/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) UsableRulesProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPSizeChangeProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(D, t) -> 1 app(D, constant) -> 0 app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y)) app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y))) app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, 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) 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(D, t) -> 1 app(D, constant) -> 0 app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y)) app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y))) app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, 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) The set Q consists of the following terms: app(D, t) app(D, constant) app(D, app(app(+, x0), x1)) app(D, app(app(*, x0), x1)) app(D, app(app(-, x0), 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) ---------------------------------------- (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(D, app(app(+, x), y)) -> APP(app(+, app(D, x)), app(D, y)) APP(D, app(app(+, x), y)) -> APP(+, app(D, x)) APP(D, app(app(+, x), y)) -> APP(D, x) APP(D, app(app(+, x), y)) -> APP(D, y) APP(D, app(app(*, x), y)) -> APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y))) APP(D, app(app(*, x), y)) -> APP(+, app(app(*, y), app(D, x))) APP(D, app(app(*, x), y)) -> APP(app(*, y), app(D, x)) APP(D, app(app(*, x), y)) -> APP(*, y) APP(D, app(app(*, x), y)) -> APP(D, x) APP(D, app(app(*, x), y)) -> APP(app(*, x), app(D, y)) APP(D, app(app(*, x), y)) -> APP(D, y) APP(D, app(app(-, x), y)) -> APP(app(-, app(D, x)), app(D, y)) APP(D, app(app(-, x), y)) -> APP(-, app(D, x)) APP(D, app(app(-, x), y)) -> APP(D, x) APP(D, app(app(-, x), y)) -> APP(D, 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(D, t) -> 1 app(D, constant) -> 0 app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y)) app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y))) app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, 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) The set Q consists of the following terms: app(D, t) app(D, constant) app(D, app(app(+, x0), x1)) app(D, app(app(*, x0), x1)) app(D, app(app(-, x0), 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. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 19 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(D, app(app(+, x), y)) -> APP(D, y) APP(D, app(app(+, x), y)) -> APP(D, x) APP(D, app(app(*, x), y)) -> APP(D, x) APP(D, app(app(*, x), y)) -> APP(D, y) APP(D, app(app(-, x), y)) -> APP(D, x) APP(D, app(app(-, x), y)) -> APP(D, y) The TRS R consists of the following rules: app(D, t) -> 1 app(D, constant) -> 0 app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y)) app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y))) app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, 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) The set Q consists of the following terms: app(D, t) app(D, constant) app(D, app(app(+, x0), x1)) app(D, app(app(*, x0), x1)) app(D, app(app(-, x0), 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) 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(D, app(app(+, x), y)) -> APP(D, y) APP(D, app(app(+, x), y)) -> APP(D, x) APP(D, app(app(*, x), y)) -> APP(D, x) APP(D, app(app(*, x), y)) -> APP(D, y) APP(D, app(app(-, x), y)) -> APP(D, x) APP(D, app(app(-, x), y)) -> APP(D, y) R is empty. The set Q consists of the following terms: app(D, t) app(D, constant) app(D, app(app(+, x0), x1)) app(D, app(app(*, x0), x1)) app(D, app(app(-, x0), 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. ---------------------------------------- (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: D1(+(x, y)) -> D1(y) D1(+(x, y)) -> D1(x) D1(*(x, y)) -> D1(x) D1(*(x, y)) -> D1(y) D1(-(x, y)) -> D1(x) D1(-(x, y)) -> D1(y) R is empty. The set Q consists of the following terms: D(t) D(constant) D(+(x0, x1)) D(*(x0, x1)) D(-(x0, 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. ---------------------------------------- (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]. D(t) D(constant) D(+(x0, x1)) D(*(x0, x1)) D(-(x0, 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) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: D1(+(x, y)) -> D1(y) D1(+(x, y)) -> D1(x) D1(*(x, y)) -> D1(x) D1(*(x, y)) -> D1(y) D1(-(x, y)) -> D1(x) D1(-(x, y)) -> D1(y) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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: *D1(+(x, y)) -> D1(y) The graph contains the following edges 1 > 1 *D1(+(x, y)) -> D1(x) The graph contains the following edges 1 > 1 *D1(*(x, y)) -> D1(x) The graph contains the following edges 1 > 1 *D1(*(x, y)) -> D1(y) The graph contains the following edges 1 > 1 *D1(-(x, y)) -> D1(x) The graph contains the following edges 1 > 1 *D1(-(x, y)) -> D1(y) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (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) The TRS R consists of the following rules: app(D, t) -> 1 app(D, constant) -> 0 app(D, app(app(+, x), y)) -> app(app(+, app(D, x)), app(D, y)) app(D, app(app(*, x), y)) -> app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y))) app(D, app(app(-, x), y)) -> app(app(-, app(D, x)), app(D, 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) The set Q consists of the following terms: app(D, t) app(D, constant) app(D, app(app(+, x0), x1)) app(D, app(app(*, x0), x1)) app(D, app(app(-, x0), 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) 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. ---------------------------------------- (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) R is empty. The set Q consists of the following terms: app(D, t) app(D, constant) app(D, app(app(+, x0), x1)) app(D, app(app(*, x0), x1)) app(D, app(app(-, x0), 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. ---------------------------------------- (19) 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 ---------------------------------------- (20) YES