# Recursive Path Orderings can also be Incremental

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1 Recursive Path Orderings can also be Incremental Mirtha-Lina Fernández 1, Guillem Godoy 2, and Albert Rubio 2 1 Universidad de Oriente, Santiago de Cuba, Cuba 2 Universitat Politècnica de Catalunya, Barcelona, España Abstract. In this paper the Recursive Path Ordering is adapted for proving termination of rewriting incrementally. The new ordering, called Recursive Path Ordering with Modules, has as ingredients not only a precedence but also an underlying ordering B. It can be used for incremental (innermost) termination proofs of hierarchical unions by defining B as an extension of the termination proof obtained for the base system. Furthermore, there are practical situations in which such proofs can be done modularly. 1 Introduction Term rewriting provides a simple (but Turing-complete) model for symbolic computation. A term rewrite system (TRS) is just a binary relation over the set of terms of a given signature. The pairs of the relation are used for computing by replacements until an irreducible term is eventually reached. Hence, the absence of infinite sequences of replacements, called termination, is a fundamental (though undecidable) property for most applications of rewriting in program verification and automated reasoning. For program verification, the termination of a particular rewriting strategy called innermost termination has special interest. In this strategy the replacements are performed inside-out, i.e. arguments are fully reduced before reducing the function. Therefore, it corresponds to the call by value computation rule of programming languages. This strategy is also important because for certain classes of TRSs, innermost termination and termination coincide [12]. Term rewrite systems are usually defined in hierarchies. This hierarchical structure is very important when reasoning about TRS properties in an incremental manner. Roughly, a property P is proved incrementally for a hierarchical TRS R = R 0 R 1 if we can prove it by using information from the proof of P for the base system R 0. The simplest form of incrementality is modularity, i.e. proving P for R just by proving P for R 0 and R 1 independently. However, termination is not a modular property even for disjoint unions of TRSs [21]. A stronger form of termination, called C E -termination, and innermost termination, are indeed modular for a restricted class of hierarchical unions [17, 14], but not

2 in general. Therefore, it is of great importance to tackle (innermost) termination of hierarchical systems using an incremental approach. Regardless the previous facts, the problem of ensuring termination of a hierarchical union without finding (if possible) an alternate proof for the base system has received quite few attention. The first and important step was done by Urbain in [22]. He showed that from the knowledge that a base system is C E -terminating, the conditions for the termination proof of a hierarchical union can be relaxed. In the context of the Dependency Pair method (DP) [1] (the most successful for termination of rewriting) this entails a significant reduction in the number and the strictness of the DP-constraints. Very recently, Urbain s contribution was used for improving the application of the Size-Change Principle (SCP) [15] to C E -termination of rewriting [10]. In the latter paper it was shown that a termination measure for a base system R 0 can be used for proving size-change termination of a hierarchical extension R 1, and this guarantees R 0 R 1 is C E -terminating. Using this result, the next TRS is easily (and even modularly) proved simply terminating. Example 1. The following hierarchical union (R plus is taken from [19]) can be used for computing Sudan s function 3. plus(s(s(x)), y) s(plus(x, s(y))) plus(x, s(s(y))) s(plus(s(x), y)) R plus = plus(s(0), y) s(y) plus(0, y) y F (0, x, y) plus(x, y) R F = F (s(n), x, 0) x F (s(n), x, s(y)) F (n, F (s(n), x, y), s(plus(f (s(n), x, y), y))) In order to prove termination of R = R plus R F (when using the DPapproach) the whole union must be included in some (quasi-) ordering. But R plus requires semantical comparisons while R F needs lexicographic ones. Therefore, no (quasi-) ordering traditionally used for automated proofs serves for this purpose. However, simple termination of R plus is easy to prove e.g. using the Knuth- Bendix Ordering (KBO) [3]. Besides, every size-change graph of R F decreases w.r.t. the lexicographic extension of KBO. Thus, R F is size-change terminating w.r.t. KBO and we conclude R is simply terminating. SCP provides a more general comparison than lexicographic and multiset ones. But it has as main drawback that it cannot compare defined function symbols (i.e. those appearing as root of left-hand sides) syntactically. 3 Chronologically, Sudan s function [7] is the first example of a recursive but not primitive recursive function. Sudan s function F (p, m, n) is greater than or equal to Ackermann s function A(p, m, n) except at the single point (2, 0, 0). The latter was used in [19] combined with R plus.

3 Example 2. Let R F = R F {F (s(n), F (s(n), x, y), z) F (s(n), x, F (n, y, z))}. The new rule [6][Lemma 6.7, page 47] can be used for computing an upper bound of the left-hand side while decreasing the size of the term. But now SCP fails in proving termination of R plus R F. This is due to the new rule which demands a lexicographic comparison determined by a subterm rooted with the defined symbol F. When dealing with defined symbols, SCP cannot compete with classical syntactical orderings like the Recursive Path Ordering [8]. Therefore, it would be nice to adapt RPO in order to prove termination of R plus R F and other hierarchical systems incrementally. In this paper we present a new RPO-like ordering which can be used for these purposes, called the Recursive Path Ordering with Modules (RPOM). It has as ingredients not only a precedence, but also an underlying ordering B. Actually RPOM defines a class of orderings that can be partitioned into three subclasses, RPOM-STAB, RPOM-MON and RPOM-IP-MON, where, under certain conditions, the first one contains stable orderings, the second one contains monotonic orderings (or a weak form of monotonocity related to B ), and the third one contains IP-monotonic orderings. We use these orderings for proving C E -termination and innermost termination of a hierarchical union R = R 0 R 1 incrementally as follows. The system R 0 is known terminating, and perhaps an ordering B including the relation R0 on terms of T (F 0, X ) is given. An ordering B is then constructed, perhaps as an extension of B to T (F, X ), or perhaps independently of the possible B. Three orderings from RPOM-STAB, RPOM-MON and RPOM-IP-MON are then obtained from B, satisfying that the one in RPOM-STAB is included into the one in RPOM-MON under some conditions on B and R 0, and into the one in RPOM-IP-MON under weaker conditions. Including R 1 in RPOM- STAB will then allow to prove C E -termination or innermost termination of R depending on the original properties of B and R 0. Note that, in the case of innermost termination, no condition at all is imposed on B and R 0. Our results are a first step towards the definition of a general framework for combining and extending different termination proof methods (this idea of combining ordering methods was early considered in [18]), and thus obtain termination proofs of hierarchical unions of TRS s whose modules have been proved using different techniques. As a first step, since based on RPO, these results are still weak to compete with the recent refinements of the DP method in [16, 20]. However, we believe that the extension of these results to more powerful path orderings, like the Monotonic Semantic Path Ordering in [5], will provide fairer comparison. The remainder of the paper is organized as follows. In Section 2 we review basic notation, terminology and results. In Section 3 we define RPOM and prove its properties, and the ones corresponding to every subclass RPOM-STAB, RPOM- MON and RPOM-IP-MON. In Section 4 (resp. Section 5) we show how to use RPOM for proving C E -termination (resp. innermost termination) incrementally. We present some concluding remarks in Section 6.

4 2 Preliminaries We assume familiarity with the basics of term rewriting (see e.g. [2]). The set of terms over a signature F is denoted as T (F, X ), where X represents a set of variables. The symbol labelling the root of a term t is denoted as root(t). The root position is denoted by λ. The set of positions of t is denoted by Pos(t). The subterm of t at position p is denoted as t p and t t p denotes the proper subterm relation. A context, i.e. a term with a hole, is denoted as t[ ]. The term t with the hole replaced by s is denoted as t[s], and the term t[s] p obtained by replacing t p by s is defined in the standard way. For example, if t is f(a, g(b, h(c)), d), then t = c, and t[d] 2.2 = f(a, g(b, d), d). We denote t[s 1 ] p1 [s 2 ] p2... [s n ] pn by t[s 1, s 2,..., s n ] p1,p 2,...,p n. We write p 1 > p 2 (or, p 2 < p 1 ) if p 2 is a proper prefix of p 1. In this case we say that p 2 is above p 1, or that p 1 is below p 2. We will usually denote a term f(t 1,..., t n ) by the simplified form ft 1... t n. The notation t is ambiguously used to denote either the tuple (t 1,..., t n ) or the multiset {t 1,..., t n }, even in case of t = f(t 1,..., t n ). The number of symbols of t is denoted as t while t denotes the number of elements in t. Substitutions are denoted with the letter σ. A substitution application is written in postfix notation. We say that a binary relation on terms is variable preserving if s t implies that every variable in t occurs in s. It is said that is non-duplicating if s t implies that every variable in t occurs at most as often as in s. If s t implies sσ tσ then is stable. If for every function symbol f, s t implies f(... s...) f(... t...) then is monotonic. It is said that a relation is wellfounded if there is no infinite sequence s 1 s 2 s The transitive and the reflexive-transitive closure of are denoted as + and resp. The union of and the syntactical equality is denoted as. We say that is compatible with if and. A (strict partial) ordering on terms is an irreflexive transitive relation. A reduction ordering is a monotonic, stable and well-founded ordering. A simplification ordering is a reduction ordering including the strict subterm relation. A precedence F over F is the union of a well-founded ordering F and a compatible equivalence relation F. We say that a precedence F is compatible with a partition of F if f F g implies that f and g belongs to the same part of F. The multiset extension of an ordering on terms to multisets, denoted as mul, is defined as s mul t iff there exists ū s such that ū t and for all t t ū there is some s s ū s.t. s t. The lexicographic extension of to tuples, denoted as lex, is defined as s lex t iff s i t i for some 1 i s and s j t j for all 1 j < i. These extensions preserve irreflexivity, transitivity, stability and well-foundedness. If is defined on T (F 0, X ) and F 0 F then F = {(sσ, tσ) s t, x X, xσ T (F, X )} is called the stable extension of to F. The stable extension of a stable (and well-founded) ordering is also a stable (and well-founded) ordering [19].

5 A term rewrite system over F is denoted as R. Here, we deal with variable preserving TRSs. Regarding termination, this restriction is not a severe one. A rewriting step with R is written as s R t. The notation s λ,r t is used for a rewriting step at position λ. A TRS R is terminating iff + R is well-founded. It is said that R is simply terminating iff R Emb F is terminating where Emb F = (F, {f(x 1,..., x n ) x k f F, 1 k n}) and x 1,..., x n are pairwise distinct variables. It is said that R is C E -terminating iff R E = R C E is terminating, where C E = (G, {G(x, y) x, G(x, y) y}) and G = F {G}. Given a TRS R, f(t 1,..., t n ) is said to be argument normalized if for all 1 k n, t k is a normal form. A substitution σ is said to be normalized if xσ is a normal form for all x X. An innermost redex is an argument normalized redex. A term s rewrites innermost to t w.r.t. R, written s i t, iff s t at position p and s p is an innermost redex. A term s rewrites innermost in parallel to t w.r.t. R, written s i,r t, iff s + i,r t and either s i,r t at position λ (denoted as s i,λ,r ) or s = f( s), t = f( t) and for all 1 k s either s k i,r t k or s k = t k is a normal form (denoted as s i,r t). A binary relation is IP-monotonic w.r.t. R iff i,r [11]. A TRS R is innermost terminating iff + i,r is well-founded. Alternatively, we have the following characterization for innermost termination. Theorem 1. [11] A TRS R is innermost terminating iff there exists a wellfounded relation which is IP-monotonic w.r.t. R. The defined symbols of a TRS R are D = {root(l) l r R} and the constructors are C = F D. The union R 0 R 1 is said to be hierarchical if F 0 D 1 =. 3 RPOM In this section we define RPOM in terms of an underlying ordering B and show that it is an ordering. Moreover, we prove that well-foundedness of B implies well-foundedness of RPOM. Actually RPOM defines a class of orderings that depends on three parameters. These are the underlying ordering B, the (usual in RPO and other orderings) statusses of the symbols in the signature, and a last parameter mc {rmul, mul, set}. Due to mc, this class of orderings can be partitioned into three subclasses, RPOM-STAB, RPOM-MON and RPOM-IP-MON, where, under certain conditions, the first one contains stable orderings, the second one contains monotonic orderings (or a weak form of monotonocity related to B ), and the third one contains IP-monotonic orderings. At the end of this section we prove the corresponding properties to every subclass. Before going into the definition of RPOM, we need some additional notation. Apart from the multiset extension mul of an ordering defined in the preliminaries we need two other extensions of orderings to multisets: the set extension and the rmul extension.

6 Definition 1. Let be an arbitrary ordering. Given two multisets S and T, S set T if S mul T, where S and T are obtained from S and T, respectively, by removing repetitions. S rmul T if S and for all t T there is some s S such that s t. It is easy to see that the relation rmul is included into mul and set, and that it preserves irreflexivity, transitivity, stability and well-foundedness, whereas set preserves all these properties except for stability. We will use the notation set and mul for denoting the inclusion in the sense of sets and multisets, respectively, in the cases where alone is not clear by the context. For facility of notations, we identify rmul with set. The ordering RP OM is defined as the union of the underlying ordering B, and a RPO-like ordering. Hence, we need a definition of not in contradiction (or even more, compatible) with B. Since B will be generally obtained as an extension of an ordering B on the base signature B = F 0, it seems natural to demand this ordering to relate pairs of terms where at least one is rooted by a base symbol (i.e. a symbol in B), but as we see as follows, a more strict condition is needed for B. The definition of s t differs depending on if the roots of s and t are or not in B. If no root is in B, then we use a classical RPO-like recursive definition. If some root is in B, we eliminate all the context containing symbols of B, resulting in two multisets, and compare them with the corresponding extension rmul, mul or set. Definition 2. Given a signature B, we say that p is a frontier position and t p is a frontier term occurrence of t if root(t p ) / B and root(t p ) B, for all p < p. The multiset of all frontier subterm occurrences of t is denoted as frt B (t) 4. For example, if B = {f}, then frt B (f(g(a), f(g(f(g(a), g(b))), g(a))) is {g(a), g(f(g(a), g(b))), g(a)}. If we want frt B (s) rmul frt B (t) or frt B (s) mul frt B (t) or frt B (s) set frt B (t) to be not in contradiction with B, it is necessary to demand that s B t implies frt B (s) rmul frt B (t) or frt B (t) mul frt B (t) or frt B (s) set frt B (t), depending on the case. We call frontier preserving (w.r.t. rmul, mul or set) to this property. Definition 3. Let mc {mul, rmul, set}, B F and F be a precedence over F B compatible with the partition of F B, F Mul F Lex. Moreover, let B be an ordering on T (F, X ) s.t. s B t implies frt B (s) mc frt B (t). Then, the Recursive Path Ordering with Modules (RPOM) is defined as rpom = B where s = f( s) t iff one of the following conditions holds: 1. f, root(t) / B and s t for some s s. 2. t = g( t), f F g and s t, for all t t. 4 Note that these multisets include not only maximal subterms of t rooted by non-base function symbols, but also variables.

7 3. t = g( t), f F g, f F Mul and s mul t. rpom 4. t = g( t), f F g, f F Lex, s lex t rpom and s t, for all t t. 5. f B or root(t) B, s T (B, X ), and frt B (s) mc frt B (t). We define rpom stab, rpom mon and rpom IP mon to be rpom in the cases where mc is rmul, mul and set, respectively. Analogously, stab, mon and IP mon refer to. It is not difficult to show (using induction on the size of s and t) that RPOM is well-defined. In order to prove that RPOM is an ordering, first we show that is compatible with B, and then, it suffices to show that is transitive and irreflexive. Lemma 1. s t iff s T (B, X ) and frt B (s) mc frt B (t). Proof. The result holds by definition if root(s) B or root(t) B. Otherwise, root(s), root(t) / B and {s} = frt B (s) mc frt B (t) = {t} iff s t. Lemma 2. is compatible with B. Proof. It has to be shown that u B s t B v implies u v. By the frontier preserving condition of B and Lemma 1 we have frt B (u) mc frt B (s) mc frt B (t) mc frt B (v). This implies u T (B, X ) and frt B (u) mc frt B (v) by definition of mc. Therefore, using again Lemma 1, u v holds. Lemma 3. If root(s) B and s t rpom u then s u. Proof. Either t B u and hence frt B (t) mc frt B (u), or t u and hence, by Lemma 1, frt B (t) mc frt B (u). In both cases, for all u frt B (u), there exists t frt B (t) s.t. s t u holds, and we obtain s u by case 1. Thereby, frt B (s) = {s} mc frt B (u) and the required result follows by Lemma 1. Lemma 4. is transitive. More generally, s t( )u implies s u. Proof. Assuming that s 1 s 2 ( )s 3 we prove that s 1 s 3, and we do it by induction on the multiset { s 1, s 2, s 3 } and the multiset extension of the usual ordering on naturals. First, note that if s 2 s 3 for some s 2 frt B (s 2 ), then, s 2 s 3, and hence, by Lemma 1 frt B (s 2 ) mc frt B (s 2) mc frt B (s 3 ), which implies frt B (s 2 ) mc frt B (s 3 ), and s 2 s 3 by Lemma 1 again. Therefore, in general we have that either s 2 s 3 or frt B (s 2 ) mc frt B (s 3 ). Assume that some of the symbols root(s 1 ), root(s 2 ) or root(s 3 ) are in B. By Lemma 1 and previous observation, frt B (s 1 ) mc frt B (s 2 )( mc mc )frt B (s 3 ). By induction hypothesis, transitivity holds for smaller terms, and since the extension mc preserves transitivity and is compatible with mc, we can conclude that frt B (s 1 ) mc frt B (s 3 ). Again by Lemma 1, s 1 s 3. Hence, from now on we can assume that all root(s 1 ), root(s 2 ) or root(s 3 ) are not in B, and therefore case 5 of the definition of RPOM does not apply any more and, moreover, by our first observation, s 2 s 3.

8 If s 1 s 2 by case 1, then there exists a proper subterm s 1 of s 1 satisfying s 1 s 2. Either because s 1 s 2 or by induction hypothesis, s 1 s 3, and s 1 s 3 holds by case 1. Hence, from now on assume that s 1 s 2 is not due to case 1. At this point it is easy to show that s 1 s 2 for any proper subterm s 2 of s 2. Note that for such s 2 there is some s 2 in s 2 that contains s 2 as subterm. If s 1 s 2 is due to case 2 or 4, then s 1 s 2. Otherwise, if it is due to case 3, for some s 1 in s 1, s 1 rpom s 2, and by Lemma 3, we obtain s 1 s 2 again. In any case s 1 s 2, and either s 2 is s 2 and hence s 1 s 2 directly, or s 2 s 2 and by induction hypothesis on s 1 s 2 s 2 we obtain s 1 s 2 again. If s 2 s 3 by case 1, then there exists a proper subterm s 2 of s 2 satisfying s 2 s 3. By the previous observation, s 1 s 2, and by induction hypothesis, s 1 s 3. Hence, from now on we can assume that case 1 does not apply in s 1 s 2 s 3. Reasoning analogously as before, it is easy to show that s 2 s 3 for any proper subterm s 3 of s 3. Moreover, by induction hypothesis on s 1 s 2 s 3, we obtain s 1 s 3 for any of such s 3 s. Hence, if root(s 1 ) F root(s 3 ), then s 1 s 3 by case 2. On the other hand root(s 3 ) F root(s 1 ) can not happen since case 1 does not apply in s 1 s 2 and s 2 s 3. Therefore, from now on we can assume that root(s 1 ) F root(s 3 ). Again since case 1 does not apply, we have root(s 1 ) F root(s 2 ) F root(s 3 ). If such a root symbol is from F Mul (F Lex ) then, since the mul (lex) extension preserves transitivity, s 1 mul rpom s 3 ( s 1 lex rpom s 3 ): note that rpom is transitive on smaller subterms since, by induction hypothesis, is, and, moreover, it is compatible with B, which is transitive too. Hence (using that s 1 s 3 for any proper subterm s 3 of s 3 in the case where the root symbol is from F Lex ) we conclude that s 1 s 3. Lemma 5. is irreflexive. Proof. Obviously, s s for all s X. Hence, we proceed by contradiction, using induction on the size of s. Depending on the case s s holds we consider 3 cases. If s s holds by case 1 then, root(s) F B and for some s s, s s holds. But by Lemma 3, s s, and by transitivity s s s implies s s contradicting the induction hypothesis. The irreflexivity of F is contradicted if s s holds by case 2. Finally, s s holding by case 3, 4 or 5 implies either s mul rpom s, s lex rpom s or frt B (s) mc frt B (s). But B is irreflexive and, by the induction hypothesis, is irreflexive for the subterms of s. Hence, since the multiset and lexicographic extensions preserve irreflexivity we obtain s mul rpom s, s lex rpom s or frt B (s) mc frt B (s) which is a contradiction. Well-foundedness of RPO follows from the fact that it is a monotonic ordering which includes the subterm relation. This is not the case of when mc mul: for example, even if B = {f} and a F b, faab fabb. Therefore, we prove its well-foundedness directly by contradiction. Lemma 6. If B is well-founded then is well-founded.

9 Proof. Proceeding by contradiction, suppose there is an infinite sequence with. We choose a minimal one w.r.t. the size of the terms involved; that is, the infinite sequence S = s 1, s 2, s 3,... satisfies that for any other sequence t 1, t 2, t 3,... with different sequence of sizes, i.e. with s 1, s 2, s 3,... t 1, t 2, t 3,..., there exists an i > 0 such that t i > s i and t j = s j for all j < i. If there exists a step in S s.t. s i s i+1 holds by case 1, then the minimality of S is contradicted. Note that if so, by definition of and Lemma 3 we have s i s s i+1 for some s i s. Hence, by transitivity we obtain the sequence S = s 1, s 2,..., s i 1, s, s i+2,..., which is smaller than S. This also applies when s i s i+1 holds by case 5 and root(s i+1 ) / B. In this case s s i+1 holds for some s frt B (s i ) and when i > 1 by Lemma 4 we have s i 1 s. Therefore, there is at most one step in S s.t. root(s i ) / B and root(s i+1 ) B. Thus, any other step in S holding by case 5 involves terms which are both rooted by a base symbol. By the previous facts and since F is a precedence, we conclude that there is some i 1 satisfying that for all j > i, s j s j+1 holds by the same case 3, 4 or 5. In cases 3 and 4, by definition of the multiset and lexicographic extensions, from the infinite sequence s i+1, s i+2, s i+3,... with mul rpom or lex rpom we extract another infinite sequence t 1, t 2, t 3,... with rpom with t 1 s i+1. Since B is wellfounded and is compatible with B, from the latter sequence we construct another infinite sequence s i+1, s i+2, s i+3,... with and where s i+1 = t 1. In case 5, from the infinite sequence frt B (s i+1 ),frt B (s i+2 ),frt B (s i+3 ),... with mc we construct another infinite sequence s i+1, s i+2, s i+3,... with and where s i+1 frt B (s i+1 ). Thus, we have s i+1 s i+1 and s i s i+1 holds by Lemma 4. Therefore, we construct the infinite sequence s 1, s 2,..., s i,s i+1,s i+2,s i+3,... with which again contradicts the minimality of S. Corollary 1. rpom is an ordering. If B is well-founded then rpom is wellfounded. 3.1 A stable subclass of RPOM In this subsection we show that rpom stab preserves the stability of B. Proposition 1. If B is stable, then rpom stab is stable. Proof. We just need to show that s stab t implies sσ stab tσ for every substitution σ. We use induction on the size of s and t. If s stab t holds by a case different from 5 then sσ stab tσ is easily obtained by the same case using the induction hypothesis and the stability of, B and the multiset and lexicographic extensions. In the case where s stab t holds by case 5, note that s F(B, X ) implies sσ F(B, X ), and hence frt B (sσ) is not empty. Besides, every term in frt B (tσ) is either at a position p such that t p is in frt B (t) and root(t p ) B, or at a position of the form p.p such that t p is a variable x, and tσ p.p frt B (xσ). In the first case, there is a term s frt B (s) such that s t p and root(s ) B. Hence, s σ frt B (sσ) and by induction

10 hypothesis s σ stab t p σ. In the second case, there is a term s frt B (s) with root(s) B that has x as proper subterm, and hence s σ frt B (s) and s σ stab t p.p σ by case 1. Altogether shows that frt B (sσ) rmul stab frt B(tσ), and hence sσ stab tσ holds by case 5. The orderings rpom mon and rpom IP mon are not stable. This is due to the terms rooted by base function symbols which are compared by using the frontier subterms and the multiset extension. Note that, after applying a substitution, some frontier positions (corresponding to variables) may disappear and thus a strict superset relation (which is included in the multiset extension) may become equality. For example, for B = {g, a}, we have s = g(a, h(x), y) mon g(a, a, h(x)) = t but sσ mon tσ if yσ is a ground term of T (B, X ). The same and more complex situations hold for IP mon. 3.2 A monotonic subclass of RPOM In this subsection we show that mon is monotonic, and rpom mon preserves the monotonicity of B. Moreover, even if B is not monotonic, there is a monotonic relation between mon and B that we define as follows. Definition 4. A relation on terms is monotonic on an other relation w.r.t. a set of symbols F 1 if for all f F 1, s t implies f(..., s,...) f(..., t,...). Proposition 2. mon is monotonic, and B is monotonic on mon w.r.t. F B. Proof. Let u = f(..., s,...), v = f(..., t,...). If f B and s mon t, then frt B (s) mul mon frt B (t) by Lemma 1, and hence frt B (u) mul mon frt B (v), which implies u mon v by case 5. If f / B and either s mon t or s B t, then s rpom mon t. Hence ū mul rpom mon v and ū lex rpom mon v. If f F Mul then u mon v holds by case 3. If f F Lex then u mon v holds by case 4 because by Lemma 3, u mon v holds for all v v. Corollary 2. If B is monotonic then rpom mon is monotonic. 3.3 An IP-monotonic subclass of RPOM In this subsection we show that, for a given hierarchical TRS R = R 0 R 1 and under certain conditions, IP-monotonicity of B w.r.t. R 0 (on terms of T (F 0, X )) implies IP-monotonicity of rpom IP mon w.r.t. R (on terms of T (F, X )). Since B will usually be an extension from an ordering orienting R 0, it is not expectable to be IP-monotonic on terms on the extended signature F = F 0 F 1. Even more, including i,r0 applied to terms on T (F, X ) into B is not possible because then the condition stating that s B t implies frt B (s) set frt B (t) is violated for terms rooted by f / B. Instead of including the whole relation i,r0 in B we demand a weaker condition based on the following definition.

11 Definition 5. Let s and t be terms in T (F, X ). Then we write s i,r0,f 0 t if s i,r0 t and all innermost redexes in s are at positions p such that for all p p, root(s p ) F 0. Proposition 3. Let R = R 0 R 1 be a hierarchical TRS, B = F 0 and B be an ordering on T (F, X ) s.t. s B t implies frt B (s) set frt B (t), and i,r0,b B. Let i,λ,r1 IP mon. Then rpom IP mon is IP-monotonic w.r.t. R. For proving the previous lemma we need the following basic facts concerning the set extension of any ordering. Proposition 4. Let be any ordering. S set T, S set T and S S = imply S S set T T. {s 1 } T 1,..., {s n } T n implies {s 1,..., s n } set T 1... T n. Proof. (of Proposition 3) To prove that s i,r t implies s rpom IP mon t, we prove, by induction on term structure, a more general statement: s i,r t implies s rpom IP mon t and if root(s) / B then s IP mon t. We distinguish two cases depending on whether or not root(s) is in B. Assume that root(s) / B. If s i,λ,r1 t then trivially s IP mon t by the assumptions of the lemma. Otherwise, s and t are of the form f(s 1... s m ) and f(t 1... t m ), respectively, every s j is either an R-normal form or s j i,r t j, and for some j {1... m}, s j i,r t j. By induction hypothesis, every s j is either a normal form or s j rpom IP mon t j, and for some j {1... m}, s j rpom IP mon t j. If f F Mul, s IP mon t by case 3. If f F Lex, then s IP mon t holds by case 4 because by Lemma 3 we have s IP mon t j for all j {1... m}. Assume now that root(s) B. We consider the set containing only the minimal positions from {p s p frt B (s) or s p is an innermost redex}, i.e. the ones in this set such that no other is above them. This set is of the form {p 1,..., p n, p 1,..., p m} where the p j s are frontier positions without innermost redexes above them, and the p j s are redex positions satisfying root(s p ) B for every p p j. Hence, s and t can be written as s[s 1,..., s n, s 1,..., s m] p1,...,p n,p 1,...,p m and s[t 1,..., t n, t 1,..., t m] p1,...,p n,p, respectively, where every s 1,...,p m j satisfies root(s j ) / B and either s j i,r t j or s j is a normal form and s j = t j, and every s j satisfies s j i,λ,r 0 t j. Moreover, frt B(s) = {s 1,..., s n } frt B (s 1)... frt B (s m), and frt B (t) = frt B (t 1 )... frt B (t n ) frt B (t 1)... frt B (t m). If all the s j s are normal forms, then s i,r0,bt, and by our assumptions s B t, and hence s rpom IP mon t holds. Hence, assume that for some j {1... n}, s j i,r t j. By induction hypothesis s j IP mon t j, and by Lemma 1, {s j } set IP mon frt B(t j ). Similarly, for the rest of j {1,..., n} we have that either {s j } set IP mon frt B(t j ) or s j = t j, depending on whether or not s j is a normal form. Since every s j satisfies s j i,λ,r 0 t j, frt B(s j ) set frt B (t j ), and hence frt B (s j ) set IP mon frt B(t j ). By propositon 4, {s 1,..., s n } frt B (s 1)... frt B (s m) set IP mon frt B(t 1 )... frt B (t n ) frt B (t 1)... frt B (t m), and hence s IP mon t by case 5, and s rpom IP mon t.

12 4 Proving C E -termination incrementally with RPOM Assume we have a hierarchical system R = R 0 R 1, and we want to prove it terminating using RPOM. We will usually have a reduction ordering B defined on T (F 0, X ) orienting R 0, or more generally, a well founded ordering B including R0 on T (F 0, X ), and we will want to obtain from it an ordering rpom stab orienting R 1. A first simple idea is to extend B to some B for terms on T (F, X ) including R0 on T (F, X ). But this will not be useful with RPOM since rewriting with R 0 on a term rooted by a symbol f / B does not preserve the frontier. An alternative idea is then to restrict the extension of B to rewriting steps with R 0 not below a symbol f / B. Then, we can take B = F B to this end, which is not monotonic, but preserves well foundedness of B (recall that F B is the stable extension of B to F). Definition 6. Let s and t be terms in T (F, X ). Then we write s R0,F 0 t if s R0 t and the involved redex is at a position p such that for all p p, root(s p ) F 0. The following theorem combines the use of stab and mon constructed from B. The monotonicity of the second requires B to be frontier preserving in the sense of multisets. Therefore, when B is defined as F B, we also need B to be non-duplicating. Theorem 2. Let R = R 0 R 1 be a hierarchical union, B = F 0 and B be a stable, well-founded on T (F, X ), such that s B t implies frt B (s) mul frt B (t), and R0,B B. If R 1 stab then R is C E -terminating. Proof. Recall that R E = R C E. We prove that RE is included in rpom mon because then the well-foundedness of rpom mon implies termination of R E. First note that R 1 C E stab, and since stab is stable, stab mon and mon is monotonic we conclude that R1 C E is included in mon. Now, let s, t T (F, X ) be s.t. s R0 t at position p. If every position above p is rooted by a symbol in F 0 then we have s B t by the assumptions of the theorem. It remains to see the case where there exist a context u[ ], a symbol f / F 0 and a position q < p s.t. s = u[f(..., s,...)] q, t = u[f(..., t,...)] q and s R0 t with every position between q and p rooted by a symbol in F 0. In this case s B t holds by the assumptions of the theorem, and s mon t is obtained by Proposition 2. Corollary 3. A hierarchical TRS R = R 0 R 1 is C E -terminating if there is a non-duplicating reduction ordering B s.t. R 0 B and R 1 stab. In addition, if B is a simplification ordering then R is simply terminating. Example 3. Simple termination of R plus R F in Example 2 is easily obtained using RPOM. Since R plus is simply terminating, B can be defined as the nonduplicating part of any simplification ordering including R plus. The rules of the

13 extension R F (listed below) are oriented using stab with F Lex = {F }. F (0, x, y) plus(x, y) F (s(n), x, 0) x F (s(n), x, s(y)) F (n, F (s(n), x, y), s(plus(f (s(n), x, y), y))) F (s(n), F (s(n), x, y), z) F (s(n), x, F (n, y, z)) Note that the first rule, here denoted as l 1 r 1, holds by case 5 of the definition of RPOM. This is because l 1 stab x and l 1 stab y hold by case 1 and therefore we have frt(l 1 ) = {l 1 } rmul stab {x, y} = frt(r 1 ). The second rule trivially holds by case 1. The last two hold by case 4. We detail the proof for the third one, denoted as l 3 r 3. First note that s(t) B t for every term t. Thereby, we have l 3 lex B r 3. By the former fact and using case 1 we obtain l 3 stab F (s(n), x, y) by case 4. Finally, l 3 stab s(plus(f (s(n), x, y), y)) holds by case 5 since frt(l 3 ) = {l 3 } rmul stab {F (s(n), x, y), y} = frt(s(plus(f (s(n), x, y), y))). Analogously to the case of SCP, there are situations where the proofs with RPOM can be done modularly. Theorem 3. Let R = R 0 R 1 be a hierarchical TRS where R 0 is nonduplicating and terminating. Let B be F 0 where 0 is on T (F 0, X ), and let R 1 stab. Then R is C E -terminating. Proof. The result can be obtained by Theorem 2 if we use B = ( R 0,F 0 B ) + and the corresponding stab instead of B and stab. Trivially R 1 stab and R0,F 0 B. We just need to show that B is frontier preserving, stable and well-founded. The first two properties follow from the fact that R0,F 0 and 0 are non-duplicating and stable, and the stable extension preserves these properties. Well-foundedness of B follows from the fact that R 0 is terminating, and that any derivation with ( R0,F 0 B ) can be commuted to a derivation with ( R0,F 0 ) followed by a derivation with F 0, preserving the number of rewrite steps. Example 4. Actually, R F in Example 2 is included in stab with B defined as F 0. Hence, by Theorem 3, the hierarchical union of R F and any non-duplicating base system R plus is C E -terminating whenever R plus is so. Example 5. Consider the following system which describes some properties of the conditional operator. if(0, y, z) z if(s(x), y, z) y if(x, y, y) y R if = if(if(x, y, z), x 1, x 2 ) if(x, if(y, x 1, x 2 ), if(z, x 1, x 2 )) if(x, if(x, y, x 1 ), z) if(x, y, z) if(x, y, if(x, x 1, z)) if(x, y, z) if(x, plus(y, x 1 ), plus(z, x 2 )) plus(if(x, y, z), if(x, x 1, x 2 ))

14 The rules of R if are included in RPOM with F Lex = {if} and B defined as F 0. The first three rules hold by case 1 and the three next by case 4. The last rule holds by case 5. Note that plus(x, y) B x and plus(x, y) B y hold. Hence, using case 4 we obtain if(x, plus(y, x 1 ), plus(z, x 2 )) stab if(x, y, z) and if(x, plus(y, x 1 ), plus(z, x 2 )) stab if(x, x 1, x 2 ). Therefore, by Theorem 3 we conclude that the hierarchical union of R if and any base system R plus is C E - terminating whenever R plus is non-duplicating and C E -terminating. We stress that R F and R if are hierarchical extensions which are not proper and where SCP cannot be used. Hence, no previous modularity result can be applied to these examples. 5 Proving innermost termination incrementally with RPOM This section proceeds analogously to the previous one. The main difference is that, for proving innermost termination, B needs to be frontier preserving only in the sense of sets. Hence, if B is constructed from B, the non-duplicating requirement on B disappears. Theorem 4. Let R = R 0 R 1 be a hierarchical union, B = F 0 and B be a stable, well-founded on T (F, X ), such that s B t implies frt B (s) set frt B (t), and i,r0,b B. If R 1 stab then R is innermost terminating. Proof. By the assumptions and Proposition 1, rpom stab is stable. Hence, it includes i,λ,r. Since rpom stab rpom IP mon, it follows that i,λ,r rpom IP mon. By the assumptions and Proposition 3, rpom IP mon is IPmonotonic w.r.t. R, and by Lemma 6, it is well-founded. Altogether with Theorem 1 imply that R is innermost terminating. Theorem 5. Let R = R 0 R 1 be a hierarchical TRS where R 0 is innermost terminating. Let B be F 0 where 0 is on T (F 0, X ), and let R 1 stab. Then R is innermost terminating. Proof. We use B = ( i,r 0,F 0 B ) + and the corresponding IP mon. Note that B and stab are not necessarily stable whereas B and stab are, the second one by Proposition 1, and hence, stab includes i,λ,r1. Since rpom stab rpom IP mon rpom IP mon, it follows that i,λ,r rpom IP mon. By the definition of B, it is IP-monotonic w.r.t. R 0 in T (F 0, X ). It is also well-founded since any derivation with i,r0,f 0 B can be commuted to a derivation with i,r0,f 0 followed by a derivation with F 0, with the same number of rewrite steps, and the fact that R 0 is innermost terminating. By the assumptions and Proposition 3, rpom IP mon is IP-monotonic w.r.t. R, and by Lemma 6, it is well-founded. Altogether with Theorem 1 imply that R is innermost terminating.

15 Example 6. Recall the systems R F in Example 2 and R if in Example 5 are included in stab with B defined as F 0. Hence, by Theorem 5, the hierarchical union of R F R if and any (possibly duplicating) base system R plus is innermost terminating whenever R plus is innermost terminating. 6 Conclusions The stable subclass of the RPOM is suitable for proving termination automatically. It is more powerful than RPO since it allows the reuse of termination proofs. But at the same time it inherits from its predecessor the simplicity and all the techniques for the automated generation of the precedence. The two main differences between RPO and RPOM-STAB are the use of B and the treatment of terms rooted by base function symbols. But these difference can be easily handled: frontier subterms can be computed in linear time and the decision between applying case 5 or B is deterministic. Besides, if B is defined as F B, we can prove s B t just by proving s r B t r, where s r and t r are obtained by replacing each occurrence of a frontier subterm of s by the same fresh variable. As future work we plan to investigate more deeply the use of RPOM for proving innermost termination incrementally, since for this particular case no condition need to be imposed on the base TRS. In particular, we will consider the combination of RPOM with the ideas from [9, 11]. Furthermore, we are interested in extending the given results to the monotonic semantic path ordering [5, 4] which will provide a much more powerful framework for combining orderings and prove termination incrementally. Finally we are also interested in extending these results to the higher-order recursive path ordering [13], which will provide necessary results for hierarchical unions for the higher-order case. References 1. T. Arts and J. Giesl. Termination of term rewriting using dependency pairs. Theoretical Computer Science, 236(1-2): , F. Baader and T. Nipkow. Term Rewriting and All That. Cambridge University Press, P. B. Bendix and D. E. Knuth. Simple word problems in universal algebras. In Computational Problems in Abstract Algebra, pages Pergamon Press, C. Borralleras. Ordering-based methods for proving termination automatically. PhD thesis, Dpto. LSI, Universitat Politècnica de Catalunya, España, C. Borralleras, M. Ferreira, and A. Rubio. Complete monotonic semantic path orderings. In Proc. CADE, volume 1831 of LNAI, pages , C. Calude. Theories of Computational Complexity. North-Holland, C. Calude, S. Marcus, and I. Tevy. The first example of a recursive function which is not primitive recursive. Historia Math., 9: , N. Dershowitz. Orderings for term rewriting systems. Theoretical Computer Science, 17(3): , 1982.

16 9. M.L. Fernández. Relaxing monotonicity for innermost termination. Information Processing Letters, 93(3): , M.L. Fernández. On proving C E-termination of rewriting by size-change termination. Information Processing Letters, 93(4): , M.L. Fernández, G. Godoy, and A. Rubio. Orderings for innermost termination. In Proc. RTA, volume 3467 of LNCS, pages 17 31, B. Gramlich. On proving termination by innermost termination. In Proc. RTA, volume 1103 of LNCS, pages , J.P. Jouannaud and A. Rubio. The higher-order recursive path ordering. In Proc. LICS, pages , M.R.K. Krishna Rao. Modular proofs for completeness of hierarchical term rewriting systems. Theoretical Computer Science, 151(2): , C. S. Lee, N. D. Jones, and A. M. Ben-Amram. The size-change principle for program termination. In Proc. POPL, pages 81 92, N. Hirokawa, and A. Middeldorp. Dependency Pairs Revisited. In Proc. 15th RTA, volume 3091 of LNCS, pages , E. Ohlebusch. Hierarchical termination revisited. Information Processing Letters, 84(4): , A. Rubio. Extension Orderings. In Proc. ICALP, volume 944 of LNCS, pages , R. Thiemann and J. Giesl. Size-change termination for term rewriting. In Proc. RTA, volume 2706 of LNCS, pages , R. Thiemann, J. Giesl, and P. Schneider-Kamp: Improved Modular Termination Proofs Using Dependency Pairs. In Proc. 2nd IJCAR, volume 3097 of LNCS, pages 75 90, Y. Toyama. Counterexamples to termination for the direct sum of term rewriting systems. Information Processing Letters, 25: , X. Urbain. Modular and incremental automated termination proofs. Journal of Automated Reasoning, 32: , 2004.

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