key: cord-0046137-g88etd34
authors: Kristiansen, Lars; Murwanashyaka, Juvenal
title: On Interpretability Between Some Weak Essentially Undecidable Theories
date: 2020-06-24
journal: Beyond the Horizon of Computability
DOI: 10.1007/978-3-030-51466-2_6
sha: a75f93c9fe0d26136669987e9c72911f835b825a
doc_id: 46137
cord_uid: g88etd34

We introduce two essentially undecidable first-order theories [Formula: see text] and [Formula: see text]. The intended model for the theories is a term model. We prove that [Formula: see text] is mutually interpretable with Robinson’s [Formula: see text]. Moreover, we prove that Robinson’s [Formula: see text] is interpretable in [Formula: see text].

A first-order theory T is undecidable if there is no algorithm for deciding if T φ. If every consistent extension of an undecidable theory T also is undecidable, then T is essentially undecidable.

We introduce two first-order theories, WT and T, over the language L T = {⊥, ·, · , } where ⊥ is a constant symbol, ·, · is a binary function symbol and is a binary relation symbol. The intended model for these theories is a term model: The universe is the set of all variable-free L T -terms. Each term is interpreted as itself, and is interpreted as the subterm relation (s is a subterm of t iff s = t or t = t 1 , t 2 and s is a subterm of t 1 or t 2 ).

The non-logical axioms of WT are given by the two axiom schemes:

where s and t are distinct variable-free terms.

where t is a variable-free term and S(t) is the set of all subterms of t. There are no other non-logical axioms except those given by these two simple schemes, and at a first glance WT seems to be a very weak theory. Still it turns out that Robinson's essentially undecidable theory R is interpretable in WT, and thus it follows that also WT is essentially undecidable. The theory T is given by the four axioms:

It is not difficult to see that T is a consistent extension of WT. Thus, since WT is essentially undecidable, we can conclude right away that also T is essentially undecidable. Furthermore, since every model of the finitely axiomatizable theory T is infinite, T cannot be interpretable in WT, and the obvious conjecture would be that T is mutually interpretable with Robinson's Q.

The Axioms of R R1 n + m = n + m ; R2 n × m = nm ; R3 n = m for n = m ; The seminal theories R and Q are theories of arithmetic. The theory R is given by axiom schemes, and Q is a finitely axiomatizable extension of R, see Fig. 1 (Q is also known as Robinson arithmetic and is more or less Peano arithmetic without the induction scheme). It was proved in Tarski et al. [9] that R and Q are essentially undecidable. Another seminal essentially undecidable first-order theory is Grzegorcyk's TC. This is a theory of concatenation. The language is { * , α, β} where α and β are constant symbols and * is a binary function symbol. The standard TC model is the structure where the universe is {a, b} + (all finite nonempty strings over the alphabet {a, b}), * is concatenation, α is the string a and β is the string b. It was proved in Grzegorzyk and Zdanowski [3] that TC is essentially undecidable. It was later proved that TC is mutually interpretable with Q, see Visser [10] for further references. The theory WTC − is a weaker variant of TC that has been shown to be mutually interpretable with R, see Higuchi and Horihata [4] for more details and further references. The axioms of TC and WTC − can be found in Fig. 2 .

The overall picture shows three finitely axiomatizable and essentially undecidable first-order theories of different character and nature: Q is a theory of arithmetic, TC is a theory of concatenation, and T is a theory of terms (it may also be viewed as a theory of binary trees). All three theories are mutually interpretable with each other, and each of them come with a weaker variant given by axiom schemes. These weaker variants are also essentially undecidable and mutually interpretable with each other.

The theory T has, in contrast to Q and TC, a purely universal axiomatization, that is, there are no occurrences of existential quantifiers in the axioms. Moreover, its weaker variant WT has a neat and very compact axiomatization compared to R and WTC − .

where x y is defined by

The Axioms of TC are axiom schemes where t ∈ {a, b} + and t is a term inductively defined by: a ≡ α, b ≡ β, au ≡ α * u and bu ≡ β * u.

Another interesting theory which is known to be mutually interpretable with Q, and thus also with TC and T, is the adjunctive set theory AST. More on AST and adjunctive set theory can found in Damnjanovic [2] . For recent results related to the work in the present paper, we refer the reader to Jerabek [5] , Cheng [1] and Kristiansen and Murwanashyaka [7] .

The rest of this paper is fairly technical, and we will assume that the reader is familiar with first-order theories and the interpretation techniques introduced in Tarski et al. [9] . In Sect. 2 we prove that R and WT are mutually interpretable. In Sect. 3 we prove that Q is interpretable in T. We expect that T can be interpreted in Q by standard techniques available in the literature.

The theory R − over the language of Robinson arithmetic is given by the axiom schemes

where n, m ∈ N. Recall that n denotes the n th numeral, that is, 0 ≡ 0 and n + 1 ≡ Sn.

We now proceed to interpret R − in WT. We choose the domain I(x) ≡ x = x (thus we can just ignore the domain). Furthermore, we translate the successor function S(x) as the function given by λx. x, ⊥ , and we translate the constant 0 as ⊥, ⊥ . Let n denote the translation of the numeral n. Then we have n + 1 ≡ n , ⊥ . It follows from WT 1 that the translation of each instance of R − 3 is a theorem of WT since m and n are different terms whenever m = n.

We translate x ≤ y as x y ∧ x = ⊥. It is easy to see that

where T (n) = S(n ) \ {⊥} and S(n ) denotes the set of all subterms of n . We observe that T (n) = {k | k ≤ n} and that (1) indeed is the translation of the axiom scheme R − 4 . Hence we conclude that the translation of each instance of R − 4 is a theorem of WT.

Next we discuss the translation of +. The idea is to obtain n + i through a formation sequence of length i. Such a sequence will be represented by a term of the form

Accordingly we translate x + y = z by the predicate add(x, y, z) given by the formula

Proof. First we prove that WT add(n , m , n + m ). This is obvious if m = 0. Assume m > 0. Let

and observe that S n i is of the form (2). We will argue that we can choose the W in the definition of add(x, y, z) to be the term S n m . So let W = S n m . By the axioms of WT, we have n , 0 W . Assume

By the axioms of WT, we have that Y m , Y = m and Y = ⊥ imply Y = k for some k < m. Since X, Y W , we know by WT 2 that X, Y is one of the subterms of W . By WT 1 and the form of S n m , we conclude that X = n + k . Furthermore, the form of S n m and WT 2 then ensures that X, ⊥ , Y, ⊥ W = S n m . Moreover, if Y, ⊥ = m , then by WT 1 , we must have k = m − 1, and thus, X, ⊥ = n + (m − 1) , ⊥ = n + m . This proves that we can deduce add(n , m , n + m ) from the axioms of WT, and thus we also have WT ∀z z = n + m → add(n , m , z) .

Next we prove that the converse implication add(n , m , z) → z = n + m follows from the axioms of WT (and thus the lemma follows). This is obvious when m = 0. Assume m = 0 and add(n , m , z). Then we have W such that n , 0 W and

Since n, 0 W and (3) hold, we have n + k + 1 , k + 1 W for any k < m. It also follows from (3) that z = n + k + 1 when m = k + 1.

It follows from the preceding lemma that there for any n, m ∈ N exists a unique k ∈ N such that WT add(n , m , k ). We translate x + y = z by the

The second disjunct of (4) ensures the functionality of our translation, that is, it ensures that WT ∀xy∃!xφ + (x, y, z) (the same technique is used in [6] ). By Lemma 1, we have WT φ + (n , m , n + m ). This shows that the translation of any instance of the axiom scheme R − 1 can be deduced from the axioms of WT. We can also achieve a translation of x × y = z such that the translation of each instance of R − 2 can be deduced from the axioms of WT. Such a translation claims the existence of a term S n m where S n 1 ≡ n , 1 and S n i+1 ≡ S n i , (i + 1)n , i + 1 and will more or less be based on the same ideas as our translation of x + y = z. We omit the details. Proof. We have seen how to interpret R − in WT. It follows straightforwardly from results proved in Jones and Shepherdson [6] that R − and R are mutually interpretable. Thus R is interpretable in WT. A result of Visser [11] states that a theory is interpretable in R if and only if it is locally finitely satisfiable, that is, each finite subset of the non-logical axioms has a finite model. Since WT clearly is locally finitely satisfiable, WT is interpretable in R.

The language of the arithmetical theory Q − is {0, S, M, A} where 0 is a constant symbol, S is a unary function symbol, and A and M are ternary predicate symbols. The non-logical axioms of the first-order theory Q − are the the following:

Svejdar [8] proved that Q − and Q are mutually interpretable. We will prove that Q − is interpretable in T.

The first-order theory T + is T extended by the two non-logical axioms

Proof. We simply relativize quantification to the domain

We need to show that u x 1 , x 2 . By T 4 and v x 1 , x 2 , at least one of the following three cases holds:

We have u x 1 , x 2 by an argument symmetric to the one used in

. This proves that I is closed under ·, · . It follows from T 3 that ⊥ ∈ I, and thus I satisfies the domain condition. Clearly, the translation of each non-logical axiom of T + is a theorem of T.

We now proceed to interpret Q − in T + . We choose the domain N given by

Proof. It follows from T 1 , T 3 and T 4 that (i) holds. In order to see that (ii) holds, assume N (x) (we will argue that N ( x, ⊥ ) holds). Suppose y x, ⊥ . Now, N ( x, ⊥ ) follows from

Thus it is sufficient to argue that (5) holds. By T 4 , we know that y x, ⊥ implies y = x, ⊥ ∨ y

x ∨ y ⊥. The case y = x, ⊥ : We obviously have ∃z[ y = z, ⊥ ] and thus (5) holds. The case y x: (5) holds since N (x) holds. The case y ⊥: We have y = ⊥ by T 3 , and thus (5) holds. This proves (ii).

We turn to the proof of (iii). Suppose N (y) ∧ z y (we show z = ⊥ ∨ N (z)). Assume w z. By T 6 , we have w y, moreover, since N (y) holds, we have w = ⊥ ∨ ∃u[w = u, ⊥ ]. Thus, we conclude that

Now

follows tautologically from (6).

We interpret 0 as ⊥, ⊥ . We interpret the successor function Sx as λx. x, ⊥ . To improve the readability we will occasionally write0 in place of ⊥, ⊥ ,Ṡt in place of t, ⊥ and t ∈ N in place of N (t). We will also write ∃x ∈ N [ η ] and ∀x ∈ N [ η ] in place of, respectively, ∃x[ N (x) ∧ η ] and ∀x[ N (x) → η ]. Furthermore, Qx 1 , . . . , x n ∈ N is shorthand for Qx 1 ∈ N . . . Qx n ∈ N where Q is either ∀ or ∃. Proof. The translation of Q 1 is ∀x, y ∈ N [ x = y →Ṡx =Ṡy ]. By T 2 , we have x = y →Ṡx =Ṡy for any x, y, and thus, the translation of Q 1 is a theorem of

By T 5 , we have x x. By (7) and x x, we have

and then, by a tautological inference, we also have

Before we give the translation of A, we will provide some intuition. The predicate A(a, b, c) holds in the standard model for Q − iff a + b = c. Let 0 ≡0 and n + 1 ≡Ṡ n, and observe that a + b = c iff there exists an L T -term of the form . . . ⊥, a, 0 , a + 1, 1 , a + 2, 2 . . . , a + b, b (8) where c = a + b. We will give a predicate φ A such that φ A ( a, b, w) holds in T + iff w is of the form (8). Thereafter we will use φ A to give the translation

Lemma 6.

Proof. We assume x ∈ N and prove the equivalence

The left-right direction of (9) follows straightforwardly from the definition of φ A .

To prove the right-left implication of (9), we need to prove φ A (x,0, ⊥, x,0 ). It is easy to see that

holds, and to show (10) , it suffices to show that

is a contradiction. (If (11) is a contradiction, then (10) will hold as the antecedent of θ A will be false for all x, Y, Z ∈ N and all u.) By x and x ∈ N . By Lemma 4 (iii), we have u, Z, Y = ⊥ or u, Z, Y ∈ N . Now, u, Z, Y = ⊥ contradicts T 1 . Furthermore, by our definitions, u, Z, Y ∈ N implies that

By

, and this yields a contradiction together with T 1 and T 2 . Case (c') is similar to Case (a'), but a bit simpler. This completes the proof of the lemma.

Proof. We assume

x, y ∈ N and w = w , z, y and φ A (x, y, w) .

We need to prove φ A (x,Ṡy, w, Ṡ z,Ṡy ) ≡

First we prove

Since φ A (x, y, w) holds by our assumptions (12), we have z 1 ∈ N and w 1 such that w = w 1 , z 1 , y . We have also assumed w = w , z, y . By T 2 , we have z = z 1 , and thus z ∈ N . By Lemma 4 (ii), we haveṠz ∈ N . This proves (14). The second conjunct of (13) follows straightforwardly from (14). (simply let z 0 beṠz and let w 0 be w). The first conjunct follows easily from T 2 and the assumption y ∈ N . Thus, we are left to prove the third conjunct of (13), namely

In order to do so, we assume

and prove

By our assumptions (16), we have u, Z, Y w, Ṡ z,Ṡy , and then T 4 yields three cases: (a) u, Z, Y = w, Ṡ z,Ṡy , (b) u, Z, Y w and (c) u, Z, Y Ṡ z,Ṡy . We prove that that (17) holds in each of these three cases.

Case (a): By T 2 , we have u = w, Z =Ṡz and Y =Ṡy. By (14), we have z ∈ N . By (12), we have y ∈ N . Moreover, by (12), we also have u = w = w , z, y . Thus there exist v and Y , Z ∈ N such that

If y =0, we must have v, z, y = w = ⊥, x,0 since φ A (x, y, w) holds by our assumptions (12). By T 2 , this implies z = x and v = ⊥. This proves that (17) holds in Case (a). Case (b): By our assumptions (12), we have φ A (x, y, w), and thus we also

We are dealing with a case where the antecedent of (18) holds, and thus (17) holds.

Case (c): This case is not possible. By T 4 , this case splits into the subcases: (a') u, Z, Y = Ṡ z,Ṡy , (b') u, Z, Y Ṡ z and (c') u, Z, Y Ṡ y. We prove that each of these subcases contradicts our axioms. Case (a'): Recall thatṠy is shorthand for y, ⊥ . Thus, by T 2 , we have Y = ⊥. This contradicts the assumption (12) that Y ∈ N . Case (b'): By Lemma 4 (iii), we have u, Z, Y = ⊥ ∨ N ( u, Z, Y ) . Now, u, Z, Y = ⊥ contradicts T 1 . Furthermore, N ( u, Z, Y ) implies that there is z 0 such that u, Z, Y = z 0 , ⊥ . By T 2 , we have Z, Y = ⊥. This contradicts T 1 . Case (c') is similar to Case (b'). This proves that (17) holds, and thus we conclude that the lemma holds.

Proof. Let x, y ∈ N and assume φ A (x,Ṡy, w). Thus, we have w and z ∈ N such that w = w , z,Ṡy

Use the assumptions (19) to prove that φ A (x, y, w ) ≡

holds. We omit the details. Proof. The translation of the axiom A is

Assume Ψ A (x, y, z 1 ) and Ψ A (x, y, z 2 ). Then it follows straightforwardly from the definition of Ψ A and T 2 that z 1 = z 2 . Hence the translation is a theorem of T + .

We have

by Lemma 6, and it easy to see that the translation of G 4 is a theorem of T + .

The translation of G 5 is

In order to prove that (21) can be deduced from the axioms of T + , we assume Ψ A (x, y, z) ∧ u =Ṡz. Then we need to prove Ψ A (x,Ṡy,Ṡz) ≡

By our assumption Ψ A (x, y, z) there is a unique w 1 such that φ A (x, y, w 1 ) and w 1 = w 0 , z, y for some w 0 . By Lemma 7, we have φ A (x,Ṡy, w 1 , Ṡ z,Ṡy ). Thus, we have w 2 such that φ A (x,Ṡy, w 2 ) and w 2 = w 1 , Ṡ z,Ṡy . It is easy to see that (22) holds if w 2 is unique. Thus we are left to prove the uniqueness of w 2 , more precisely, we need to prove that

In order to prove (23), we assume φ A (x,Ṡy, W 2 ) (we will prove W 2 = w 2 = w 1 , Ṡ z,Ṡy ). By our assumption φ A (x,Ṡy, W 2 ) and Lemma 8, we have u 0 ∈ N and W 1 such that W 2 = W 1 , u 0 ,Ṡy and φ A (x, y, W 1 ). We have argued that there is a unique w 1 = w 0 , z, y such that φ A (x, y, w 1 ) holds. By this uniqueness, we have W 1 = w 1 = w 0 , z, y . So far we have proved w 2 = w1 w 0 , z, y , Ṡ z,Ṡy and W 2 = W1 w 0 , z, y , u 0 ,Ṡy and then we are left to prove that u 0 =Ṡz. By our assumption φ A (x,Ṡy, W 2 ), we have v and Z , Y ∈ N such that u 0 =ṠZ ,Ṡy =ṠY and W 1 = v, Z , Y . Thus, v, Z , Y = w 0 , z, y . By T 2 , we have z = Z , and thus, u 0 =ṠZ = Sz. This proves that (23) holds.

We will now give the translation Ψ M of M . Let φ M (x, y, w) ≡

We The proof of the next lemma follows the lines of the proof of Lemma 9. We omit the details. Proof. It is proved in Svejdar [8] that Q is interpretable in Q − . It follows from the lemmas above that Q − is interpretable in T + which again is interpretable in T. Hence the theorem holds.

Finding the limit of incompleteness I

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Weak theories of concatenation and minimal essentially undecidable theories

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An interpretation of Robinson arithmetic in its Grzegorczyk's weaker variant

Growing commas. A study of sequentiality and concatenation

Foundational Adventures. Essays in Honour of Harvey Friedman