/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 66 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 0 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 4 ms] (8) QTRS (9) Overlay + Local Confluence [EQUIVALENT, 0 ms] (10) QTRS (11) DependencyPairsProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) TransformationProof [EQUIVALENT, 0 ms] (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) QReductionProof [EQUIVALENT, 0 ms] (24) QDP (25) TransformationProof [EQUIVALENT, 0 ms] (26) QDP (27) UsableRulesProof [EQUIVALENT, 0 ms] (28) QDP (29) QReductionProof [EQUIVALENT, 0 ms] (30) QDP (31) TransformationProof [EQUIVALENT, 0 ms] (32) QDP (33) NonTerminationLoopProof [COMPLETE, 0 ms] (34) 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))) zeros -> n__zeros activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(and(x_1, x_2)) = 2 + 2*x_1 + 2*x_2 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = x_1 POL(n__zeros) = 0 POL(nil) = 0 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: 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))) zeros -> n__zeros activate(n__zeros) -> zeros 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 + 2*x_2 POL(length(x_1)) = x_1 POL(n__zeros) = 0 POL(nil) = 2 POL(s(x_1)) = x_1 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))) zeros -> n__zeros activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (5) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2*x_1 POL(cons(x_1, x_2)) = x_1 + 2*x_2 POL(length(x_1)) = 2*x_1 POL(n__zeros) = 1 POL(s(x_1)) = x_1 POL(zeros) = 2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: zeros -> n__zeros ---------------------------------------- (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))) activate(n__zeros) -> zeros activate(X) -> X Q is empty. ---------------------------------------- (7) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 2 + x_1 POL(cons(x_1, x_2)) = 2 + x_1 + 2*x_2 POL(length(x_1)) = 2*x_1 POL(n__zeros) = 0 POL(s(x_1)) = x_1 POL(zeros) = 2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: activate(X) -> X ---------------------------------------- (8) 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))) activate(n__zeros) -> zeros Q is empty. ---------------------------------------- (9) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (10) 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))) activate(n__zeros) -> zeros The set Q consists of the following terms: zeros length(cons(x0, x1)) activate(n__zeros) ---------------------------------------- (11) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (12) 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) ACTIVATE(n__zeros) -> ZEROS The TRS R consists of the following rules: zeros -> cons(0, n__zeros) length(cons(N, L)) -> s(length(activate(L))) activate(n__zeros) -> zeros The set Q consists of the following terms: zeros length(cons(x0, x1)) activate(n__zeros) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (14) 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))) activate(n__zeros) -> zeros The set Q consists of the following terms: zeros length(cons(x0, x1)) activate(n__zeros) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) 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. ---------------------------------------- (16) 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: zeros length(cons(x0, x1)) activate(n__zeros) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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]. length(cons(x0, x1)) ---------------------------------------- (18) 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: zeros activate(n__zeros) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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)) ---------------------------------------- (20) 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: zeros activate(n__zeros) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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. ---------------------------------------- (22) 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 activate(n__zeros) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) 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) ---------------------------------------- (24) 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. ---------------------------------------- (25) 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))) ---------------------------------------- (26) 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. ---------------------------------------- (27) 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. ---------------------------------------- (28) 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. ---------------------------------------- (29) 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 ---------------------------------------- (30) 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. ---------------------------------------- (31) 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))) ---------------------------------------- (32) 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. ---------------------------------------- (33) 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)). ---------------------------------------- (34) NO