Oracle Computability and Turing Reducibility in the Calculus of Inductive Constructions

Abstract

We develop synthetic notions of oracle computability and Turing reducibility in the Calculus of Inductive Constructions (CIC), the constructive type theory underlying the Coq proof assistant. As usual in synthetic approaches, we employ a definition of oracle computations based on meta-level functions rather than object-level models of computation, relying on the fact that in constructive systems such as CIC all definable functions are computable by construction. Such an approach lends itself well to machine-checked proofs, which we carry out in Coq.

There is a tension in finding a good synthetic rendering of the higher-order notion of oracle computability. On the one hand, it has to be informative enough to prove central results, ensuring that all notions are faithfully captured. On the other hand, it has to be restricted enough to benefit from axioms for synthetic computability, which usually concern first-order objects. Drawing inspiration from a definition by Andrej Bauer based on continuous functions in the effective topos, we use a notion of sequential continuity to characterise valid oracle computations.

As main technical results, we show that Turing reducibility forms an upper semilattice, transports decidability, and is strictly more expressive than truth-table reducibility, and prove that whenever both a predicate p and its complement are semi-decidable relative to an oracle q, then p Turing-reduces to q.

Yannick Forster received funding from the European Union’s Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No. 101024493. Dominik Kirst is supported by a Minerva Fellowship of the Minerva Stiftung Gesellschaft fuer die Forschung mbH.

Appendices

A Glossary of Definitions

We collect some basic notations and definitions:

• $\mathbb {P}$ is the (impredicative) universe of propositions.

• Natural numbers:              $n : \mathbb {N}:\,\!:= 0 \mid \textsf{S}\; n$

• Booleans:              $b : \mathbb {B}:\,\!:= \textsf{true}\mid \textsf{false}$

• Unit type:                    $\mathbbm {1}:\,\!:= \star$

• Sum type:        $X + Y :\,\!:= \textsf{inl} x \mid \textsf{inr} y \quad (x : X, y : Y)$

• Option type:        $o : X^{?} :\,\!:= \textsf {None}\mid {\textsf {Some}\;}x \quad (x : X)$

• Lists:        $l : X^* :\,\!:= [\;] \mid x :\,\!: l \quad (x : X)$

List operations. We often rely on concatenation of of two lists $l_1 \mathbin {+\!\!\!+}l_2$:

Also, we use an inductive predicate $\textsf{Forall}_2\!:\!(X\rightarrow Y\rightarrow \mathbb {P}) \rightarrow X^* \rightarrow Y^* \rightarrow \mathbb {P}$

Characteristic Relation. The characteristic relation $\hat{p}\!:\!X\rightarrow \mathbb {B}\rightarrow \mathbb {P}$ of a predicate $p\!:\!X\rightarrow \mathbb {P}$ is introduced in Sect. 3 as

$$\hat{p}~:=~\lambda x b. {\left\{ \begin{array}{ll} px &{} \text {if } b = \textsf{true}\\ \lnot p x &{} \text {if } b = \textsf{false}. \\ \end{array}\right. }$$

Reducibility.$\preceq _{m}$ is many-one reducibility, introduced in Sect. 6. $\preceq _{\texttt {tt}}$ is truth-table reducibility, introduced in Sect. 8. $\preceq _{\textsf{T}}$ is Turing reducibility, introduced in Sect. 3.

Interrogations. The interrogation predicate $\sigma \mathbin {;} R \vdash _{} qs \mathbin {;} as$ is introduced in Sect. 2. It works on a tree $\sigma \!:\!A^* \rightarrow Q + O$. We often also use trees taking an input, i.e. $\tau \!:\!I \rightarrow A^* \rightarrow Q + O$. Given $\sigma$, we denote the subtree starting at path $l \!:\!A^*$ with $\sigma \mathbin {@}l := \lambda l'.\;\sigma (l \mathbin {+\!\!\!+}l')$.

Partial Functions. We use an abstract type of partial values over X, denoted as $\mathcal P X$, with evaluation relation $\triangleright \!:\!\mathcal P X \rightarrow X \rightarrow \mathbb {P}$. We set $X \rightharpoonup Y := X \rightarrow \mathcal P Y$ and use

• $\textsf {ret}\;\!:\!X \rightharpoonup X$ with $\textsf {ret}\;x \triangleright x$,

• $\mathbin {>\!\!\!>\!\!=}\!:\! \mathcal P X \rightarrow (X \rightarrow \mathcal P Y) \rightarrow \mathcal P Y$ with $x \mathbin {>\!\!\!>\!\!=} f \triangleright y \leftrightarrow \exists v.\;x \triangleright v \wedge f v \triangleright y$,

• $\mu \!:\!(\mathbb {N}\rightarrow \mathcal P \mathbb {B}) \rightarrow \mathcal P \mathbb {N}$ with $\mu f \triangleright n \leftrightarrow f n \triangleright \textsf{true}\wedge \forall m < n.\;f m \triangleright \textsf{false}$, and

• $\textsf {undef}\!:\!\mathcal P X$ with $\forall v.\;\textsf {undef}\not \triangleright v$.

One can for instance implement $\mathcal P X$ as monotonic sequences $f:\mathbb {N}\rightarrow X^{?}$, i.e. with $f n = {\textsf {Some}\;}x \rightarrow \forall m \ge n.\;f m = {\textsf {Some}\;}x$ and $f \triangleright x := \exists n.\;f n = {\textsf {Some}\;}x$. For any implementation it is only crucial that the graph relation $\lambda x y.f x \triangleright y$ for $f \!:\!\mathbb {N}\rightharpoonup \mathbb {N}$ is semi-decidable but cannot be proved decidable. Semi-decidability induces a function $\rho \!:\!\mathcal {P} X \rightarrow \mathbb {N}\rightarrow X^{?}$, which we write as $\rho ^n x$ with the properties that $x \triangleright v \leftrightarrow \exists n.\;\rho ^n x = {\textsf {Some}\;}v$ and $\rho ^n x = {\textsf {Some}\;}v \rightarrow \forall m \ge n.\;\rho ^m x = {\textsf {Some}\;}v$.

B Extended Forms of Interrogations

1.1 B.1 Extended Interrogations with State

As an auxiliary notion, before introducing the stalling interrogations, we first introduce extended interrogations with a state argument, but without stalling. An extended tree is a function $\sigma : S \rightarrow A^* \rightharpoonup (S \times Q) + O$. We define an inductive extended interrogation predicate $\sigma \mathbin {;} R \vdash _{} qs \mathbin {;} as \mathbin {;} s \succ s'$ by:

A functional F mapping $R \!:\!Q \rightarrow A \rightarrow \mathbb {P}$ to a relation of type $I \rightarrow O \rightarrow \mathbb {P}$ is computable via extended interrogations if there are a type S, an element $s_0 : S$, and a function $\tau \!:\!I \rightarrow S \rightarrow A^* \rightharpoonup (S \times Q) + O$ such that

$$\begin{aligned} \forall R\,i\,o.\;F R\,i\,o \leftrightarrow \exists qs \; as \; s. ~ \tau i \mathbin {;} R \vdash _{} qs \mathbin {;} as \mathbin {;} s_0 \succ s ~\wedge ~ \tau \,i\,s\; as \triangleright \textsf{out}\;o. \end{aligned}$$

Note that we do not pass the question history to the function here, because if necessary it can be part of the type S.

Lemma 37 Computable functionals are computable via extended interrogations.

Proof

Let F be computable by $\tau$. Set S to be any inhabited type with element $s_0$ and define

$$\tau '\; i\; s\; l := \tau \,i\,l \mathbin {>\!\!\!>\!\!=} \lambda x.\; {\left\{ \begin{array}{ll} \textsf {ret}\;(\textsf{ask}\;(s, q)) &{} \text {if } x = \textsf{ask}\;q \\ \textsf {ret}\;o &{} \text {if } x = \textsf{out}\;o. \end{array}\right. }. $$

Then $\tau '$ computes F via extended interrogations.    $\square$

Lemma 38 Functionals computable via extended interrogations are computable.

Proof

Let $\tau \!:\!I \rightarrow S \rightarrow A^* \rightharpoonup (S \times Q) + O$ compute F via extended interrogations. Define $\tau ' \!:\!S \rightarrow A^* \rightarrow I \rightarrow A^* \rightharpoonup Q + O$ as

Then $\tau '\,s_0\,[]$ computes F.    $\square$

1.2 B.2 Stalling Interrogations

We here give the left out proofs that stalling interrogations as described in Sect. 9 and interrogations are equivalent.

Lemma 39 Functionals computable via extended interrogations are computable via stalling interrogations.

Proof

Let F be computable using a type S and element $s_0$ by $\tau$ via extended interrogations. We use the same type S and element $s_0$ and define $\tau '$ to never use stalling:

$$\tau '\, i\, s\, l := \tau \,i\,s\,l \mathbin {>\!\!\!>\!\!=} \lambda x.\; {\left\{ \begin{array}{ll} \textsf {ret}\;(\textsf{ask}\;(s', {\textsf {Some}\;}q)) &{} \text {if } x = \textsf{ask}\;(s',q) \\ \textsf {ret}\;(\textsf{out}\;o) &{} \text {if } x = \textsf{out}\;o. \end{array}\right. } $$

Then $\tau '$ computes F via stalling interrogations.    $\square$

Lemma 40 Functionals computable via stalling interrogations are computable via extended interrogations.

Proof

Take $\tau \!:\!I \rightarrow S \rightarrow A^* \rightharpoonup (S \times Q^{?}) + O$ computing F via stalling interrogations. We construct $\tau '\,i\,s\, as$ to iterate the function $\lambda s'.\;\tau \,i\,s'\, as$ of type $S \rightharpoonup (S \times Q^{?}) + O$. If $\textsf{ask}\;(s'',\textsf {None})$ is returned, the iteration continues with $s''$. If $\textsf{ask}\;(s, {\textsf {Some}\;}q)$ is returned, $\tau '\,i\,s as$ returns $\textsf{ask}\;(s, q)$. If $\textsf{out}\;o$ is returned, $\tau '\,i\,s\, as$ returns $\textsf{out}\;o$ as well.

We omit the technical details how to implement this iteration process using unbounded search $\mu : (\mathbb {N}\rightharpoonup \mathbb {B}) \rightharpoonup \mathbb {N}$.    $\square$

1.3 B.3 Proofs of Closure Properties

We here give the proofs that executing two computable functionals one after the other, composing computable functionals, and performing an unbounded search on a computable functional are all computable operations as stated in Sect. 4. We explain the tree constructions, which are always the core of the argument. The verification of the trees are then tedious but relatively straightforward inductions, we refer to the Coq code for full detail.

Proof

(of Lemma 7). Let $\tau _1$ compute $F_1$ maping relations $R \!:\!Q\rightarrow A \rightarrow \mathbb {P}$ to relations of type $I \rightarrow O' \rightarrow \mathbb {P}$, and $\tau _2$ compute $F_2$ mapping relations $R \!:\!Q\rightarrow A \rightarrow \mathbb {P}$ to relations of type $(I \times O') \rightarrow O \rightarrow \mathbb {P}$.

To compute the functional mapping an oracle $R \!:\!Q\rightarrow A \rightarrow \mathbb {P}$ to a computation $\lambda i o.\,\exists o' \!:\!O'.\,F_1\,R\,i\,o' \wedge F_2\,R\,(i,o')\,o$ of type $I \rightarrow O \rightarrow \mathbb {P}$ we construct a stalling tree with state type $(O' \times \mathbb {N})^{?}$ and starting state $\textsf {None}$. The intuition is that the state s remains $\textsf {None}$ as long as $\tau _1$ asks questions, and once an output $o'$ is produced we save it and the number of questions that were asked until then in the state, which remains unchanged after. Then, $\tau _2$ can ask questions, but since $as$ contains also answers to questions of $\tau _1$, we drop the first n before passing it to $\tau _2$.

Formally, the tree takes as arguments the input i, state s ans answer list $as$, and returns

$$ {\left\{ \begin{array}{ll} \textsf {ret}\;(\textsf{ask}\;(\textsf {None}, {\textsf {Some}\;}q)) &{} \text {if } s = \textsf {None}, \tau _1\,i\, as \triangleright {\textsf {Some}\;}(\textsf{ask}\;q) \\ \textsf {ret}\;(\textsf{ask}\;({\textsf {Some}\;}(o', | as |), \textsf {None})) &{} \text {if } s = \textsf {None}, \tau _1\,i\, as \triangleright {\textsf {Some}\;}(\textsf{out}\;o') \\ \textsf {ret}\;(\textsf{ask}\;({\textsf {Some}\;}(o',n), {\textsf {Some}\;}q)) &{} \text {if } s = {\textsf {Some}\;}(o', n), \tau _2\,(i, o')\,( as \uparrow _n) \triangleright {\textsf {Some}\;}(\textsf{ask}\;q) \\ \textsf {ret}\;(\textsf{ask}\;({\textsf {Some}\;}(o',n), {\textsf {Some}\;}q)) &{} \text {if } s = {\textsf {Some}\;}(o', n), \tau _2\,(i, o')\,( as \uparrow _n) \triangleright {\textsf {Some}\;}(\textsf{out}\;o) \end{array}\right. } $$

where $as \uparrow _n$ drops the first n elements of $as$. Note that formally, we use bind to analyse the values of $\tau _1$ and $\tau _2$, but just write a case analysis on paper.    $\square$

Proof

(of Lemma 8). Let $\tau _1$ compute $F_1$ mapping relations $R \!:\!Q\rightarrow A \rightarrow \mathbb {P}$ to relations $X \rightarrow Y \rightarrow \mathbb {P}$, and $\tau _1$ compute $F_2$ mapping relations $R \!:\!X\rightarrow Y \rightarrow \mathbb {P}$ to relations $I \rightarrow O \rightarrow \mathbb {P}$. We construct a stalling tree $\tau$ computing a functional mapping $R \!:\!Q\rightarrow A \rightarrow \mathbb {P}$ to $\lambda i o.\;F_2\,(F_1 R)\,i\,o$ of type $I \rightarrow O \rightarrow \mathbb {P}$.

Intuitively, we want to execute $\tau _2$. Whenever it asks a question x, we record it and execute $\tau _1\,x$ to produce an answer. Since the answer list $as$ at any point will also contain answers of the oracle produces for any earlier question $x'$ of $\tau _2$, we record furthermore how many questions were already asked to the oracle to compute $\tau _1 x$.

As state type, we thus use $(X\times Y)^* \times (X \times \mathbb {N})^{?}$, where the first component remembers questions and answers for $\tau _2$, and the second component indicates whether we are currently executing $\tau _2$ (then it is $\textsf {None}$), or $\tau _1$, when it is ${\textsf {Some}\;}(x,n)$ to indicate that on answer list $as$ we need to run $\tau _1\,x\,( as \downarrow ^n)$, where $as \downarrow ^n$ contains the last n elements of $as$. The initial state is $([\,], \textsf {None})$.

We define $\tau$ to take as arguments an input i, a state (t, z), and an answer list $as$ and return

$$ {\left\{ \begin{array}{ll} \textsf{out}\;o &{} \text {if } x = \textsf {None}, \tau _2\,i\,(\textsf {map}\,\pi _2\,t) \triangleright \textsf{out}\;o \\ \textsf{ask}\;(t, {\textsf {Some}\;}(x,0), \textsf {None}) &{} \text {if } x = \textsf {None}, \tau _2\,i\,(\textsf {map}\,\pi _2\,t) \triangleright \textsf{ask}\;x \\ \textsf{ask}\;(t, {\textsf {Some}\;}(x, {\textsf {S}\;}n), {\textsf {Some}\;}q) &{} \text {if } x = {\textsf {Some}\;}(x,n), \tau _1\,x\,( as \uparrow ^n) \triangleright \textsf{ask}\;q \\ \textsf{ask}\;(t \mathbin {+\!\!\!+}[(x,y)], \textsf {None}, \textsf {None}) &{} \text {if } x = {\textsf {Some}\;}(x,n), \tau _1\,x\,( as \uparrow ^n) \triangleright \textsf{out}\;y \end{array}\right. } $$

Intuitively, when we are in the mode to execute $\tau _2$ and it returns an output, we return the output. If it returns a question x, we change mode and stall. When we are in the mode to execute $\tau _1$ to produce an answer for x, taking the last n given answers into account and it asks a question q, we ask the question and indicate that now one more answer needs to be taken into account. If it returns an output y, we add the pair [(x, y)] to the question answer list for $\tau _1$, change the mode back to execute $\tau _2$, and stall.    $\square$

Proof

(of Lemma 9). We define a tree $\tau$ computing the functional mapping $R\!:\!(I \times \mathbb {N}) \rightarrow \mathbb {B}\rightarrow \mathbb {P}$ to the following relation of type $I \rightarrow \mathbb {N}\rightarrow \mathbb {P}$: $\lambda i n.\; R\, (i,n)\, \textsf{true}\wedge \forall m < n.\; R\, (i, m)\, \textsf{false}$.

$$\tau \,i\, as := {\left\{ \begin{array}{ll} \textsf {ret}\;(\textsf{out}\;i) &{} \text {if } as [i] = \textsf{true}\\ \textsf {ret}\;(\textsf{ask}\;(i, | as |)) &{} \text {if } \forall j.\, as [j] = \textsf{false}\end{array}\right. } $$

Note that a function $\textsf{find}\, as$ computing the smallest i such that $as$ at position i is $\textsf{true}$, and else returning $\textsf {None}$ is easy to implement.

Intuitively, we just ask all natural numbers as questions in order. On answer list l with length n, this means we have asked $[0,\dots ,n-1]$. We check whether for one of these the oracle returned $\textsf{true}$, and else ask $n = |l|$.    $\square$

C Relation to Bauer’s Turing Reducibility

We show the equivalence of the modulus continuity as defined in Lemma 1 with the order-theoretic characterisation used by Bauer [3]. The latter notion is more sensible for functionals acting on functional relations, so we fix some

$$F:(Q\,\rightsquigarrow \,A)\rightarrow (I\,\rightsquigarrow \,O)$$

where $X\,\rightsquigarrow \,Y$ denotes the type of functional relations $X\rightarrow Y \rightarrow \mathbb {P}$. To simplify proofs and notation, we assume extensionality in the form that we impose $R=R'$ for all $R,R':X\,\rightsquigarrow \,Y$ with $Rxy\leftrightarrow R'xy$ for all x : X and y : Y.

To clarify potential confusion upfront, note that Bauer does not represent oracles on $\mathbb {N}$ as (functional) relations but as pairs (X, Y) of disjoint sets with $X,Y:\mathbb {N}\rightarrow \mathbb {P}$, so his oracle computation operate on such pairs. However, since such a pair (X, Y) gives rise to a functional relation $R: \mathbb {N}\,\rightsquigarrow \,\mathbb {B}$ by setting $R\,n\,b := (X\,n \wedge b = \textsf{true})\vee (Y\,n \wedge b = \textsf{false})$ and, conversely, $R: \mathbb {N}\,\rightsquigarrow \,\mathbb {B}$ induces a pair (X, Y) via $X\,n := R\,n\,\textsf{true}$ and $Y\,n := R\,n\,\textsf{false}$, Bauer’s oracle functionals correspond to our specific case of functionals $(\mathbb {N}\,\rightsquigarrow \,\mathbb {B})\rightarrow (\mathbb {N}\,\rightsquigarrow \,\mathbb {B})$. He then describes the computable behaviour of an oracle functional by imposing continuity and a computational core operating on disjoint pairs (X, Y) of enumerable sets that the original oracle functional factors through, which in our chosen approach correspond to the existence of computation trees. So while the overall setup of our approach still fits to Bauer’s suggestion, we now show that our notion of continuity is strictly stronger than his by showing the latter equivalent to modulus continuity.

Informally, Bauer’s notion of continuity requires that F preserves suprema, which given a non-empty directed set $:(Q\,\rightsquigarrow \,A)\rightarrow \mathbb {P}$ of functional relations requires that $F\,(\bigcup _{R\in S} R)=\bigcup _{R\in S}\, F\,R$, i.e. that the F applied to the union of S should be the union of F applied to each R in S. Here directedness of S means that for every $R_1,R_2\in S$ there is also $R_3\in S$ with $R_1,R_2\subseteq R_3$, which ensures that the functional relations included in S are compatible so that the union of S is again a functional relation.

Lemma 41 If F is modulus-continuous, then it preserves suprema.

Proof

First, we observe that F is monotone, given that from $F\,R\,i\,o$ we obtain some modulus $L:Q^*$ that directly induces $F\,R'\,i\,o$ for every $R'$ with $R\subseteq R'$.

So now S be directed and non-empty, we show both inclusions separately. First $\bigcup _{R\in S}\, F\,R \subseteq F\,(\bigcup _{R\in S} R)$ follows directly from monotonicity, since if $F\,R\,i\,o$ for some $R\in S$ we also have $F\,(\bigcup _{R\in S} R)\,i\,o$ given $R\subseteq \bigcup _{R\in S} R$.

Finally assuming $F\,(\bigcup _{R\in S} R)\,i\,o$, let $L:Q^*$ be a corresponding modulus, so in particular $L\subseteq \textsf{dom}(\bigcup _{R\in S} R)$. Using directedness (and since S is non-empty), by induction on L we can find $R_L\in S$ such that already $L\subseteq \textsf{dom}(R_L)$. But then also $F\,R_L\,i\,o$ since L is a modulus and $R_L$ agrees with $\bigcup _{R\in S} R)$ on L.    $\square$

Lemma 42 If F is preserves suprema, then it is modulous continuous.

Proof

Again, we first observe that F is monotone, given that for $R\subseteq R'$ the (non-empty) set $S:=\{R,R'\}$ is directed and hence if $F\,R\,i\,o$ we obtain $F\,R'\,i\,o$ since $R'=\bigcup _{R\in S} R$.

Now assuming $F\,R\,i\,o$ we want to find a corresponding modulus. Consider

$$S:=\{R_L \mid L \subseteq \textsf{dom}(R) \}$$

where $R_L\,q\,a := q\in L \wedge R\,q\,a$, so S contains all terminating finite subrelations of R. So by construction, we have $R=\bigcup _{R\in S} R$ and hence $F\,(\bigcup _{R\in S} R)\,i\,o$, thus since F preserves suprema we obtain $L \subseteq \textsf{dom}(R)$ such that already $F\,R_L\,i\,o$. The remaining part of L being a modulus for $F\,R\,i\,o$ follows from monotonicity.    $\square$