/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 63 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 17 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 0 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) AND (11) QDP (12) UsableRulesProof [EQUIVALENT, 0 ms] (13) QDP (14) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (15) QDP (16) DependencyGraphProof [EQUIVALENT, 0 ms] (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) QDPSizeChangeProof [EQUIVALENT, 0 ms] (21) YES (22) QDP (23) UsableRulesProof [EQUIVALENT, 0 ms] (24) QDP (25) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (26) QDP (27) MRRProof [EQUIVALENT, 0 ms] (28) QDP (29) UsableRulesProof [EQUIVALENT, 0 ms] (30) QDP (31) MRRProof [EQUIVALENT, 0 ms] (32) QDP (33) MNOCProof [EQUIVALENT, 0 ms] (34) QDP (35) TransformationProof [EQUIVALENT, 0 ms] (36) QDP (37) UsableRulesProof [EQUIVALENT, 0 ms] (38) QDP (39) QReductionProof [EQUIVALENT, 0 ms] (40) QDP (41) TransformationProof [EQUIVALENT, 0 ms] (42) QDP (43) UsableRulesProof [EQUIVALENT, 0 ms] (44) QDP (45) QReductionProof [EQUIVALENT, 0 ms] (46) QDP (47) TransformationProof [EQUIVALENT, 0 ms] (48) QDP (49) NonTerminationLoopProof [COMPLETE, 0 ms] (50) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: zeros -> cons(0, n__zeros) and(tt, X) -> activate(X) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(and(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(length(x_1)) = x_1 POL(n__take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(n__zeros) = 0 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(tt) = 0 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: and(tt, X) -> activate(X) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(nil) -> 0 length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(length(x_1)) = 1 + x_1 POL(n__take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(n__zeros) = 0 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: length(nil) -> 0 ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(cons(N, L)) -> s(length(activate(L))) take(0, IL) -> nil take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(length(x_1)) = x_1 POL(n__take(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(n__zeros) = 0 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: take(0, IL) -> nil ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(cons(N, L)) -> s(length(activate(L))) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. ---------------------------------------- (7) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) LENGTH(cons(N, L)) -> ACTIVATE(L) TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) ACTIVATE(n__zeros) -> ZEROS ACTIVATE(n__take(X1, X2)) -> TAKE(activate(X1), activate(X2)) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(cons(N, L)) -> s(length(activate(L))) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (10) Complex Obligation (AND) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__take(X1, X2)) -> TAKE(activate(X1), activate(X2)) TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(cons(N, L)) -> s(length(activate(L))) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__take(X1, X2)) -> TAKE(activate(X1), activate(X2)) TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) take(X1, X2) -> n__take(X1, X2) zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) The following rules are removed from R: take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(ACTIVATE(x_1)) = x_1 POL(TAKE(x_1, x_2)) = 2*x_1 + x_2 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + x_2 POL(n__take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(n__zeros) = 0 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 POL(zeros) = 0 ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__take(X1, X2)) -> TAKE(activate(X1), activate(X2)) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X take(X1, X2) -> n__take(X1, X2) zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X take(X1, X2) -> n__take(X1, X2) zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X2) The graph contains the following edges 1 > 1 *ACTIVATE(n__take(X1, X2)) -> ACTIVATE(X1) The graph contains the following edges 1 > 1 ---------------------------------------- (21) YES ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(cons(N, L)) -> s(length(activate(L))) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> n__zeros take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) take(X1, X2) -> n__take(X1, X2) zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. No dependency pairs are removed. The following rules are removed from R: take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(LENGTH(x_1)) = x_1 POL(activate(x_1)) = x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(n__take(x_1, x_2)) = x_1 + x_2 POL(n__zeros) = 0 POL(s(x_1)) = 2 + 2*x_1 POL(take(x_1, x_2)) = x_1 + x_2 POL(zeros) = 0 ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) activate(X) -> X take(X1, X2) -> n__take(X1, X2) zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented rules of the TRS R: activate(n__take(X1, X2)) -> take(activate(X1), activate(X2)) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(LENGTH(x_1)) = 2*x_1 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(n__take(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(n__zeros) = 0 POL(take(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(zeros) = 0 ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(X) -> X take(X1, X2) -> n__take(X1, X2) zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: activate(n__zeros) -> zeros activate(X) -> X zeros -> cons(0, n__zeros) zeros -> n__zeros Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented rules of the TRS R: activate(X) -> X zeros -> n__zeros Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(LENGTH(x_1)) = 2*x_1 POL(activate(x_1)) = 1 + x_1 POL(cons(x_1, x_2)) = 1 + x_1 + x_2 POL(n__zeros) = 0 POL(zeros) = 1 ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: activate(n__zeros) -> zeros zeros -> cons(0, n__zeros) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(N, L)) -> LENGTH(activate(L)) The TRS R consists of the following rules: activate(n__zeros) -> zeros zeros -> cons(0, n__zeros) The set Q consists of the following terms: activate(n__zeros) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule LENGTH(cons(N, L)) -> LENGTH(activate(L)) at position [0] we obtained the following new rules [LPAR04]: (LENGTH(cons(y0, n__zeros)) -> LENGTH(zeros),LENGTH(cons(y0, n__zeros)) -> LENGTH(zeros)) ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(y0, n__zeros)) -> LENGTH(zeros) The TRS R consists of the following rules: activate(n__zeros) -> zeros zeros -> cons(0, n__zeros) The set Q consists of the following terms: activate(n__zeros) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(y0, n__zeros)) -> LENGTH(zeros) The TRS R consists of the following rules: zeros -> cons(0, n__zeros) The set Q consists of the following terms: activate(n__zeros) zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. activate(n__zeros) ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(y0, n__zeros)) -> LENGTH(zeros) The TRS R consists of the following rules: zeros -> cons(0, n__zeros) The set Q consists of the following terms: zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule LENGTH(cons(y0, n__zeros)) -> LENGTH(zeros) at position [0] we obtained the following new rules [LPAR04]: (LENGTH(cons(y0, n__zeros)) -> LENGTH(cons(0, n__zeros)),LENGTH(cons(y0, n__zeros)) -> LENGTH(cons(0, n__zeros))) ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(y0, n__zeros)) -> LENGTH(cons(0, n__zeros)) The TRS R consists of the following rules: zeros -> cons(0, n__zeros) The set Q consists of the following terms: zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(y0, n__zeros)) -> LENGTH(cons(0, n__zeros)) R is empty. The set Q consists of the following terms: zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. zeros ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(y0, n__zeros)) -> LENGTH(cons(0, n__zeros)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule LENGTH(cons(y0, n__zeros)) -> LENGTH(cons(0, n__zeros)) we obtained the following new rules [LPAR04]: (LENGTH(cons(0, n__zeros)) -> LENGTH(cons(0, n__zeros)),LENGTH(cons(0, n__zeros)) -> LENGTH(cons(0, n__zeros))) ---------------------------------------- (48) Obligation: Q DP problem: The TRS P consists of the following rules: LENGTH(cons(0, n__zeros)) -> LENGTH(cons(0, n__zeros)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (49) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = LENGTH(cons(0, n__zeros)) evaluates to t =LENGTH(cons(0, n__zeros)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from LENGTH(cons(0, n__zeros)) to LENGTH(cons(0, n__zeros)). ---------------------------------------- (50) NO