/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 29 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) ATransformationProof [EQUIVALENT, 0 ms] (13) QDP (14) QReductionProof [EQUIVALENT, 0 ms] (15) QDP (16) QDPSizeChangeProof [EQUIVALENT, 0 ms] (17) YES (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QReductionProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) nonzero -> app(filter, app(neq, 0)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) nonzero -> app(filter, app(neq, 0)) The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) nonzero ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y) APP(app(neq, app(s, x)), app(s, y)) -> APP(neq, x) APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f) APP(app(filter, f), app(app(cons, y), ys)) -> APP(filtersub, app(f, y)) APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y) APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(cons, y), app(app(filter, f), ys)) APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(filter, f) APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(filter, f) NONZERO -> APP(filter, app(neq, 0)) NONZERO -> APP(neq, 0) The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) nonzero -> app(filter, app(neq, 0)) The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) nonzero We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y) The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) nonzero -> app(filter, app(neq, 0)) The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) nonzero We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) 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. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y) R is empty. The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) nonzero We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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]. nonzero ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y) R is empty. The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: neq1(s(x), s(y)) -> neq1(x, y) R is empty. The set Q consists of the following terms: neq(0, 0) neq(0, s(x0)) neq(s(x0), 0) neq(s(x0), s(x1)) filter(x0, nil) filter(x0, cons(x1, x2)) filtersub(true, x0, cons(x1, x2)) filtersub(false, x0, cons(x1, x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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]. neq(0, 0) neq(0, s(x0)) neq(s(x0), 0) neq(s(x0), s(x1)) filter(x0, nil) filter(x0, cons(x1, x2)) filtersub(true, x0, cons(x1, x2)) filtersub(false, x0, cons(x1, x2)) ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: neq1(s(x), s(y)) -> neq1(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) 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: *neq1(s(x), s(y)) -> neq1(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (17) YES ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y) APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) nonzero -> app(filter, app(neq, 0)) The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) nonzero We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y) APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) nonzero We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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]. nonzero ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y) APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) The TRS R consists of the following rules: app(app(neq, 0), 0) -> false app(app(neq, 0), app(s, y)) -> true app(app(neq, app(s, x)), 0) -> true app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys) app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys)) The set Q consists of the following terms: app(app(neq, 0), 0) app(app(neq, 0), app(s, x0)) app(app(neq, app(s, x0)), 0) app(app(neq, app(s, x0)), app(s, x1)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(filtersub, true), x0), app(app(cons, x1), x2)) app(app(app(filtersub, false), x0), app(app(cons, x1), x2)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) 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: *APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys)) The graph contains the following edges 2 >= 2 *APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) The graph contains the following edges 2 > 2 *APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys) The graph contains the following edges 2 > 2 ---------------------------------------- (24) YES