NO proof of Transformed_CSR_04_LengthOfFiniteLists_nosorts_noand_GM.trs # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 75 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 6 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 0 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 0 ms] (10) QTRS (11) Overlay + Local Confluence [EQUIVALENT, 0 ms] (12) QTRS (13) DependencyPairsProof [EQUIVALENT, 0 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) AND (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) QReductionProof [EQUIVALENT, 0 ms] (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES (24) QDP (25) UsableRulesProof [EQUIVALENT, 0 ms] (26) QDP (27) QReductionProof [EQUIVALENT, 0 ms] (28) QDP (29) MRRProof [EQUIVALENT, 0 ms] (30) QDP (31) TransformationProof [EQUIVALENT, 0 ms] (32) QDP (33) UsableRulesProof [EQUIVALENT, 0 ms] (34) QDP (35) QReductionProof [EQUIVALENT, 1 ms] (36) QDP (37) TransformationProof [EQUIVALENT, 0 ms] (38) QDP (39) UsableRulesProof [EQUIVALENT, 0 ms] (40) QDP (41) QReductionProof [EQUIVALENT, 0 ms] (42) QDP (43) TransformationProof [EQUIVALENT, 0 ms] (44) QDP (45) TransformationProof [EQUIVALENT, 0 ms] (46) QDP (47) NonTerminationLoopProof [COMPLETE, 0 ms] (48) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(nil) -> 0 a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(U11(x_1, x_2)) = 1 + x_1 + x_2 POL(U12(x_1, x_2)) = 1 + x_1 + x_2 POL(a__U11(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(a__length(x_1)) = 2 + x_1 POL(a__zeros) = 2 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(length(x_1)) = 1 + x_1 POL(mark(x_1)) = 2*x_1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(tt) = 0 POL(zeros) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__length(nil) -> 0 a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(U11(x_1, x_2)) = 2*x_1 + 2*x_2 POL(U12(x_1, x_2)) = 2*x_1 + 2*x_2 POL(a__U11(x_1, x_2)) = x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = x_1 + 2*x_2 POL(a__length(x_1)) = x_1 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = 1 + x_1 POL(mark(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = 2*x_1 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: mark(length(X)) -> a__length(mark(X)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(U11(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(U12(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(a__U11(x_1, x_2)) = x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = x_1 + 2*x_2 POL(a__length(x_1)) = x_1 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(mark(x_1)) = x_1 POL(nil) = 2 POL(s(x_1)) = 2*x_1 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: mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil Q is empty. ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__U11(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(a__length(x_1)) = 1 + 2*x_1 POL(a__zeros) = 1 POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(mark(x_1)) = 1 + x_1 POL(nil) = 2 POL(s(x_1)) = x_1 POL(tt) = 1 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: mark(0) -> 0 mark(tt) -> tt mark(nil) -> nil ---------------------------------------- (8) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) Q is empty. ---------------------------------------- (9) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a__U11(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(a__length(x_1)) = x_1 POL(a__zeros) = 1 POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(mark(x_1)) = 1 + 2*x_1 POL(s(x_1)) = x_1 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: mark(cons(X1, X2)) -> cons(mark(X1), X2) ---------------------------------------- (10) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) Q is empty. ---------------------------------------- (11) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (12) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) ---------------------------------------- (13) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, L) -> A__LENGTH(mark(L)) A__U12(tt, L) -> MARK(L) A__LENGTH(cons(N, L)) -> A__U11(tt, L) MARK(zeros) -> A__ZEROS MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes. ---------------------------------------- (16) Complex Obligation (AND) ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) 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. ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) R is empty. The set Q consists of the following terms: a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) 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]. a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (22) 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: *MARK(s(X)) -> MARK(X) The graph contains the following edges 1 > 1 ---------------------------------------- (23) YES ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, L) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__length(cons(N, L)) -> a__U11(tt, L) mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) The set Q consists of the following terms: a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) 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. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, L) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) The TRS R consists of the following rules: mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) 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]. a__U11(tt, x0) a__U12(tt, x0) a__length(cons(x0, x1)) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, L) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) The TRS R consists of the following rules: mark(zeros) -> a__zeros mark(s(X)) -> s(mark(X)) a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) 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: mark(s(X)) -> s(mark(X)) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__LENGTH(x_1)) = x_1 POL(A__U11(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(A__U12(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(a__zeros) = 1 POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(mark(x_1)) = 1 + 2*x_1 POL(s(x_1)) = 2 + 2*x_1 POL(tt) = 0 POL(zeros) = 0 ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, L) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) The TRS R consists of the following rules: mark(zeros) -> a__zeros a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule A__U12(tt, L) -> A__LENGTH(mark(L)) at position [0] we obtained the following new rules [LPAR04]: (A__U12(tt, zeros) -> A__LENGTH(a__zeros),A__U12(tt, zeros) -> A__LENGTH(a__zeros)) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(a__zeros) The TRS R consists of the following rules: mark(zeros) -> a__zeros a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) 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. ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(a__zeros) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros mark(zeros) mark(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) 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]. mark(zeros) mark(s(x0)) ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(a__zeros) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule A__U12(tt, zeros) -> A__LENGTH(a__zeros) at position [0] we obtained the following new rules [LPAR04]: (A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)),A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros))) ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) The set Q consists of the following terms: a__zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) 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. ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) R is empty. The set Q consists of the following terms: a__zeros We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) 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]. a__zeros ---------------------------------------- (42) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (43) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule A__LENGTH(cons(N, L)) -> A__U11(tt, L) we obtained the following new rules [LPAR04]: (A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros),A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros)) ---------------------------------------- (44) Obligation: Q DP problem: The TRS P consists of the following rules: A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (45) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule A__U11(tt, L) -> A__U12(tt, L) we obtained the following new rules [LPAR04]: (A__U11(tt, zeros) -> A__U12(tt, zeros),A__U11(tt, zeros) -> A__U12(tt, zeros)) ---------------------------------------- (46) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros) A__U11(tt, zeros) -> A__U12(tt, zeros) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (47) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = A__LENGTH(cons(0, zeros)) evaluates to t =A__LENGTH(cons(0, zeros)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros) with rule A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros) at position [] and matcher [ ] A__U11(tt, zeros) -> A__U12(tt, zeros) with rule A__U11(tt, zeros) -> A__U12(tt, zeros) at position [] and matcher [ ] A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) with rule A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (48) NO