1 Introduction We study classical rst-order logic with equality but without any other relation symbols. The letters ' and are reserved for quantier-fr

Transcription

1 UPMAIL Technical Report No. 138 March 1997 (revised July 1997) ISSN 1100{0686 Some Undecidable Problems Related to the Herbrand Theorem Yuri Gurevich EECS Department University of Michigan Ann Arbor, MI USA Margus Veanes Computing Science Department Uppsala University Box 311, S Uppsala Sweden Abstract We improve upon a number of recent undecidability results related to the so-called Herbrand Skeleton Problem, the Simultaneous Rigid E-Unication Problem and the prenex fragment of intuitionistic logic with equality. Partially supported by grants from NSF, ONR and the Faculty of Science and Technology of Uppsala University.

2 1 Introduction We study classical rst-order logic with equality but without any other relation symbols. The letters ' and are reserved for quantier-free formulas. The signature of a syntactic object S (a term, a set of terms, a formula, etc.) is the collection of function symbols in S augmented, in the case when S contains no constants, with a constant c. The language of S is the language of the signature of S. Any syntactic object is ground if it contains no variables. A substitution is ground if its range is ground, and it is said to be in a given language if the terms in its range are in that language. A set of substitutions is ground if each member is ground. A ground substitution corroborates a formula ' (or is a corroborator for ') if the formula ' is provable. Given a positive integer m, a set of m ground substitutions f 1 ; : : : ; m g is an m-corroborator for ' if the disjunction ' 1 ' m is provable. One popular form of the classical Herbrand theorem [27] is this: An existential formula 9~x'(~x) is provable if and only if there exist a positive integer m and an m-corroborator for ' in the language of '. The minimal appropriate number m will be called the multiplicity of '. The multiplicity may exceed one. Here is a formula of multiplicity two suggested by Erik Palmgren in a dierent but similar context; we use `' for the formal equality sign. '(x) = (c 0 ) x 1) ^ (c 1 ) x 0) The Herbrand theorem plays a fundamental role in automated theorem proving methods known as the rigid variable methods [45]. We can identify the following procedure underlying such methods. We call it the principal procedure of rigid variable methods. Let 9~x'(~x) be a closed formula that we wish to prove. Step I Choose a positive integer m. Step II Check if there exists an m-corroborator for '. Step III If Step II succeeds then 9~x'(~x) is provable, otherwise increase m and return to Step II. The kernel of the principal procedure is of course Step II or: The Herbrand Skeleton Problem Instance: A quantier free formula ' and a positive integer m. Question: Is the multiplicity of ' bounded by m? 1

3 We refer the reader to [10] for a detailed discussion of the problem. It is important to us here that the Herbrand Skeleton Problem is intimately related to the Existential Intuitionistic Problem and the Simultaneous Rigid E-Unication Problem [22]. The rst of these problems is easy to formulate: The Existential Intuitionistic Problem Instance: An existential formula 9~x'(~x). Question: Is the formula provable in intuitionistic logic? The second requires auxiliary denitions. A rigid equation is an expression E `8 e where E is a nite set of equations and e is an equation. A ground substitution solves a rigid equation E `8 e if E ` e. A system (that is a nite set) of rigid equations is solvable if there is one substitution that solves all rigid equations in the system. The Simultaneous Rigid E-Unication Problem (SREU) Instance: A system of rigid equations. Question: Is the system solvable? The SREU problem has an interesting history [10]. Several false decidability claims have been published until nally it has been proved undecidable by Degtyarev and Voronkov [12, 14, 16, 17]. Later Plaisted has shown that the fragment of the SREU problem with ground left-hand sides is undecidable [36] (the left-hand side of a rigid equation E `8 e is E). It is easy to see that SREU is essentially a special case of the Herbrand skeleton problem with Horn formulas and multiplicity one. It follows that the Herbrand skeleton problem is undecidable even in this very special case. Voronkov suggested the following generalizaton of the Herbrand Skeleton Problem. Let f be a function that assignes a positive integer to every pair (k; ') where k is a positive integer and ' a formula in our logic. Moreover, it is assumed that k < l implies that f(k; ') < f(l; '). Such a function is called a strategy for multiplicity. The intended meaning of the rst argument of a strategy is the number of times that Step II of the principal procedure has been executed. The Herbrand f-skeleton Problem Instance: A quantier free formula ' and a positive integer k. Question: Is the multiplicity of ' bounded by f(k; ')? In case f(k; ') = m for all ' and k, the Herbrand f-skeleton Problem will be called the Herbrand m-skeleton Problem, or simply the m-skeleton problem. Thus the undecidability result of Degtyarev and Voronkov implies that the 1-Skeleton problem is undecidable. Voda and Komara have proved that, for each xed m, the m-skeleton problem is undecidable [42]. One important conclusion for automated theorem proving, drawn in [42], is that there is no m for which there exists an eective decision procedure that 2

4 would tell us whether m substitutions suce to establish the provability of a given quantier free formula. Actually, we had hard time to understand the proof of Voda and Komara until, nally, we convinced ourselves that they have a proof. We wondered if there is a way to derive their result from the Degtyarev{Voronkov theorem. It turns out that indeed there is such a way. In order to formulate our results, we need to recall a few denitions and give denitions of our own. Recall that a Horn clause is a disjunction of negated atomic formulas and at most one non-negated atomic formula; a Horn clause is often represented as a set of its disjuncts. Here we restrict attention to Horn clauses that contain exactly one non-negated atom. A Horn formula is a conjuction of Horn clauses. Since the equality sign is the only relation symbol in our logic, every Horn clause is equivalent to an implication E ) s t where E is a conjunction of equalities. We say that a collection of formulas is constant-disjoint if there is no constant that occurs in two or more of the given formulas. Call a Horn formula ' guarded if, for every variable x that occurs in ', there exists a clause E ) s t in ' where E and s are ground and x occurs in t. Finally, call a corroborator of a disjunction ' partisan if it corroborates one of the disjuncts of '. Now we are ready to formulate our rst result. Partisan Corroboration Theorem Every corroborator for a disjunction of constant-disjoint guarded Horn formulas is partisan. This theorem is proved in Section 3. We believe it is of independent interest. It allows us an easy derivation of the Voda{Komara result from the Degtyarev{Voronkov theorem in Section 4. In fact, we strengthen the Voda{Komara theorem in several ways. For each m, we eectively reduce SREU to the Herbrand m-skeleton problem in such a way that the positive-arity part of the signature remains unchanged. In particular, for every m, the monadic (all function symbols are of arity 1) SREU reduces to the monadic Herbrand m-skeleton problem; this reduction is of interest because the decidability of monadic SREU is an open problem [26]. In Section 5, we improve upon a construction in Veanes [41] and show the undecidability of a fragment of SREU with only two variables and three rigid equations with ground left-hand sides. Using this fact, we show, for each positive integer m, the undecidability of the m-skeleton problem where each formula is a conjunction of 3m Horn clauses with 2m variables and ground negative literals; the negative literals can even be xed. In Section 7 we obtain some undecidability results related to the prenex fragment of intuitionistic logic with equality and proof search in intuitionistic 3

5 logic with equality. Finally, in Section 8 we describe the current status of SREU and related results and list some open problems. 2 Preliminaries We will rst establish some notation and terminology. We follow Chang and Keisler [4] regarding rst order languages and structures. For the purposes of this paper it is enough to assume that the rst order languages that we are dealing with are languages with equality and contain only function symbols and constants, so we will assume that from here on. We will in general use, possibly with an index, to stand for a signature, i.e., is a collection of function symbols with xed arities. A function symbol of arity 0 is called a constant. We will always assume that contains at least one constant. 2.1 Terms and Formulas Terms and formulas are dened in the standard manner and are called - terms and -formulas respectively whenever we want be precise about the language. We refer to terms and formulas collectively as expressions. In the following let X be an expression or a set of expressions or a sequence of such. We write (X) for the signature of X: the set of all function symbols that occur in X, V(X) for the set of all free variables in X and C(X) for the set of all constants in X. We write X(x 1 ; x 2 ; : : : ; x n ) to express that V(X) fx 1 ; x 2 ; : : : ; x n g. Let t 1 ; t 2 ; : : : ; t n be terms, then X(t 1 ; t 2 ; : : : ; t n ) denotes the result of replacing each (free) occurence of x i in X by t i for 1 i n. By a substitution we mean a function from variables to terms. We will use to denote substitutions. We write X for X((x 1 ); (x 2 ); : : : ; (x n )). We say that X is closed or ground if V(X) = ;. By T or simply T we denote the set of all ground -terms. A substitution is called ground if its range consists of ground terms. A closed formula is called a sentence. Since there are no relation symbols all the atomic formulas are equations, i.e., of the form t s where t and s are terms and `' is the formal equality sign. Atomic formulas and negated atomic formulas are called positive and negative literals respectively. A clause is a disjunction of literals. By a Horn clause we mean a clause with exactly one positive literal. 1 A Horn clause can be written as E ) s t where E is a conjunction of equations, and s and t are terms. By a Horn formula we understand a conjunction of Horn clauses. 2.2 First Order Structures First order structures will (in general) be denoted by capital gothic letters like A and B and their domains by corresponding capital roman letters like 1 By a Horn clause we mean thus a strict Horn clause. 4

6 A and B respectively. A rst order structure in a signature is called a -structure. For F 2 we write F A for the interpretation of F in A. If A is a -structure and 0 then A 0 is the 0 -structure that is the reduction of A to signature 0. Let A and B be -structures, A is a substructure of B, in symbols A B, if A B and for each n-ary F 2, F A = F B A n. For X a sentence or a set of sentences, A j= X means that the structure A is a model of or satises X according to Tarski's truth denition. A set of sentences is called satisable if it has a model. If X and Y are (sets of) sentences then X j= Y means that Y is a logical consequence of X, i.e., that every model of X is a model of Y. We write j= X to say that X is valid, i.e., true in all models. One easily establishes, by induction on terms and formulas that, if A B then for all quantier free sentences ', A j= ' i B j= '. By the free algebra over we mean the -structure A, with domain T, such that for each n-ary F 2 and t 1 ; : : : ; t n 2 T, F A (t 1 ; : : : ; t n ) = F (t 1 ; : : : ; t n ). We let T also stand for the free algebra over. Let E be a set of ground equations. Dene the equivalence relation = E on T by s = E t i E j= s t. By T =E (or simply T =E ) we denote the quotient of T over = E. Thus, for all s; t 2 T, T =E j= s t, E j= s t: We call T =E the canonical model of E. 2.3 Term Rewriting In some cases it is convenient to consider a system of ground equations as a rewrite system. We will assume that the reader is familiar with basic notions regarding ground term rewrite systems [18]. We will only use very elementary properties. In particular, in the next section we will use Bircho's completeness theorem for equational logic [2]. In the case of ground equations it states simply that, given a ground set of equations E and and a ground equation s t, E j= s t i s can be reduced to t by using the equations in E as rewrite rules in both directions. In Section 5 we will use the following property of canonical (or convergent) rewrite systems (cf [18, Section 2.4]). Let R be a ground and canonical rewrite system. Then for any two ground terms t and s, the equation t s follows logically from R (seen as a set of equations) i the normal forms of t and s with respect to R coincide, i.e., R j= t s, t# R = s# R : 3 Some Logical Tools In this section we will prove some logical properties that will be used in the next section. The main result is Theorem 5. The following lemma is 5

7 actually a consequence of Los-Tarski theorem. 2 We say that two (sets of) expressions X and Y are constant-disjoint if C(X) \ C(Y ) = ;. Lemma 1 Let ' i for i 2 I, be pairwise constant-disjoint quantier free sentences. Then j= W i2i ' i implies j= ' i for some i 2 I. Proof. For i 2 I, let i = (' i ) and let = S i i. Assume by contradiction that j= 6 ' i for all i 2 I. Then there is (for each i 2 I) a i -structure A i such that A i j= :' i. Without loss of generality, take all the A i to be pairwise disjoint. We now construct a -structure A such that A i A i for i 2 I. First let A = S i2i A i. For each i 2 I and constant c 2 L i let c A = c A i. For each n-ary function symbol F in dene F A as follows. For all ~a = a 1 ; : : : ; a n 2 A, F A F A i (~a); if ~a 2 A (~a) = i ; a 1 ; otherwise. It is clear that A is well dened because of the disjointness criteria and that A i A i for i 2 I. Hence A i j= :' i, and thus A j= :' i for each i 2 I. But this contradicts that j= W i2i ' i. If we drop the constant-disjointness criterion in Lemma 1, then of course the lemma is false. A simple counterexample is j= 0 1 _ :(0 1): We will state now some other obvious but useful lemmas. Lemma 2 is an easy corollary of Birkho's completeness theorem. Lemma 2 Let t and s be ground terms and let E and E 0 be ground sets of equations such that C(E 0 ) \ C(E; s) = ;. The following is true. 1. If E 0 [ E j= t s then E j= t s. 2. If E j= t s then (t) (E; s). Proof. Assume that E 0 [ E j= t s. By Bircho's completeness theorem we know that s can be rewritten to t by using E 0 [E as a set of rewrite rules. So there is a sequence of terms s 0 ; s 1 ; : : : ; s n 1 ; s n where s 0 = s, s n = t and s i is rewritten to s i+1 by using some rule in E 0 [ E, for 0 i < n. By induction on i (for i n) follows that (s i ) (E; s) and only a rule from E can be used to rewrite s i. Part 1 follows by Bircho's completeness theorem and part 2 follows immediately (take E 0 = ;). 2 Existential sentences are preserved under extensions. 6

8 For a nite set E of equations we will write E also for the corresponding conjunction of equations and let the context determine whether a set or a formula is meant. Lemma 3 Let t and s be ground terms and E 0 and E ground sets of equations such that E is nite and C(E 0 ) \ C(E; s) = ;. Then T =E0 [E j= (E ) t s) ) j= (E ) t s): Proof. From T =E0 [E j= (E ) t s) follows immediately that T =E0 [E j= t s and thus E 0 [ E j= t s. Hence E j= t s by Lemma 2, i.e., j= (E ) t s). We will use the following denitions. Let ' be a quantier free formula and m a positive integer. I A set of m ground substitutions is an m-corroborator for ' if _ j= ': 2 When = fg we say that is a corroborator for ' or corroborates '. The m-skeleton problem is the problem of existence of m-corroborators for given formulas. I For x 2 V('), a guard for x in ', if it exists, is a clause E ) t s in ' such that E and s are ground and x occurs in t. We say that ^ x2v(') is a guard of ' if each x is a guard for x in '; ' is is called guarded if it has a guard. Intuitively, in the light of the second part of Lemma 2, the notion of a Horn formula being guarded is a sucient condition to guarantee that if there is a corroborator for ' then the range of V(') is T ('), i.e., (') = ('). SREU is, by denition, the problem of existence of corroborators for Horn formulas. However, we only need to consider guarded Horn formulas. To see that consider a Horn formula '; let be its signature expanded with a x 7

9 constant if ' has no constants and let c be a constant in. Let ' 0 (x) be the Horn clause E ) x c where E = f f(c; : : : ; c) c j f 2 g: Let now be the guarded Horn formula ^ ( ' 0 (x)) ^ ': x2v(') Clearly, has a corroborator i ' has one. Note that, for all terms t, j= (E ) t c), t 2 T : Example 4 A simple example of a guarded Horn formula is = (A 1 ) c 0 1 x c 1) ^ (A 2 ) c 0 2 y c 2) ^ ( 1 ) x y) ^ ( 2 ) x t y) where A 1, A 2, 1, 2 and t are ground, c 1, c 0 1, c 2 and c 0 2 is a binary function symbol. The guard of is are constants and (A 1 ) c 0 1 x c 1) ^ (A 2 ) c 0 2 y c 2): An example of a Horn formula with a common guard for all variables is ' = (A ) x y c) ^ ( 1 ) x y) ^ ( 2 ) x t y); where A, 1, 2 and t are ground and c is a constant. The guard of ' is A ) x y c: Both formulas are of particular interest for us, see Section 5. 2 We will use the following denition. I A corroborator of a disjunction ' is partisan, if it corroborates some disjunct of '. The main result of this section is the following theorem. Theorem 5 (Partisan Corroboration Theorem) Every corroborator of a disjunction of constant-disjoint guarded Horn formulas is partisan. 8

10 Proof. Let ' = W i2i ' i where all the ' i 's are constant-disjoint guarded Horn formulas. Let be a corroborator for '. We must prove that corroborates ' i for some i 2 I. We can assume (without loss of generality) that there exist positive integers m and n such that each ' i has the following form: ^ ^ ' i = (Ei k ) s k i t k i ) ^ (Di k ) u k i vi k ); 1km {z } i 1kn where i is a guard of ' i, i.e., each E k i and s k i is ground and V(' i) = V( i ), for all i 2 I. Let C i = C(' i ) for i 2 I. We have that C i \ C j = ; (8i; j 2 I; i 6= j): (1) Let = ('). For i 2 I let K i denote the class of all -structures that satisfy ' i, i.e, K i = f -structure A j A j= ' i g: From the validity of ' follows that each -structure belongs to some K i. Let now J be any subset of I such that (Take for example J = ;.) So j= i (8i 2 J): (2) C(' i ) = C i (8i 2 J): (3) To see that, suppose (by contradiction) that C(' i ) contains some c =2 C i. Clearly, c belongs to some x where x occurs in the guard i. By the second part of Lemma 2, every constant in x belongs to C i. This gives the desired contradiction. If I = J then the theorem follows by Lemma 1. Assume that I 6= J. Now we prove the following statement: If 6j= ' i for all i 2 J then j= i for some i 2 I n J. (4) Proof of (4) Assume 6j= ' i for all i 2 J. follows. If J = ; let D = ;. Form an equation set D as If J 6= ; then there is for each i 2 J a clause in ' i that is not valid and by (2) this clause is not in i. In other words, there is a mapping f : J! f1; 2; : : : ; ng such that j= 6 (D f(i) i ) u f(i) i v f(i) i ) (8i 2 J): (5) Let f be xed and let D = S i2j Df(i) i. 9

11 For each mapping g : I n J! f1; 2; : : : ; mg let E g denote the following set of equations: [ E g = E g(i) i ; i2inj and let A g be the canonical model of D [ E g, i.e., A g = T =Eg[D: We will now prove the following statement. (*) Fix g : I n J! f1; 2; : : : ; mg. There exists i 2 I n J such that A g 2 K i. Proof of (*) Assume that (*) does not hold. (Assume also that J 6= ; or else (*) holds trivially.) Then A g 2 K j for some j 2 J. Fix such an appropriate j. So A g satises each clause in ' j and in particular Let D 0 = D f(j) j and So, by (1), A g j= (D f(j) j, u 0 = u f(j) j ) u f(j) j v f(j) j ): and v 0 = v f(j) j. By (3) follows that C(D 0 ; u 0 ; v 0 ) C j C(E g ; D n D 0 ) = C(E g ) [ C(D n D 0 [ ) = C(E g ) [ C(D f(i) i ) It follows, by Lemma 3, that But this contradicts (5). [ = i2inj [ i2i;i6=j i2j;i6=j C i [ [ C i : i2j;i6=j C i C(D 0 ; u 0 ; v 0 ) \ C(E g ; D n D 0 ) = ;: j= (D f(j) j ) u f(j) j v f(j) j ): By using (*) we can now prove the following statement (**) There exists i 2 I n J such that j= i. 10

12 Proof of (**) Assume that the claim is wrong. Then there is for each i 2 I n J a clause in i that is not valid, i.e., there is a mapping g : I n J! f1; 2; : : : ; mg such that j= 6 E g(i) i ) s g(i) i (t g(i) i ) (8i 2 I n J): (Note that only the t i 's can be nonground.) Fix such an appropriate g. By using (*) we know that A g 2 K i for some i 2 I n J. Choose such an i. So A g satises each clause in ' i and in particular A g j= E g(i) i But, by (3) and (1), C(E g(i) i Lemma 3, j= E g(i) i So we have contradiction. ; s g(i) i ) s g(i) i (t g(i) i ): ) \ C(E g n E g(i) i ; D) = ;. Hence, by ) s g(i) i (t g(i) i ): This proves statement (4). Let now J be the maximal subset of I such that (2) holds. In other words, for all i 2 I nj, 6j= i. By the contrapositive of (4) we conclude that for some i 2 J, j= ' i and the theorem follows. Remark Theorem 5, as well as its proof, remain correct if the disjunction is innite. We will not use this generalization. The following example illustrates why the conditions of being constantdisjoint and guarded are important and cannot in general be discarded. In each case there is a counterexample to the theorem. Example 6 Let us rst consider an example where the disjuncts are guarded but not constant-disjoint. Let '(x) be the following guarded Horn formula: (c 0 ) x 1) ^ (c 1 ) x 0) where c, 0 and 1 are contants, and let ' 1 = '(x 1 ), ' 0 = '(x 0 ) and = ' 1 _ ' 0 where x 1 and x 0 are distinct variables. Consider now any ground substitution such that (x 1 ) = 1 and (x 0 ) = 0. It is easy to show by case analysis that corroborates, i.e., that j= ((c 0 ) 1 1) ^ (c 1 ) 1 0)) _ ((c 0 ) 0 1) ^ (c 1 ) 0 0)): However, corroborates neither ' 1 nor ' 0. 11

13 Let us now consider the case when constant-disjointness is not violated but the disjuncts are not guarded. Let ' 1 (y; x 1 ; y 1 ) be the formula and let ' 0 (x 0 ; y 0 ) be the formula ((y 0 ) x 1 y 1 ) ^ (y y 1 ) x 1 0)) ((c y 0 ) x 0 1) ^ (c 1 ) x 0 y 0 )) where c, 0 and 1 are constants and x 1 ; x 0 ; y 1 ; y 0 ; y distinct variables. Let = ' 1 _ ' 0. Let be a ground substitution such that (x 1 ) = 1, (x 0 ) = 0, (y) = c, (y 1 ) = 1 and (y 0 ) = 0. Then j= but 6j= ' 1 and 6j= ' 0 (the situation is exaclty the same as in the previous case). 2 4 Reduction of 1-Skelton Problem to n-skeleton Problem The 1-Skeleton problem is undecidable. This follows from the undecidability of SREU by Degtyarev and Voronkov [14, 17]. We can formulate their result in the current setting as follows (cf [17, Theorem 1]). Theorem 7 (Degtyarev{Voronkov) The 1-Skeleton problem of guarded Horn formulas is undecidable. Under certain restrictions on the language and the structure of formulas, the 1-Skeleton problem becomes decidable. It is known, however, that it is already undecidable in the presence of one binary function symbol (in addition to constants); moreover, two variables suce for undecidability [41]. For a summary over what is known to be decidable or undecidable under various restrictions see Section 8. For technical reasons it will be convenient to assume in the following that we have a xed signature with fc 1 ; c 2 ; : : :g as the set of distinct constants in it. may also have other function symbols of arity 1. Let us also be precise about the variables that we allow in -expressions, by assuming that all variables come from the collection fx 1 ; x 2 ; : : :g. For each natural number n, constant c and variable x, let c (n) denote a new constant and let x (n) denote a new variable. We dene by induction on any -expression X the corresponding expression X (n) as the one obtained from X by replacing in it each variable x with x (n) and each constant c with c (n). For any substitution of -variables with -terms we let (n) denote a substitution that takes the variable x (n) t the term (x) (n). So, for any -expression X and natural number n, (X) (n) = X (n) (n) : The following property is immediate. For any -sentence ' and natural number n, j= ', j= ' (n) : 12

14 Theorem 8 Let ' be a guarded Horn formula and n a positive integer. Then ' has a corroborator i V n i=1 '(i) has an n-corroborator. Proof. The `)' direction is trivial. We prove the `(' direction as follows. Let I = f1; 2; : : : ; ng and let be the formula V i2i '(i). Assume that has an n-corroborator f i j i 2 I g. So _ j= (^ ' (j) i ): i2i j2i By the distributive law this is equivalent to ^ _ j= ' f:i!i( (f(i)) i ): i2i From this follows in particular that j= _ i2i ' (i) i : Let X i = V(' (i) ) for i 2 I. Since all the X i 's are pairwise disjoint we can let 0 be a substitution such that 0 X i = i X i for i 2 I, and it follows that j= _ i2i ' (i) 0 : By Theorem 5 follows now that j= ' (i) 0 for some i 2 I. Fix such an appropriate i. But then, by Lemma 2, the range of 0 X i is T (' (i) ), and thus there is a substitution with range T such that (i) X i = 0 X i. Hence j= ' (i) (i) and so j= ' by above. Corollary 9 (Voda{Komara) For all n 1, n-skeleton problem of guarded Horn formulas is undecidable. Proof. The reduction in Theorem 8 is trivially eective. So, if we had a decision procedure (for some n) for nding n-corroborators, we could use it to nd corroborators, but this would contradict Theorem 7. Assume that we are using an automated theorem proving method that is based on the Herbrand theorem. Roughly, this involves a search for terms, for a given bound m on multiplicity. Corollary 9 (Voda and Komara [42]) tells us that there is no m for which we could eectively decide when to stop our search for such terms in case they do not exist. By using the fact that SREU is undecidable already with ground left-hand sides [36], (i.e., variables occur only in positive literals in the corresponding Horn formulas) and two variables [40, 41] we obtain a sharper version of the above corollary: 13

15 Corollary 10 For all n 1, n-skeleton problem of guarded Horn formulas is undecidable already if there are 2n variables and all variables occur in positive literals. The decidability of monadic SREU is currently one of the problems related to SREU that is still open [26]. An eectively equivalent problem is the decidability of the prenex fragment of intuitionistic logic with equality with unary function symbols [15]. Some evidence speaks in favour of that the problem is decidable although with very high computational complexity (e.g., many subcases are decidable, see Section 8). From Theorem 8 follows that: Corollary 11 If the 1-Skeleton problem is undecidable in the monadic case then so is the n-skeleton problem for n > 1, or equivalenty, if the n-skeleton problem is decidable in the monadic case for some n > 1 then so is the 1- Skeleton problem. 5 Undecidability of SREU: Minimal case W show that three rigid equations with ground left-hand sides and two variables in a signature with one binary function symbol an no other nonconstant function symbols, already imply undecidability. In fact, we give a uniform representation of all the recursively enumerable sets by using just three rigid equations with these properties. As a corollary we get that the undecidability of SREU holds already in very restricted cases. We generalize the construction in Veanes [41] and improve the lower bound on the number of rigid equations from four to three by using nite tree automata techniques. We then use this result to improve the undecidability result of the n-skeleton problem. The main idea behind our proof is based on a technique that was used by Plaisted [36] in a similar context, who called the technique shifted pairing. The idea is to express repetition explicitly by a sequence of strings (like IDs of a TM). The rst string of the sequence fullls some initial conditions, the last string some nal conditions and another sequence is used to check that the consequtive strings of the rst sequence satisfy some relationship (like validity of a computation step). A similar technique was used already by Goldfarb in the proof of the undecidability of second-order unication [24], which is by reduction of Hilbert's tenth problem, and later, adopted from that proof, also in a proof of the undecidability of SREU by Degtyarev and Voronkov [16], which is also by reduction of Hilbert's tenth problem. In this proof the key point is to explicitly represent the \history of a multiplication process". We note also that shifted pairing bears certain similarities to the technique that is used to prove that any recursively enumerable set of strings is 14

16 given by the intersection of two (deterministic) context free languages [28, Lemma 8.6]. Finite Tree Automata Finite tree automata, or simply tree automata from here on, are a generalization of classical automata. Tree automata were introduced, independently, in Doner [19] and Thatcher and Wright [39]. The main motivation was to obtain decidability results for the weak monadic second-order logic of the binary tree. Here we adopt the following denition of tree automata, based on rewrite rules [5, 6]. I A tree automaton or TA A is a quadruple (Q; ; R; F ) where { Q is a nite set of constants called states, { is a signature that is disjoint from Q, { R is a set of rules of the form f(q 1 ; : : : ; q n )! q, where f 2 has arity n 0 and q; q 1 ; : : : ; q n 2 Q, { F Q is the set of nal states. A is called a deterministic TA or DTA if there are no two dierent rules in R with the same left-hand side. Note that if A is deterministic then R is a reduced set of ground rewrite rules and thus canonical [37]. Tree automata as dened above are usually also called bottom-up tree automata. Acceptance for tree automata or recognizability is dened as follows. I The set of terms recognized by a TA A = (Q; ; R; F ) is the set T (A) = f 2 T j (9q 2 F )! R q g: A set of terms is called recognizable if it is recognized by some TA. 5.1 Main Idea We consider a xed Turing machine M = (Q M ; in ; tape ; ; q 0 ; b; fq acc g); and assume, without loss of generality, that the nal ID of M is simply q acc i.e., the tape is always empty when M enters the nal state, and that q 0 6= q acc. Let also v be a string over the input alphabet of M. We eectively construct a system S M v (x; y) of three rigid equations: S M v (x; y) = f S 0 (x; y); S 1 (x; y); S 2 (x; y) g 15

17 where S 0 (x; y) = E 0 `8 x y c 0 ; S 1 (x; y) = 1 `8 x y; S 2 (x; y) = 2 `8 x t v y where E 0, 1 and 2 are ground, c 0 is a constant, `' is the only nonconstant function symbol in the system and t v is a ground term that represents the initial ID of M with input string v. We prove that M accepts v i Sv M is solvable. This establishes the undecidability result because all the steps in the construction are eective. The main idea behind the rigid equations is roughly as follows. Assume that there is a substitution that solves the system. From being a solution of S 0 (x; y), it follows that { x represents a sequence (v 0 ; v 1 ; : : : ; v m ) of IDs of M, and v m is the nal ID of M, and { y represents a sequence ((w 0 ; w + 0 ); (w 1; w + 1 ); : : : ; (w n; w + n )) of moves of M, i.e., w i `M w + i for 0 i n. From being a solution of S 1 (x; y) it follows that n = m and v i = w i for 0 i m. And nally, from being a solution of S 2 (x; y) it follows that v 0 = v and v i = w + i 1 for 1 i m. The combination of the last two points is the so-called \shifted pairing" technique. This is illustrated by Figure 1. The outcome of this shifted pairing is that x is a valid computation of M with input v, and thus M accepts v. Conversely, if M accepts v then it is easy to construct a solution of the system. We now give a formal construction of the above idea. 5.2 Words and Trains Words are certain terms that we choose to represent strings with, and trains are certain terms that we choose to represent sequences of strings with. We use the letters v and w to stand for strings of constants. Let be a binary function symbol. We write it in inx notation and assume that it associates to the right. For example t 1 t 2 t 3 stands for the term (t 1 ; (t 2 ; t 3 )). 16

18 v 0 v 1 v n 2 v n 1 v n v 0 v 1 v 2 v n 1 v n hv 0 ; v 1 i hv 1 ; v 2 i hv n 2 ; v n 1 i hv n 1 ; v ni hv n; i Figure 1: (hv 0 ; v 1 i; hv 1 ; v 2 i; : : : ; hvn; i) is a \shifted pairing" of (v 0 ; v 1 ; : : : ; vn). I We say that a (ground) term t is a c-word if it has the form a 1 a 2 a n c for some n 0 where each a i and c is a constant. A word is a c-word for some constant c. We use the following convenient shorthand notation for words. Let t be the word a 1 a 2 a n c and v the string a 1 a 2 a n. We write v c for t and say that t represents v. I A term t is called a c-train if it has the form t 1 t 2 t n c for some n 0 where each t i is a word and c is a constant. If n = 0 then t is said to be empty. The t i 's are called the words of t. A train is a c-train for some constant c. By the pattern of a train (v 1 c 1 ) (v 2 c 2 ) (v n c n ) c we mean the string c 1 c 2 c n. Let V = fv i g i2i be a nite family of regular sets of strings over a nite set of constants, where I is a set of constants disjoint from. Let U be a regular set of strings over I and let c be a constant not in or I. I We let Tn(V; U; c) denote the set of all c-trains t such that the pattern of t is in U and, for i 2 I, each i-word of t represents a string in V i. Example 12 Consider the set Tn(fV a ; V b ; V c g; ab c; ). This is the set of all -trains t such that the rst word of t is an a-word representing a string in V a, the last word of t is a c-word representing a string in V c and the middle ones (if any) are b-words representing strings in V b. 2 We say that a set of trains has a regular pattern if it is equal to some set Tn(V; U; c) with V, U and c as above. The main result of this section is the following theorem. 17

19 Theorem 13 (Train Theorem) Any set of trains with a regular pattern is recognizable and a DTA that recognizes this set can be obtained eectively. As we shall see, the construction of the rigid equation S 0 follows easily from the Train Theorem and some basic properties of tree automata. We believe that this theorem is of independent interest. For example, several theorems that are used in a similar context in Plaisted [36, Theorems 8.2{8.11], can be stated as corollaries of Theorem 13. Before we prove the theorem we state the following simple lemma. This lemma follows from the wellknown fact that all regular sets of strings are recognizable (cf [23]), assuming an appropriate representation of strings. 3 For any string v, we write v r for v in reverse and for a set of strings V we let V r = f v r j v 2 V g. Lemma 14 Let V be a regular set of strings over a set of constants and c a constant not in. Then f v c j v 2 V g is recognizable and a DTA is obtained eectively from V. Proof. Let M = (Q; ; ; q 0 ; F ) be a DFA that accepts the reverse of V, or V r, (clearly M exists, cf [28, p 281]). For each a 2 let ~a be a new state. Let A be the DTA (Q A ; ; R A ; F A ) where Q A = Q [ f ~a j a 2 g; = [ f; cg; R A = f ~a q! p j (q; a) = p g [ f a! ~a j a 2 g [ fc! q 0 g; F A = F: We must prove that, for all t 2 T, t! RA q for some q 2 F, t = v c for some v 2 L(M) r : Let us consider the direction `(' rst. So assume that v = a n 1 a n 2 a 0 2 L(M) r ; i.e, a 0 a n 2 a n 1 2 L(M). So, there exist q 1 ; q 2 : : : ; q n 2 Q, such that q n 2 F and the following holds: (q 0 ; a 0 ) = q 1 ; : : : ; (q n 2 ; a n 2 ) = q n 1 ; (q n 1 ; a n 1 ) = q n : But then, by the denition of R A, we can construct the following reduction: v c = a n 1 a n 2 a 0 c! ~a n 1 ~a n 2 ~a 1 ~a 0 q 0! ~a n 1 ~a n 2 ~a 1 q 1! ~a n 1 q n 1! q n 2 F; 3 Traditionally a string a 1a 2 an is represented by a term an( a 2(a 1(q 0))), i.e., the symbols of the alphabet are treated as unary function symbols, and the term is written using the reverse notation q 0a 1a 2 an. 18

20 which shows that v c 2 T (A). The direction `)' follows also easily. First note that any term t in T that reduces to a nal state q with respect to R A must be a c-word that represents some string v over. From the denition of R A follows then, like above, that v must be in V. We now prove the Train Theorem. Proof. Let V,, U, I, and c be like above. For each i 2 I, let i = [f; ig and let A i = (Q i ; i ; R i ; F i ) be a DTA given by Lemma 14 such that T (A i ) = f v i j v 2 V i g: Let c = I[f; cg and let A c = (Q c ; c ; R c ; F c ) be a DTA given by Lemma 14 such that T (A c ) = f u c j u 2 U g: Assume, without loss of generality, that all the DTAs have mutually disjoint sets of states, except for the states ~a for a 2 that are the same in all the A i 's for i 2 I. In fact, one can think of any constant a 2 and the corresponding state ~a as being the same element. Let now R 0 be the set of rules obtained from R c by relpacing, for all i 2 I, each rule ~{ p 1! p 2 in it with the set of rules f q p 1! p 2 j q 2 F i g, and discarding the rule i! ~{. Let now R be the following set of rules: R = [ i2i R i [ R 0 : Note that R is a reduced set of rewrite rules due to the disjointness assumptions and the assumption that the states ~a for a 2 are the same in all the DTAs. We are now ready to dene A as the DTA (Q; ; R; F c ) where = [ I [ f; cg and Q = [ i2i Q i [ (Q c n f~{ j i 2 I g): We can now prove that T (A) = Tn(V; U; c): Let use consider the direction `' rst. Assume that t 2 T (A), i.e., t reduces to some state q in F c via the rules in R. This reduction is only possible if it has (in principle) the following form: 4 t! R q 1 q 2 q n c! R 0 q: 4 A formal argument can be given by using induction and proving some lemmas rst [40, Chapter 3]. 19

21 where each q k is in F ik for some i k 2 I. Furthermore, by denition of R 0 and A c, we know that i 1 i 2 i n 2 U. The rst part of the reduction is possible only if t = t 1 t 2 t n c; where each t k reduces to q k. Note that, due to the disjointness properties of the DTAs, only the rules in R ik can be used in the reduction t k! q k, and thus t k 2 T (A ik ). Hence each t k has the form v i k for some v 2 V ik, and the pattern of t is i 1 i 2 i n, which we know is in U. This proves that t 2 Tn(V; U; c). Let us now consider the direction `'. So assume that t = t 1 t 2 t n c where each t k is in T (A ik ) for some i k 2 I and i 1 i 2 i n 2 U. It follows that each t k reduces with R ik to some q k 2 F ik and thus t reduces to q 1 q 2 q n c. By denition of R 0, q 1 q 2 q n c reduces to some q 2 F c. It follows that t! R q for some q 2 F c and thus t 2 T (A). The following example illustrates the construction that is used in the proof of the Train Theorem. Example 15 Let = f0; 1g, I = fa; bg and let be a new constant. Let V = fv i g i2i where V a = 0 1 and V b = Let U = bab a. We construct a DTA that recognizes the set Tn(V; U; ). Consider the following transition diagrams of a DFA for V r a : 1 q 1 q 2 0 and of a DFA for V b r : 0 1 q 3 q 4 0 By following the construction in Lemma 14 we get that the rules of A a and A b are as follows: R a = f1! ~ 1; 0! ~ 0; a! q1 ; ~ 1 q1! q 2 ; ~ 0 q2! q 2 g; R b = f1! ~ 1; 0! ~ 0; b! q3 ; ~ 0 q3! q 3 ; ~ 1 q3! q 4 ; ~ 0 q4! q 4 g: For the set U r we can consider a DFA with the following transition diagram: a b a p 1 p 2 p 3 p 4 b 20

22 From this we can extract the DTA A with the following set of rules: R = fa! ~a; b! ~ b;! p1 ; ~a p 1! p 2 ; ~ b p2! p 2 ; ~a p 2! p 3 ; ~ b p3! p 4 g: Now, following the construction in the Train Theorem, we get that the DTA A has the following set of rules. First, a set R 0 is constructed by removing the rst two rules in R and replacing ~a and ~ b with q2 and q 4, respectively. Second, R is taken as the union of R a, R b and R 0. So R is the following set of rules: R = f1! ~ 1; 0! ~ 0; a! q1 ; ~ 1 q1! q 2 ; ~ 0 q2! q 2 g [ fb! q 3 ; ~ 0 q3! q 3 ; ~ 1 q3! q 4 ; ~ 0 q4! q 4 g [ f! p 1 ; q 2 p 1! p 2 ; q 4 p 2! p 2 ; q 2 p 2! p 3 ; q 4 p 3! p 4 g: Let us consider a reduction in R. Let us write a -train t 1 t 2 t n as [t 1 ; t 2 ; : : : ; t n ]. Take for example t = [010 b; 001 a; 1 b; 01 a]: The pattern of t is baba which is in U. Let us see how t reduces to p 4. t! R [~ 01 ~ 0 ~ q3 ; 0 ~ 0 ~ 1 ~ q1 ; 1 ~ q3 ; 0 ~ 1 ~ q1 ]! R [~ 01 ~ q3 ; 0 ~ 0 ~ q2 ; q 4 ; 0 ~ q2 ]! R [~ 0 q4 ; 0 ~ q2 ; q 4 ; q 2 ]! R [q 4 ; q 2 ; q 4 ; q 2 ]! R 0 q 4 q 2 q 4 q 2 p 1! R 0 q 4 : 5.3 Representing IDs and Moves We show how to construct the rigid equation S 0 (x; y). Our main tool in doing so is the Train Theorem. We use also the following simple observation, that relates rigid E-unication with recognizability. Let us, for simplicity, consider a set of rules also as a set of equations. Lemma 16 Let A = (Q; ; R; fqg) be a DTA. Then, for all with range T, solves R `8 x q i x 2 T (A). Proof. Since R is a canonical rewrite system and q is irreducible in R, we have (for all ground ) that R j= x q i x! R q. But for with range T, by denition of recognizability, x 2 T (A) i x! R q. 21 2

23 Let us assign arity 0 to all the tape symbols ( tape ) and all the states (Q M ) of M. Let be the following signature: = tape [ Q M [ fe 0 ; e 1 ; ; g; where e 0, e 1 and are new constants Representing ID Sequences Recall that an ID of M is any string in tape Q M tape that does not end with a blank (b). We represent IDs by e-words, where e is one of e 0 or e 1. In particular, the nal ID is represented by the word q acc e 1 and IDs in general are represented by corresponding e 0 -words. I Any train of the form (v 0 e 0 ) (v 1 e 0 ) (v 2 e 0 ) (v n e 0 ) (q acc e 1 ) ; where n 0 and each v i is an ID of M, is called an ID-train. It is clear that the set of all IDs and the set consisting of just the nal ID are regular sets. The set of patterns of the ID-trains is given by the regular expression e 0 e 0 e 1. By using the Train Theorem, let A id = (Q id ; ; R id ; F id ) be a DTA that recognizes the set of all ID-trains Representing Move Sequences Let c ab be a new constant for each pair of constants a and b in the set tape [ Q M. Let also e 2 and 0 be new constants. Let now be the following signature: = f c ab j a; b 2 tape [ Q M g [ fe 2 ; 0 ; g Note that is the only symbol that occurs in both and. For and ID w of M we let w + denote the successor of w with respect to the transition function of M. For technical reasons it is convenient to let q + acc =, i.e., the successor of the nal ID is the empty string. The pair (w; w + ) is called a move. Let w = a 1 a 2 a m and w + = b 1 b 2 b n for some m 1 and n 0. Note that n 2 fm 1; m; m + 1g. Let k = max(m; n). If m < n let a k = b and if n < m let b k = b, i.e., pad the shorter of the two strings with a blank at the end. I We write hw; w + i for the string c a1 b 1 c a2 b 2 c ak b k and say that the e 2 -word hw; w + i e 2 represents the move (w; w + ). By a move-train we mean any 0 -train t = t 0 t 1 t n 0 ; such that each t i represents a move and n 1. 22

24 Example 17 Take in = f0; 1g, and let q; p 2 Q M. Assume that the transition function of M is such that, when the tape head points to a blank and the state is q then 1 is written to the tape, the tape head moves left-and M enters state p, i.e., (q; b) = (p; 1; L). Imagine that the current ID is 00q, i.e., the tape contains the string 00 and the tape head points to the bank following the last 0. So (00q; 0p01) is a move. This move is represented by the word c 00 c 0p c q0 c b1 e 2 = h00q; 0p01i e 2. 2 It is straightforward to see that the set of all strings hw; w + i where w is an ID, is a regular set. The patterns of all move-trains are given by the regular expression e 2 e 2 e 2. By using the Train Theorem let A mv = (Q mv ; ; R mv ; F mv ) be a DTA that recognizes the set of all move-trains. Assume also that the states of A mv are new constants Construction of S 0 We are now ready to construct S 0. First, let A 0 = (Q 0 ; 0 ; R 0 ; F 0 ) be the following DTA. Q 0 = Q id [ Q mv [ fc 0 g; 0 = [ ; R 0 = R id [ R mv [ f q 1 q 2! c 0 j q 1 2 F id ; q 2 2 F mv g; F 0 = fc 0 g: By the disjointness conditions between A id and A mv it follows that A 0 is indeed a deterministic tree automaton. It follows by elementary properties of tree automata that T (A 0 ) = f t s j t 2 T (A id ); s 2 T (A mv ) g: Let now E 0 = R 0 in the rigid equation S Final Construction In this section we nish the construction of Sv M and prove the undecidability results. The only essential components that we have not dened yet are 1 and 2. We let 1 and 2 be the following rewrite systems. The dierences between 1 and 2 are indicated with frames. 1 = f c ab! a j a; b 2 tape [ Q M g [ f e 1! e 0 ; e 2! e 0 ; 0! ; b e 0! e 0 g 2 = f c ab! b j a; b 2 tape [ Q M g [ f e 1! e 0 ; e 2! e 0 ; 0! ; b e 0! e 0 ; e 0! g 23

25 It is easy to see that both sets are in fact reduced sets of ground rewrite rules and thus canonical. For any input string v for M let the term t v in the system Sv M be the word q 0 v e 0, i.e., t v represents the initial ID of M with input v. We can now state the main theorem of this section. Theorem 18 S M v (x; y) is solvable i M accepts v. Before proving the theorem we state and prove some useful lemmas. Lemma 19 If solves S 1 (x; y) and S 2 (x; y) then x; y 2 T [. Proof. We prove by induction on the size of x that if solves the following system, where t 0 is any term in T [, then x; y 2 T [. f 1 `8 x y; 2 `8 x t 0 y g The statement follows then by choosing t 0 = q 0 v e 0. So consider a xed t 0 and assume that solves the above system. If x is a constant then so is its normal form in 2, say x# 2 = c, and so t 0 y! 2 c. But then c 2 and consequently x; y 2 T [. The cases when x is not a constant, but either x# 1 or x# 2 is a constant, are also immediate. So assume that x = t 1 t and (t 1 t)# = i t 1# i t# i for i 2 f1; 2g. So t 1 # 2 = t 0 # 2 and thus t 1 2 T [ since t 0 2 T [ ; also 2 j= t y: It follows from 1 j= t 1 t y that y = s 1 s for some terms s 1 and s such that 1 j= t s and 1 j= s 1 t 1. From the latter follows that s 1 2 T [ because t 1 2 T [. Let now 0 be such that x 0 = t and y 0 = s. So 0 solves the system f 1 `8 x y; 2 `8 x s 1 y g; and it follows by the induction hypothesis that t and s are in T [, and consequently, so are t 1 t = x and s 1 s = y. Lemma 20 If solves S M v (x; y) then x is an ID-train and y is a movetrain. Proof. Assume that solves Sv M (x; y). By Lemma 19, the range of is T [. But then, by denition of S 0 (x; y) and Lemma 16, x y 2 T (A 0 ), and thus x 2 T (A id ) and y 2 T (A mv ). 24

26 We can now prove Theorem 18. Proof. We prove that S M v (x; y) is solvable, M accepts v. Proof of `)' Let be a substitution that solves S M v (x; y). By using Lemma 20 we get that x and y have the following form: x = (v 0 e 0 ) (v 1 e 0 ) (v m 1 e 0 ) (v m e 1 ) y = (hw 0 ; w + 0 i e 2) (hw 1 ; w + 1 i e 2) (hw n ; w n + i e 2 ) 0 where m 1, n 1 and all the v i 's and w i 's are IDs of M and v m = q acc. Since solves S 1 (x; y), it follows that the normal forms of x and y under 1 must coincide. But x# 1 = (v 0 e 0 ) (v 1 e 0 ) (v m 1 e 0 ) (v m e 0 ) ; y# 1 = (w 0 e 0 ) (w 1 e 0 ) (w n 1 e 0 ) (w n e 0 ) : Note that each term hw i ; w + i i e 2 reduces rst to w 0 i e 0 where w 0 i = w i or w 0 i = w ib. The extra blank at the end is removed with the rule b e 0! e 0. So n = m; v n = q acc ; v i = w i (0 i n): (6) Since solves S 2 (x; y) it follows that the normal forms of x and (q 0 v e 0 )y under 2 must coincide. But x# 2 = x# 1 because x does not contain any constants from and the rule e 0! is not applicable. Moreover, since w n = q acc, it follows that w + n = and thus hw n ; w + n i e 0 = c qaccb e 0. But (c qaccb e 0 )! 2 (b e 0 )! 2 e 0! 2 : The normal form of (q 0 v e 0 ) y under 2 is thus (q 0 v e 0 ) (w + 0 e 0) (w + 1 e 0) (w + n 1 e 0) : It follows that v 0 = q 0 v, i.e., v 0 is the initial ID of M with input v, and w + i = v i+1 (0 i < n): (7) From (6) and (7) follows now that (v 0 ; v 1 ; : : : ; v n ) is a valid computation of M, and thus M accepts v. Proof of `(' Assume that M accepts v. So there exists a valid computation (v 0 ; v 1 ; : : : ; v n ) of M where v 0 = q 0 v, v n = q acc and v i + = v i+1 for 0 i < n. Let be such that x is the corresponding ID-train and y the corresponding move-train. It follows easily that solves S M (x; y). 25

27 The shifted pairing technique that is used in Theorem 18 is illustrated in Figure 1. The Degtyarev{Voronkov theorem is an immediate consequence of Theorem 18, because all the constructions in it are eective. Furthermore, the following result due to Plaisted [36] (that we used to prove Corollary 10). is an immediate consequence. Corollary 21 (Plaisted) SREU is undecidable even if the left-hand sides are ground. Furthermore, we can sharpen this result as follows. Corollary 22 SREU is undecidable if the left-hand sides are ground, there are only two variables and three rigid equations and one binary function symbol. The undecidability with two variables and three rigid equations may seem like an artical extra condition, but in fact, it turns out to be an important special case. One implication is that the provability problem for the 99-fragment of intuitionistic logic with equality is undecidable. Another important fact is that two variables are necessary to get undecidability. If there is only one variable then SREU is decidable [9]. Remark We can also note that one constant suces. One can easily simulate any number of constants with one constant and a binary function symbol. 5.5 Undecidability Proofs of SREU The rst proof of the udecidability of SREU [14] was by reduction of the monadic semi-unication [1] to SREU. This proof was followed by two alternative (more transparent) proofs by the same authors, rst by reducing second order unication to SREU [13, 17], and then by reducing Hilbert's tenth problem to SREU [16]. The undecidability of second order unication was proved by Goldfarb [24]. Reduction of second order unication to SREU is very simple, showing how close these problem are to each other. Plaisted took the Post's Correspondence Problem and reduced it to SREU [36]. From his proof follows that SREU is undecidable already with ground left-hand sides. Veanes improves the construction of Plaisted by using the membership problem for Turing machines and shows that two variables and one binary function symbol is enough to obtain undecidability [40, 41]. 6 Minimal Undecidable Case of the n-skeleton Problem Let ' M v (x; y) stand for the following formula: ' M v (x; y) = (E 0 ) x y c 0 ) ^ 26