\sigma ]_T$ has infinite width. This would imply that v has at least two children, allowing us to conclude that $T'$ has infinite width. The inductive step proceeds as follows: Suppose $v \in T'$ and $\sigma \in A_v$ are such that $[\sigma ]_T$ has infinite width. Say the children of v are labelled by the antichains $A_0,A_1,\dots ,A_k$ . Then $A_0 \cup \dots \cup A_k$ is not extendible to an infinite antichain in $\bigcup _{\rho \in A_v} [\rho ]_T$ (else v would have additional children), but the latter contains $[\sigma ]_T$ and thus has infinite width. So there is some i and some $\rho \in A_i$ such that $[\rho ]_T$ has infinite width. The base case proceeds similarly; simply consider the children of the root in $T'$ and use the assumption that T has infinite width.