/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 49 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) ATransformationProof [EQUIVALENT, 0 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPOrderProof [EQUIVALENT, 87 ms] (15) QDP (16) PisEmptyProof [EQUIVALENT, 0 ms] (17) YES (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) ---------------------------------------- (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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(:, app(app(:, x), z)) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, x), z) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(:, x) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, app(app(:, x), y)), z)), u) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(:, app(app(:, app(app(:, x), y)), z)) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, x), y)), z) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(:, app(app(:, x), y)) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, x), y) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) The TRS R consists of the following rules: app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) 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 14 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, x), z) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, app(app(:, x), y)), z)), u) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, x), y)), z) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, x), y) The TRS R consists of the following rules: app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) 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(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, x), z) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, app(app(:, x), y)), z)), u) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, app(app(:, x), y)), z) APP(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> APP(app(:, x), y) The TRS R consists of the following rules: app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) The set Q consists of the following terms: app(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: :1(:(:(:(C, x), y), z), u) -> :1(x, z) :1(:(:(:(C, x), y), z), u) -> :1(:(x, z), :(:(:(x, y), z), u)) :1(:(:(:(C, x), y), z), u) -> :1(:(:(x, y), z), u) :1(:(:(:(C, x), y), z), u) -> :1(:(x, y), z) :1(:(:(:(C, x), y), z), u) -> :1(x, y) The TRS R consists of the following rules: :(:(:(:(C, x), y), z), u) -> :(:(x, z), :(:(:(x, y), z), u)) The set Q consists of the following terms: :(:(:(:(C, x0), x1), x2), x3) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) 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]. map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: :1(:(:(:(C, x), y), z), u) -> :1(x, z) :1(:(:(:(C, x), y), z), u) -> :1(:(x, z), :(:(:(x, y), z), u)) :1(:(:(:(C, x), y), z), u) -> :1(:(:(x, y), z), u) :1(:(:(:(C, x), y), z), u) -> :1(:(x, y), z) :1(:(:(:(C, x), y), z), u) -> :1(x, y) The TRS R consists of the following rules: :(:(:(:(C, x), y), z), u) -> :(:(x, z), :(:(:(x, y), z), u)) The set Q consists of the following terms: :(:(:(:(C, x0), x1), x2), x3) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. :1(:(:(:(C, x), y), z), u) -> :1(x, z) :1(:(:(:(C, x), y), z), u) -> :1(:(x, z), :(:(:(x, y), z), u)) :1(:(:(:(C, x), y), z), u) -> :1(:(:(x, y), z), u) :1(:(:(:(C, x), y), z), u) -> :1(:(x, y), z) :1(:(:(:(C, x), y), z), u) -> :1(x, y) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. :1(x1, x2) = :1(x1) :(x1, x2) = :(x1, x2) C = C Recursive path order with status [RPO]. Quasi-Precedence: C > :_2 > :1_1 Status: :1_1: multiset status :_2: [1,2] C: multiset status The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: :(:(:(:(C, x), y), z), u) -> :(:(x, z), :(:(:(x, y), z), u)) ---------------------------------------- (15) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: :(:(:(:(C, x), y), z), u) -> :(:(x, z), :(:(:(x, y), z), u)) The set Q consists of the following terms: :(:(:(:(C, x0), x1), x2), x3) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (17) YES ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(:, app(app(:, app(app(:, app(app(:, C), x)), y)), z)), u) -> app(app(:, app(app(:, x), z)), app(app(:, app(app(:, app(app(:, x), y)), z)), u)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) 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(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) R is empty. The set Q consists of the following terms: app(app(:, app(app(:, app(app(:, app(app(:, C), x0)), x1)), x2)), x3) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) The graph contains the following edges 1 >= 1, 2 > 2 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 ---------------------------------------- (22) YES