/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(quot, 0), app(s, y)), app(s, z)) -> 0 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 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(quot, 0), app(s, x0)), app(s, x1)) app(app(app(quot, app(s, x0)), app(s, x1)), x2) app(app(app(quot, x0), 0), app(s, x1)) 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) ---------------------------------------- (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(app(quot, app(s, x)), app(s, y)), z) -> APP(app(app(quot, x), y), z) APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(quot, x), y) APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(quot, x) APP(app(app(quot, x), 0), app(s, z)) -> APP(s, app(app(app(quot, x), app(s, z)), app(s, z))) APP(app(app(quot, x), 0), app(s, z)) -> APP(app(app(quot, x), app(s, z)), app(s, z)) APP(app(app(quot, x), 0), app(s, z)) -> APP(app(quot, x), app(s, z)) 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(quot, 0), app(s, y)), app(s, z)) -> 0 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 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(quot, 0), app(s, x0)), app(s, x1)) app(app(app(quot, app(s, x0)), app(s, x1)), x2) app(app(app(quot, x0), 0), app(s, x1)) 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. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 14 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(quot, x), 0), app(s, z)) -> APP(app(app(quot, x), app(s, z)), app(s, z)) APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(app(quot, x), y), z) The TRS R consists of the following rules: app(app(app(quot, 0), app(s, y)), app(s, z)) -> 0 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 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(quot, 0), app(s, x0)), app(s, x1)) app(app(app(quot, app(s, x0)), app(s, x1)), x2) app(app(app(quot, x0), 0), app(s, x1)) 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. ---------------------------------------- (6) 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. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(quot, x), 0), app(s, z)) -> APP(app(app(quot, x), app(s, z)), app(s, z)) APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(app(quot, x), y), z) R is empty. The set Q consists of the following terms: app(app(app(quot, 0), app(s, x0)), app(s, x1)) app(app(app(quot, app(s, x0)), app(s, x1)), x2) app(app(app(quot, x0), 0), app(s, x1)) 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) 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: quot1(x, 0, s(z)) -> quot1(x, s(z), s(z)) quot1(s(x), s(y), z) -> quot1(x, y, z) R is empty. The set Q consists of the following terms: quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) quot(x0, 0, s(x1)) 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. ---------------------------------------- (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]. quot(0, s(x0), s(x1)) quot(s(x0), s(x1), x2) quot(x0, 0, s(x1)) 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) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: quot1(x, 0, s(z)) -> quot1(x, s(z), s(z)) quot1(s(x), s(y), z) -> quot1(x, y, z) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) 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: *quot1(s(x), s(y), z) -> quot1(x, y, z) The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3 *quot1(x, 0, s(z)) -> quot1(x, s(z), s(z)) The graph contains the following edges 1 >= 1, 3 >= 2, 3 >= 3 ---------------------------------------- (13) YES ---------------------------------------- (14) 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(quot, 0), app(s, y)), app(s, z)) -> 0 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 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(quot, 0), app(s, x0)), app(s, x1)) app(app(app(quot, app(s, x0)), app(s, x1)), x2) app(app(app(quot, x0), 0), app(s, x1)) 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. ---------------------------------------- (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: 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(quot, 0), app(s, x0)), app(s, x1)) app(app(app(quot, app(s, x0)), app(s, x1)), x2) app(app(app(quot, x0), 0), app(s, x1)) 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. ---------------------------------------- (17) 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 ---------------------------------------- (18) YES