/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 64 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 19 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 0 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 11 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) UsableRulesReductionPairsProof [EQUIVALENT, 8 ms] (30) QDP (31) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (32) QDP (33) QReductionProof [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(X1, X2) 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*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__take(x_1, x_2)) = x_1 + 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) = 2 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(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (3) 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__take(x_1, x_2)) = x_1 + x_2 POL(n__zeros) = 1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 2*x_1 + 2*x_2 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 ---------------------------------------- (4) 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))) take(X1, X2) -> n__take(X1, X2) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) 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)) = 2*x_1 + 2*x_2 POL(length(x_1)) = 2*x_1 POL(n__take(x_1, x_2)) = 1 + x_1 + x_2 POL(n__zeros) = 0 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 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: take(0, IL) -> nil take(X1, X2) -> n__take(X1, X2) ---------------------------------------- (6) 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(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) 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)) = x_1 POL(n__take(x_1, x_2)) = x_1 + x_2 POL(n__zeros) = 0 POL(nil) = 2 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 ---------------------------------------- (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))) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) activate(X) -> X Q is empty. ---------------------------------------- (9) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(activate(x_1)) = 1 + 2*x_1 POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(length(x_1)) = x_1 POL(n__take(x_1, x_2)) = x_1 + x_2 POL(n__zeros) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(zeros) = 1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: activate(X) -> X ---------------------------------------- (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))) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) 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: 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))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) The set Q consists of the following terms: zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) ---------------------------------------- (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: 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(X1, 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))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) The set Q consists of the following terms: zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) 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: ACTIVATE(n__take(X1, X2)) -> TAKE(X1, X2) TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) 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))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) The set Q consists of the following terms: zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) 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: ACTIVATE(n__take(X1, X2)) -> TAKE(X1, X2) TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) R is empty. The set Q consists of the following terms: zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) 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]. zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) ---------------------------------------- (21) Obligation: Q DP problem: The TRS P consists of the following rules: ACTIVATE(n__take(X1, X2)) -> TAKE(X1, X2) TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) 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: *TAKE(s(M), cons(N, IL)) -> ACTIVATE(IL) The graph contains the following edges 2 > 1 *ACTIVATE(n__take(X1, X2)) -> TAKE(X1, X2) The graph contains the following edges 1 > 1, 1 > 2 ---------------------------------------- (23) YES ---------------------------------------- (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: 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))) activate(n__zeros) -> zeros activate(n__take(X1, X2)) -> take(X1, X2) The set Q consists of the following terms: zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) 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: 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(X1, X2) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> cons(0, n__zeros) The set Q consists of the following terms: zeros length(cons(x0, x1)) take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) 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]. length(cons(x0, x1)) ---------------------------------------- (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(n__take(X1, X2)) -> take(X1, X2) take(s(M), cons(N, IL)) -> cons(N, n__take(M, activate(IL))) zeros -> cons(0, n__zeros) The set Q consists of the following terms: zeros take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) 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 + 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 ---------------------------------------- (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(n__take(X1, X2)) -> take(X1, X2) zeros -> cons(0, n__zeros) The set Q consists of the following terms: zeros take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) 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: activate(n__take(X1, X2)) -> take(X1, X2) 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 + 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 ---------------------------------------- (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) The set Q consists of the following terms: zeros take(s(x0), cons(x1, x2)) activate(n__zeros) activate(n__take(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) 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]. take(s(x0), cons(x1, x2)) ---------------------------------------- (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: zeros activate(n__zeros) activate(n__take(x0, x1)) 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: zeros activate(n__zeros) activate(n__take(x0, x1)) 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: zeros activate(n__zeros) activate(n__take(x0, x1)) 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) activate(n__take(x0, x1)) ---------------------------------------- (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