/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 53 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) ATransformationProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (16) QDP (17) MRRProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE (21) QDP (22) QDPSizeChangeProof [EQUIVALENT, 0 ms] (23) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z))) app(app(times, x), 0) -> 0 app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x) app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) 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) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z))) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(plus, y), app(app(times, app(s, z)), 0)) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, app(s, z)), 0) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(times, app(s, z)) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(s, z)) APP(app(times, x), app(s, y)) -> APP(app(plus, app(app(times, x), y)), x) APP(app(times, x), app(s, y)) -> APP(plus, app(app(times, x), y)) APP(app(times, x), app(s, y)) -> APP(app(times, x), y) APP(app(plus, x), app(s, y)) -> APP(s, app(app(plus, x), y)) APP(app(plus, x), app(s, y)) -> APP(app(plus, 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(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z))) app(app(times, x), 0) -> 0 app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x) app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) 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. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 17 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(plus, x), app(s, y)) -> APP(app(plus, x), y) The TRS R consists of the following rules: app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z))) app(app(times, x), 0) -> 0 app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x) app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) 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. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(plus, x), app(s, y)) -> APP(app(plus, x), y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: plus(x, s(y)) -> plus(x, y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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: *plus(x, s(y)) -> plus(x, y) The graph contains the following edges 1 >= 1, 2 > 2 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(s, z)) APP(app(times, x), app(s, y)) -> APP(app(times, x), y) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))) The TRS R consists of the following rules: app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z))) app(app(times, x), 0) -> 0 app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x) app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) 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. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(s, z)) APP(app(times, x), app(s, y)) -> APP(app(times, x), y) APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))) The TRS R consists of the following rules: app(app(times, x), 0) -> 0 app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesReductionPairsProof (EQUIVALENT) First, we A-transformed [FROCOS05] the QDP-Problem. Then we obtain the following A-transformed DP problem. The pairs P are: times1(x, plus(y, s(z))) -> times1(x, s(z)) times1(x, s(y)) -> times1(x, y) times1(x, plus(y, s(z))) -> times1(x, plus(y, times(s(z), 0))) and the Q and R are: Q restricted rewrite system: The TRS R consists of the following rules: times(x, 0) -> 0 plus(x, 0) -> x plus(x, s(y)) -> s(plus(x, y)) Q is empty. 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. The following dependency pairs can be deleted: times1(x, plus(y, s(z))) -> times1(x, s(z)) The following rules are removed from R: app(app(times, x), 0) -> 0 app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) Used ordering: POLO with Polynomial interpretation [POLO]: POL(0) = 0 POL(plus(x_1, x_2)) = 1 + 2*x_1 + 2*x_2 POL(s(x_1)) = x_1 POL(times(x_1, x_2)) = x_1 + 2*x_2 POL(times1(x_1, x_2)) = x_1 + 2*x_2 ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: times1(x, s(y)) -> times1(x, y) times1(x, plus(y, s(z))) -> times1(x, plus(y, times(s(z), 0))) The TRS R consists of the following rules: times(x, 0) -> 0 plus(x, s(y)) -> s(plus(x, y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) 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 dependency pairs: times1(x, s(y)) -> times1(x, y) Strictly oriented rules of the TRS R: plus(x, s(y)) -> s(plus(x, y)) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(plus(x_1, x_2)) = 2*x_1 + 2*x_2 POL(s(x_1)) = 2 + x_1 POL(times(x_1, x_2)) = x_1 + x_2 POL(times1(x_1, x_2)) = x_1 + 2*x_2 ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: times1(x, plus(y, s(z))) -> times1(x, plus(y, times(s(z), 0))) The TRS R consists of the following rules: times(x, 0) -> 0 Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (20) TRUE ---------------------------------------- (21) 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(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z))) app(app(times, x), 0) -> 0 app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x) app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) 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. 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: *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(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) The graph contains the following edges 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 ---------------------------------------- (23) YES