2. Preliminaries
We will briefly outline any relevant notation and related fundamental results. We work in the framework of Zermelo–Fraenkel set theory with the axiom of choice, ZFC, denote by the set of all natural numbers, and use as a variable for ordinals.
We will say that that a chain has length if it contains exactly n elements, and that a partial order P is of depth if every chain in P is of length at most n and there is at least one chain in P of length n.
2.1. First-Order Languages and Logic
The first-order languages we discuss will have a countably infinite set of individual variables and will contain a symbol ≐ which will be interpreted as formal equality. We will only be interested in relational languages, i.e. containing no function (or constant) symbols. We use standard first-order logic semantics based on variable assignments, using gothic letters to denote structures (we might sometimes alternatively say models for the language, [5]) and capital latin letters to denote formulas.
To recall, atomic formulas of a relational first-order language are of the form or where are individual variables and p is a relation symbol in ; the formulas of are either atomic formulas or of the form , , . Other propositional connectives and the universal quantifier are introduced as abbreviations in terms of the symbols already introduced; as usual, we sometimes omit parentheses for brevity. We define free and bound variables in a formula in the usual manner and say that A is a sentence if none of its variables are free. We denote by the quantifier rank of A, i.e. the largest depth of nesting of quantifiers within A.
A structure for consists of a set (the universe of ) together with a corresponding relation of an appropriate arity for each relation symbol p of . Given a variable assignment V that maps individual variables of to elements of , the Tarski satisfaction relation ⊧ is defined by recursion on formulas:
iff
iff
iff it is not the case that (we write )
iff and
iff there exists an assignment such that for all variables except maybe x, such that
Whenever is a structure, A is a first-order formula with free variables among , and are elements of , we shall write for where V is any variable assignment such that . If A is a sentence, then we write for where V is any variable assignment.
We say that validates a sentence A if . The theory of a class of structures is the set of those sentences valid in each structure in .
We assume that the reader is acquainted with fundamental notions and results such as the compactness theorem and the Löwenheim-Skolem theorem, substructures and elementary substructures, isomorphisms and embeddings. The reader may consult [5] for further references.
For a formula A with free variables among and a formula B having no common variables with A, we define the relativization B with respect to A and y and denote it by , by recursion on B:
if B is atomic
where is the formulas obtained by replacing every free occurence of y in A by z.
Given structures and , formula A with free variables among and elements , we say that is the relativized reduct of A with respect to A and if is the substructure of with universe those elements such that . The relativized reduct of A with respect to A and exists iff .
The relativization theorem connects the notion of relativized reducts and relativizations of formulas:
Theorem 1. (Relativization theorem)
If is the relativized reduct of with respect to A and and B is a formula with free variables among that has no common variable with A, then for every it holds that iff .
Proof. The reader may consult Theorem 5.1.1 of [7]. □
2.2. Monadic Second-Order Languages and Logic
We obtain a relational monadic second-order language by extending a first-order language with a countably infinite set of set variables which we shall denote by capital latin letters, and a new logical symbol ∈. We introduce new atomic formulas of the form for each individual variable x and set variable Z, and we obtain the formulas of by extending the first-order definition with the additional clause that if X is a set variable and A is a formula, then is a formula.
Structures for a monadic second-order language are defined in the same way as in the first-order case. For a structure we extend variable assignments to map set variables to subsets of . We then define the satisfaction relation by adding the following clauses to the first-order definition:
iff
iff there exists an assignment such that for all individual variables and for all set variables except maybe X, such that
Similarly to the first-order case, if A is a monadic second-order formula with free variables among , and and , then we write for where V is any variable assignment such that and .
We shall use abbreviations for usual set-theoretic properties, e.g. we write for and write for .
2.3. Intuitionistic Propositional Logic
The formulas of the intuitionistic propositional logic are the same syntactic objects as the formulas of classical propositional logic, i.e., we have a countably infinite set of propositional variables, together with the propositional connectives and the propositional constant ⊥. Any propositional constant or variable is a propositional formula and if and are propositional formulas, then so are and ; we often omit parentheses. We introduce the connective ¬ and the constant ⊤ as abbreviations, i.e. is an abbreviation for and ⊤ is an abbreviation for .
We recall the standard relational Kripke semantics (see [8]). A Kripke frame is where F is a non-empty set (the universe of the frame) and is a partial order on F. A variable assignment for maps propositional variables to upward closed subsets of F, i.e. if and , then for every propositional variable p and elements . A Kripke model is a Kripke frame together with a variable assignment. For a Kripke model consisting of a frame together with a variable assignment V the pointwise satisfaction relation ⊧ at points is defined by recursion on formulas:
iff
iff and
iff or
iff for every such that it holds that if then
We say that a frame validates a propositional formula and write if for every variable assignment V and every ; if is a class of frames we write if for every . We denote by the minimal intuitionistic logic. We say that validates a superintuitionistic logic L and write if validates each of its axioms. We say that a logic L is complete with respect to a class of frames if iff for every . The logic of a class of frames is the logic .
We define a certain set of formulas for each natural number n by recursion:
For each natural number n, a frame validates the formula iff the depth of is at most n.
We will assume the reader is acquainted with standard constructions and properties of Kripke frames, such as generated subframes, disjoint unions of frames and p-morphic images, isomorphisms. For further reference the reader may consult [8].
2.4. Definability by Intuitionistic Formulas
There is a natural duality between Kripke frames and first-order models for the language of order, i.e. the language containing a single nonlogical relation symbol ≤ which is interpreted as the partial order of the frame.
Every Kripke frame is a model for the first-order theory of partial orders. Consider a non-empty class of Kripke frames. We say that a propositional formula defines a first-order sentence A with respect to if for every Kripke frame we have that iff . If defines A with respect to we also say is a propositional definition of A with respect to and that A defines with respect to .
The problem is the following algorithmic task: when given as input a first-order sentence A, output ’yes’ if there exists a propositional definition of A, and output ’no’ otherwise.
For our purposes, we shall deal with algorithms by describing them in natural language pseudocode, and assume that the reader is able to translate it to either a formal mathematical model such as a Turing machine, or to a program in a desired programming language.
3. Decidable Instances of Definability
We consider classes based on linear orders, in particular:
The class of all linear orders.
The class of all finite linear orders.
The class of all unions of families of pairwise disjoint linear orders.
The class of all finite frames in .
We note that the classes
and
are finitely axiomatizable, while the classes
and
are not. A possible axiomatization for
is to take as its axiom the conjunction of the axiom for partial orders and the sentence
The above classes are prospective candidates for a positive resolution to the algorithmic definability problem since the linear intuitionistic Kripke frames have simple structure. We denote by the superintuitionistic logic obtained by adding to the additional axiom . This logic is well studied, and in the following proposition we remind the reader of some of its properties that will be useful later:
Proposition 1. The logic has the following properties:
-
(i)
A frame validates iff every generated subframe of is linear. Since a frame validates a formula iff each of its generated subframes validate it, is complete with respect to and with respect to .
-
(ii)
For any propositional formula φ with , iff where is a linear order with elements.
-
(iii)
is complete with respect to and with respect to .
-
(iv)
Any finite linear order is a p-morphic image of any infinite linear order.
-
(v)
is complete with respect to any infinite linear order.
-
(vi)
For any propositional formula φ, either or there is a natural number n such that for every frame for it holds that iff .
Proof. The properties in are well-known standard results.
-
(ii)
-
The left to right direction is immediate, for right to left:
Suppose that and take any linear order , variable assignment V in and point , we will show that .
For a point denote by the set of those propositional variables among such that . Consider the finite partial order , where . Since variable assignments are upward closed, is a linear order. Moreover, G contains at most elements since any contains at most n elements and the sets in G are ordered by inclusion.
One verifies by straightforward induction on the formula with that for every such that we have that where . But is isomorphic to a generated subframe of and since we conclude that , hence .
-
(iii)
Immediate corollary to and .
-
(iv)
-
Take any infinite linear order and natural number , we will show that is a p-morphic image of . Since is infinite, we can pick such that for . Now the following function is a p-morphism of onto :
For such that define .
For such that define , where m is the least natural number such that and .
-
(v)
Take an infinite linearly ordered frame . For any propositional formula , if we have by and the properties of p-morphic images that is valid in all finite linearly ordered frames, therefore by .
-
(vi)
-
Suppose that . Then by , where and is a linear order with elements. If is a finite linear order with elements, then since is isomorphic to a generated subframe of . Therefore either is valid in no finite linear frame (in which case satisfies the desired property) or there is a greatest natural number m such that and . Since is isomorphic to a generated subframe of for , we have that the finite linear frames that validate are exactly those of depth at most m.
Now if , then every chain in contains at most m elements, in particular every generated subframe of is a linear order with at most m elements and thus validates , so .
Conversely, if , then every generated subframe of is a linear order that validates , therefore contains at most m elements. Therefore any chain in must have at most m elements, thus .
□
Notice that property is quite illuminating: it states that the formulas ⊤ and for natural n exhaust all possible definitions expressible in the propositional language. More broadly, Proposition 1 provides enough clarity about the properties of the logic to be able to devise a general procedure that can be used to resolve the propositional definability problem for each of the classes .
Throughout the remainder of this section, unless explicitly specified, we shall write for any of the classes mentioned above. We now describe the procedure for solving the problem with respect to and argue about the correctness of the steps involved:
- (1)
-
(Validity) Decide whether .
If true, then ⊤ clearly is a propositional definition of A with respect to .
If false:
- (2)
-
(Finiteness) Decide whether A is valid in a structure from with an infinite chain.
If true, then A is undefinable with respect to .
Proof. Suppose that there is some such that and contains an infinite chain. Assume for contradiction that defines A with respect to . Then in particular . By property , this means that so . But then since defines A, this means that , which we ruled out in the previous step and is therefore a contradiction. □
If false:
- (3)
-
(Boundedness) Decide whether there exists a uniform bound m of the depth of allmodels of A in .
If false, then A is undefinable with respect to .
Proof. Assume for contradiction that A has a propositional definition . Then since (Boundedness) gives a negative answer, there are frames of arbitrary depth validating . In particular, there are arbitrarily long linear orders (as generated subframes of the frames in ) validating and therefore by property we have that , thus . But since defines A, this means that which we ruled out in step (Validity). □
If true:
- (4)
-
(Least bound) Find the least uniform bound m of the depth of all models of A in .
Proof. Such uniform bound exists by the positive answer given in the previous step. □
- (5)
-
(Bound-completeness) Decide whether for all frames of depth at most m.
If true, then defines A with respect to .
Proof. Take .
Suppose first that . Then by (Boundedness), all chains in are of length at most m. Therefore .
Now suppose that . Then by the positive answer of (Bound-completeness), . □
If false, then A is undefinable with respect to .
Proof. Assume for contradiction that there exists a propositional definition of A with respect to . The uniform bound m cannot be 0, since then hence (Bound-completeness) gives a positive answer. Since m is the least uniform bound, this means that there is frame of depth m such that , otherwise m would not be least. Since defines A, . But then the linear order with m elements is a generated subframe of so and therefore for . Take any frame of depth at most m. Then any generated subframe of is isomorphic to for some . So any generated subframe of validates , so . But since was arbitrary of depth at most m, (Bound-completeness) gives a positive answer — contradiction. □
Observe that the procedure we just described not only recognizes the definable first-order sentences but also explicitly points out a propositional definition when it exists. Now, in order to prove that the propositional definability problem with respect to is decidable, it is sufficient to show the problems (Validity), (Finiteness), (Boundedness), (Least bound) and (Bound-completeness) can be computably solved for the class . This will allow us to prove the main result of this section, namely the following theorem:
Theorem 2. The instances of the definability problem with respect to any of the classes is decidable.
Proof. The result follows from the procedure described above and the following lemmas:
Lemma 2 shows that (Validity) is decidable.
Lemma 1 deals with (Bound-completeness) and (Least bound).
Lemma 3 shows that (Finiteness) is decidable.
Lemmas 4 and 5 show that (Boundedness) is decidable.
□
We now proceed to show that the above problems are decidable. First, we notice that solving (Validity) and (Boundedness) immediately allows us to solve (Bound-completeness) and (Least bound).
Lemma 1. If the first-order theory of is decidable, then the problem (Bound-completeness) and the restriction of the problem (Least bound) to only those input sentences for which (Boundedness) gives a positive answer are decidable.
Proof. Suppose that the first-order theory of is decidable. For any natural number m denote by the sentence which is valid in exactly the frames of depth at most m.
We now have that for all frames of depth at most m iff . Since the sentence can be computed from the parameters A and m and the first-order theory of is decidable, this gives us an effective procedure to solve (Bound-completeness).
Now suppose in addition that the instance of (Boundedness) for the class gives a positive answer for the sentence A. Then there exists a least uniform bound m on the depth of the the frames from that validate A. This m is the least natural number such that (this sentence is again computable from the parameters A and m). Since such a natural number is guaranteed to exist, an effective solution to (Least bound) is to check in increasing order for each natural number t whether the sentence is valid in , halting with answer m when in step m we get a positive answer for the first time. □
Of the problems we now need to solve, in some sense the most challenging is (Finiteness) because it deals with a property which is not expressible in the first-order language. This naturally leads us to consider the monadic second-order language which is better suited for internally analyzing this property. When dealing with second-order theories we shall be interested in the class consisting of those frames in whose universe is at most countable. This is necessitated by the fact that the monadic second-order theory of the full class is undecidable (a result due to Shelah [9]), which immediately means that the monadic second-order theory of the full class is also undecidable. Working with the restricted class will allow us to obtain decidability results on the monadic second-order side and then we will make use of the Löwenheim-Skolem theorem to reduce problems for to problems for which we can solve. A direct corollary to Rabin’s theorem [6] on the decidability of the theory is the following theorem, which we will be key to our results:
Theorem 3. (Rabin [6]) The monadic second-order theory of the class in the language of order is decidable.
Note that an immediate corollary of the above theorem is that the monadic second-order theory of
in the language of order expanded by a predicate
for finite sets is decidable. The reason is that
can be defined by the formula
which states that
X is finite iff every non-empty subset of
X has a least and a greatest element.
Now, our main task is to obtain the corresponding result for the class . We will do this by showing that we can embed frames inside frames in a way that allows us to translate properties of into properties of . To further elaborate, since is the disjoint union of some family of linear orders, we clearly have to embed the linear orders of inside , but in such a way that we can easily distinguish which elements of belong to the same chain of and which do not. We accomplish this by adding to new auxiliary elements which we will call indices of . To each linear order we designate an index and then obtain a "segment" from L by adding to it the index as a greatest element. Then we obtain from by gluing one after another all the "segments" obtained from all the linear orders in .
This clearly results in an at most countable linear order which we will call a linearization of . A linearization is inherently dependent on the order in which we glue together the segments. We can think of this order as an ordering on the set of indices of and different orderings on the set of indices may result in non-isomorphic linearizations of . Additionally, if we are given a linearization and know the set of all indices, we want to be able to distinguish which elements of belong to the same segment. The most natural way to do this is to look at the position in of the index and declare that its corresponding linear order L consists of all elements in which are less than and bigger than any other index .
Those considerations lead us to consider only sets of indices which are isomorphic to initial segments of (with the inherited order from the linearization). Such sets avoid pathological situations in which non-indices from cannot be assigned to an index, while also somewhat limiting the variance of possible linearizations of a frame , making them simpler to reason about. Clearly, we do not lose any expressiveness by imposing this restriction as any frame can be obtained by taking the union of a family of linear orders with an index set which is an initial segment of .
Definition 1. Let be a frame in . Without loss of generality, we will assume that is the union of a family of pairwise disjoint linear orders where α is an ordinal, ; and . We will say that α is the set of indices of .
We define a linearization of (with respect to the set of indices α) as the frame with universe and whose order is the unique linear order on with the following properties:
Let R be a monadic set variable and define the following mapping —translation— between formulas of the monadic second-order language of order expanded by the predicate for finite sets:
Now the translation transforms a formula A expressing a monadic second-order property of into the formula which expresses a monadic second-order property of . The key consideration in the translation is that the order of is interpreted inside as discussed before Definition 1 and that this interpretation is monadic second-order definable in terms of the set variable R which will be interpreted as the set of indices of .
Proposition 2. Let and be a linearization of . For each monadic second-order sentence A such that the variable R does not occur in A we have that iff where α is the set of indices of .
Proof. By induction on the formulas A with free variables among and having no occurrences of R we can prove that for each tuple of elements of and each tuple of subsets of we have that iff . In particular, when A is a sentence we obtain the required property. □
Proposition 3. For a frame and a subset , is isomorphic to a linearization of a frame with the set of indices of mapped to the set D precisely when D satisfies the following condition, expressible in the monadic second-order language by a formula :
-
(1)
D is either finite or isomorphic to ω under .
-
(2)
For every such that there exists some such that .
-
(3)
and for every there is some such that .
Proof. The listed conditions are immediately verified and follow the discussion before Definition 1.
We define the formula as the conjunction of the following formulas, each corresponding to one of the listed properties:
- (1)
stating that all initial segments of elements of the interpretation of R are finite. This happens precisely when the interpretation of R is either a finite linear order or has the same order type as .
- (2)
- (3)
□
Theorem 4. The monadic second-order theory of in the language of order expanded by the predicate for finite sets is decidable.
Proof. We reduce to the monadic second-order theory of . For a monadic second-order sentence A, we first replace all occurrences of the set variable R in A with a new set variable not occurring in A. We will show that iff where is the formula from Proposition 3. This immediately shows how to obtain the desired reduction.
Suppose first that . Let and be such that . Then by Proposition 3 we know that is isomorphic to a linearization of some frame with the set of indices of mapped to the set D. Since , this means that so by Proposition 2 this means that . Since D was an arbitrary subset of such that , this means that . Since was arbitrary, this in turn means that .
Now suppose that . Let be arbitrary and be a linearization of . Since this means that . Let D be the set of indices of . Then hence . Then by Proposition 2 this means that . Since was arbitrary, this means that . □
Lemma 2. Let be any of the classes , , , . Then (Validity) for , i.e. the following problem:
input:first-order sentence A
output: true, if ; and false, otherwise
is decidable.
Proof. Consequence of Theorems 3 and 4.
For a first-order sentence A we have that precisely when . This reduces the first-order theory of to the monadic second-order theory of . We reduce the first-order theory of to the monadic second-order theory of in a similar way.
By the downward Löwenheim-Skolem theorem we can argue that the first order theories of and coincide because the class is axiomatizable. But the first-order theory of is decidable because it is a restriction of the dedidable monadic second-order theory of the class. The same argument goes for the decidability of the first-order theory of the class . □
Lemma 3. Let be any of the classes , , , . Then (Finiteness) for , i.e. the following problem:
input: first-order sentence A
output: true, if A is valid in some frame which has an infinite chain; and false, otherwise
is decidable.
Proof. Clearly (Finiteness) is trivial for and .
We now argue for
. The sentence
is valid in a frame iff the frame contains an infinite chain. We will prove that (
Finiteness) for
should output
true on input
A iff
. The latter condition can be computably checked in view of Theorem 4 and the fact that the sentence
can be computed from
A.
Suppose first that has an infinite chain and . Let C be a countably infinite chain in and obtain by the downward Löwenheim-Skolem theorem a countable elementary substructure of such that . Then, since is an elementary substructure of and is countable, we have that belongs to and . Moreover, contains the infinite chain C so . Therefore, , hence .
Conversely, suppose that . Then there is a frame , such that . But this means that , and , i.e. validates A and has an infinite chain.
The argument for is similar to that for . □
Lemma 4. Let be any of , . Then (Boundedness) for , i.e. the following problem:
input: first-order sentence A
output: true, if there exists a uniform bound on the depth of the frames from that validate A; and false, otherwise
is decidable. More specifically, on input A (Boundedness) for always outputs the answer (Finiteness) did not, so (Boundedness) is decidable as the complement of a decidable problem.
Proof. Let A be a first-order sentence.
First, if (Finiteness) outputs true on input A, this means that there is a frame with an infinite chain such that . This clearly means that (Boundedness) should output false.
Conversely, suppose that (
Boundedness) outputs
false on input
A. Then no natural number
m uniformly bounds the depth the frames from
that validate
A. By application of a standard compactness argument we can then show that the set
is satisfiable where
K is the axiom for
and the
for
are distinct constant symbols. Let
, then we can conclude that
because
,
because
, and
has an infinite chain because the set
is a chain. Therefore (
Finiteness) outputs
true on input
A. □
Remark 1.
Observe that in the procedure for solving the propositional definability problem for where is or , the step in which we need to query (Boundedness) is only reached if (Finiteness) has previously output false for A. Therefore, in view of the lemma we just proved, we can skip the step that involves (Boundedness) in the procedure for
Lemma 5.(Boundedness) for and for is decidable.
Proof. Let A be a first-order sentence and .
A classical application of Ehrenfeucht-Fraïssé games shows that any two linear orders with more at least elements agree on the validity of A (the reader may consult [10] for information on Ehrenfeucht-Fraïssé games and basic results). Therefore, in there are frames of arbitrarily large depth that validate A precisely when the unique up to isomorphism linear order with elements validates A. Since and a linear order with elements are computable from the parameter A, and satisfiability in the obtained structure is computable, the result follows for .
We will now obtain a similar property for . First, from each frame obtain the frame by replacing each maximal chain of depth more than with its initial segment of length and keeping the other maximal chains intact. We claim that and agree on the validity of A and prove it by describing a winning strategy for Duplicator for the n-turn Ehrenfeucht-Fraïssé game for and . On each step:
If Spoiler picks an element from a chain that has not been modified in the construction, Duplicator picks the same element from the other frame.
If Spoiler picks an element from a maximal chain in that has been shrinked to in or an element from a maximal chain in that that has been obtained by shrinking the chain in , Duplicator chooses an element in by consulting the winning strategy for the n-turn Ehrenfeucht-Fraïssé game for the linear orders and (both chains contain at least elements so such strategy exists).
We will use the property to prove that there is no uniform bound m of the depth of the frames from that validate A iff A has a model which has a chain C with at least elements.
Suppose that A has a model which contains a chain C with at least elements and assume for contradiction that there is a uniform bound m of the depth of the frames from that validate A. Consider the frame obtained by extending with m new elements the maximal chain in which contains C. Now, since we obtain that . Clearly, is the same as , so . And finally since . But contains a chain with more than m elements which contradicts the property of m being a uniform bound for A.
Conversely, if A has no uniform bound of the depth of its models from , this means that in particular is not a uniform bound so A must have a model from which contains a chain with at least elements.
Combining the above, we conclude that (Boundedness) for reduces to checking whether A has a model with a chain of depth (at least) . This in turn reduces to (Validity), as such a model exists iff is satisfiable in . □
Remark 2. Note that we can further employ the technique of Ehrenfeucht-Fraïssé games used in the above proof to show alternative decision methods for the validity problem for and . Let A be a first-order sentence and .
For finite linear orders, we have that precisely when all (finitely many up to isomorphism) linear orders of depth at most validate A because all linear orders with at least elements agree on the validity of A.
Let be arbitrary and obtain as in the proof of 5. We then obtain the subframe of by removing some (possibly none) of the maximal chains in in the following way: for each such that there are more than n maximal chains of depth k in we remove all but n of those chains in . There is again a simple winning strategy for Duplicator for the n-turn Ehrenfeucht-Fraïssé game for and .
This means that precisely when all frames of the form for some validate A. But the frames of type are precisely those (finitely many up to isomorphism) frames whose chains are of depth at most and for each contain at most n maximal chains with k elements.
Now, to decide validity for either class we can generate and check the validity of A in a finite number of finite frames (in which satisfaction of sentences is computable) which can be computed from A.
4. Some Classes of Partial Orders with Respect to Which the Problem Is Undecidable
In this section we obtain negative solutions to the definability problem for some natural classes of partial orders. A classical result in correspondence theory is a theorem by Chagrova [2,3] stating that the definability problem with respect to the class of all partial orders is undecidable. Her method is based on reductions of undecidable problems for Minsky machines. Here we present a technique developed by Balbiani and Tinchev [4] which allows for a model theoretic approach for proving negative results for the definability problem for modal logics. First we present the technique in its straightforward adaptation for the intuitionistic case.
Definition 2.
We say that a class of partial orders isstableif there is a first-order sentence B and a first-order formula A with free variables among which satisfy the following conditions:
-
(1)
For every frame and every tuple of points in , the relativized reduct of with respect to A and , if it exists, is a frame in .
-
(2)
For every frame there are frames and such that , , and is a relativized reduct of with respect to A and some parameters .
Theorem 5. If is a stable class of partial orders, then the first-order validity problem reduces to the problem with respect to .
Proof. The proof is almost the same as the proof for the modal version of the theorem in [4], only replacing modal with intuitionistic semantics for the propositional formulas. □
The above theorem provides a very useful reduction since it allows to prove undecidability of the definability problem by showing stability of the class in consideration and undecidability of its first-order theory.
We shall apply the theorem to show that the problem with respect to the following classes is undecidable:
The class of all partial orders (a model-theoretic proof of Chagrova’s result).
For each the class of all partial orders of depth at most n.
The class of all dense partial orders.
The classes and for (where the superscript denotes the restriction of the class to finite frames).
We begin with stability of the classes:
Lemma 6. The classes , for and are stable.
Proof. Denote by any of the above classes and by K its first-order axiom (all of the above classes are finitely axiomatizable). We choose the formula and the sentence where is an abbreviation for the formula which states that z is incomparable with any other point.
We show that A and B witness the stability of .
For property in the definition of stability: suppose that and are such that the relativized reduct of with respect to A and exists. Then is the subframe of with universe . If is , then obviously is a partial order so . If is then the depth of the chains in is also bounded by n since every chain in is a chain in so . If is then to show that it suffices to check that if and or then there is some such that . But this is immediate since if for then by density there is some such that . If we are done and otherwise by density again there exists some c such that and then .
We now turn to property (2) in the definition of stability. Suppose that and take . Take the following frames:
with universe and
with universe and .
Obviously and are partial orders, if the chains in are bounded in depth by then so are the chains in and , and if is dense then so are and ; so . Since is a generated subframe of we have that . contains a single point so it certainly does not have two isolated points, i.e. . On the other hand, contains the isolated points a and b so . Finally, is obtained from by omitting the new points a and b so is the relativized reduct of with respect to A and . □
Lemma 7. The classes and for are stable.
Proof. The construction for the unrestricted classes in the previous lemma produces finite frames when is finite. □
We are now left to show undecidability of the first-order theories of the classes. A theorem of Tarski [11] gives undecidability of the first-order theories of the classes and . In order to prove that the other classes of frames have undecidable first-order theories, we will consider a certain class of frames which we will call connectivity maps. We will later see that such frames are intrinsically related to undirected graphs without loops (which we think are represented by reflexive and symmetric relations).
Definition 3. The class of connectivity maps, , contains all partial orders that satisfy the following conditions:
In the context of partial orders we will use the following abbreviations:
stands for the formula which states that x is a minimal element.
stands for the formula which states that x is a maximal element.
stands for the formula abbreviated by which states that x is neither minimal nor maximal.
Proposition 4. The class has the following properties:
-
(1)
Each connectivity map is a partial order of depth at most 2.
-
(2)
is finitely axiomatizable.
-
(3)
The first-order theory of reduces to the first-order theory of for . The same relationship holds for and for .
Proof.
- (1)
Assume for contradiction that there is and are such that . Since y has a point above it, it must have exactly two points above it. But then both x and y are strictly below and which violates the second condition for membership to .
- (2)
-
We take the axioms for partial orders together with the following axioms corresponding to the two membership conditions for :
- (3)
is an axiomatizable subclass of for so for a first-order sentence A we have that iff where C is the axiom for . The same reduction holds for the classes of finite frames.
□
The essence of the above proposition is that undecidability of the first-order theory of the class (and ) propagates to the other classes we have listed with the exception of the class . For we will need to take an intermediate step by considering "dense variants" of connectivity maps.
Definition 4.
Let be a connectivity map. We say that the partial order is adensification of if is obtained by inserting between each pair a fresh dense linear order with least element x and greatest element y. Denote by the class of all densifications of connectivity maps.
Proposition 5. The class has the following properties:
-
(1)
is finitely axiomatizable.
-
(2)
There is a translation of first-order formulas such that for each first-order sentence A it holds that iff where is any connectivity map and is any of its densifications.
-
(3)
The first-order theory of reduces to the first-order theory of
Proof.
- (1)
-
The following sentences (we use as the usual abbreviation for "there exists a unique x") provide an axiomatization for :
- (a)
The axiom for dense partial orders.
- (b)
stating that any interval between a minimal and a maximal element is linearly ordered.
- (c)
stating that each nonextremal element is comparable with a unique maximal element and a unique minimal element.
- (d)
stating that each minimal element is either maximal or is below exactly two maximal elements (the axiom roughly corresponds to the first membership condition for ).
- (e)
stating that each pair of distinct maximal elements have at most one common minimal element below them (roughly corresponds to the second membership condition for ).
An immediate verification shows that each frame from satisfies the axioms. Conversely, suppose that satisfies the axioms. We can then obtain as the subframe of whose universe consists of the extremal elements of . is a partial order so is too. Moreover, satisfies the two membership conditions for because the last two axioms above force the extremal elements of to be in an appropriate configuration, so .
By axiom we know that the extremal elements of together with the intervals between minimal and maximal elements exhaust all of and that the interiors of such intervals are two by two disjoint. Axiom together with density of means that each such interval is a dense linear order. Therefore, can be obtained as a densification of by replacing each pair in by the dense and linearly ordered interval , hence .
- (2)
Consider the formula . Define the translation that transforms a first-order formula A into its relativization with respect to the formula U and the variable y, i.e. . Now consider a connectivity map and any of its densifications . consists of the extremal elements of so is the relativized reduct of with respect to U and y. Therefore, by the relativization theorem it follows that iff .
- (3)
Denote by D the axiom for the class and consider the translation from . Then for any sentence A we have that iff for each connectivity map iff (by (2)) for each densification of any connectivity map iff .
□
Corollary 1. The first-order theory of reduces to the first-order theory of
Proof. The first-order theory of reduces to the first-order theory of by the previous proposition. But if D is the axiom for and A is an arbitrary sentence, then iff . □
We now turn to proving that the first-order theory of is undecidable. As we previously eluded, the class is closely connected to the class of graphs. A theorem by Rogers [12] shows the undecidability of the first-order theory of a reflexive and symmetric relation (which we think of as encoding a graph) and we will use it to anchor the chain of reductions.
Definition 5.
A graph is a frame for the language with sole non-logical symbol E which is interpreted as a reflexive and symmetric relation.
If is a graph (without loss of generality we assume that the elements in are not two-element sets), its canonical connectivity map is the frame with the following definition:
is the least partial order on such that and for each
We briefly elaborate on the above definition. In traditional terms, the graphs we consider are unordered and without loops, and says that there is an edge between u and v when u and v are distinct vertices. As we do not allow loops in our graphs, does not encode any significant information and for simplicity, we take E to be reflexive.
The canonical connectivity map for is then essentially a different representation of the graph: the vertices are maximal elements in and an edge is encoded in by the configuration of and . An immediate but useful property is that is finite iff its canonical connectivity map is finite. The following Definition and Proposition show how we translate first-order formulas for graphs into first-order formulas for connectivity maps which have the same meaning.
Definition 6. The translation that transforms first-order formulas for the language to first-order formulas for the language of order is defined by recursion as follows:
Proposition 6. If is a graph and is its canonical connectivity map, then for each sentence A in the language it holds that iff .
Proof. By straightforward induction on formulas A in the language of graphs with free variables we can prove that for each tuple of elements of we have that iff . In particular, when A is a sentence we obtain the desired property of the translation. □
Proposition 7. Any connectivity map is isomorphic to the canonical connectivity map of some graph .
Proof. It is immediately verified that is isomorphic to the canonical connectivity map of the graph defined as follows:
is the set of all maximal elements in
For each distinct we define iff and have a lower bound in
□
Theorem 6. The first-order theories of the classes for and are undecidable. The first-order theories of the classes and for are undecidable (not even semidecidable).
Proof. The first-order theory of graphs is undecidable by a theorem of Rogers [12]. We claim that the sentence A about graphs is valid iff . Indeed, all graphs validate A iff (by Proposition 6) the class of all canonical connectivity maps validates iff (by Proposition 7) all connectivity maps validate . Therefore, reduces the first-order theory of graphs to the first-order theory of , which we then conclude is undecidable.
Now undecidability of the first-order theory of for follows from (3) in Proposition 4 and undecidability of the first-order theory of follows from Corollary 1.
For the classes of finite frames we follow the parallel chain of reductions starting from the first-order theory of finite graphs (the relevant constructions for the reductions construct finite frames when applied to finite frames). Its undecidability follows from a theorem in [13] which states that the first-order theory of graphs and the complement of the first-order theory of finite graphs are recursively inseparable, in particular this means that the first-order theory of finite graphs is undecidable. A finer analysis shows that the first-order theory of finite graphs is not even semidecidable since its complement is semidecidable (given a sentence A start enumerating all finite graphs and search for a countermodel for A). Then untractability propagates down the chain of reductions. □
Theorem 7. The problem with respect to each of the classes , , , and for is undecidable.
Proof. By Lemma 6 and Lemma 7 each of the listed classes is stable. The first-order theory of and are undecidable by [11] and the first-order theories of the other classes are undecidable by Theorem 6. By Theorem 5 the first-order theories reduce to the respective instances of with respect to the classes hence the result follows. □