/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 122 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) ATransformationProof [EQUIVALENT, 0 ms] (20) QDP (21) QReductionProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) ATransformationProof [EQUIVALENT, 0 ms] (29) QDP (30) QReductionProof [EQUIVALENT, 0 ms] (31) QDP (32) QDPOrderProof [EQUIVALENT, 11 ms] (33) QDP (34) QDPOrderProof [EQUIVALENT, 0 ms] (35) QDP (36) PisEmptyProof [EQUIVALENT, 0 ms] (37) YES (38) QDP (39) QDPSizeChangeProof [EQUIVALENT, 0 ms] (40) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), 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(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x) APP(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y) APP(app(le, app(s, x)), app(s, y)) -> APP(le, x) APP(perfectp, app(s, x)) -> APP(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) APP(perfectp, app(s, x)) -> APP(app(app(f, x), app(s, 0)), app(s, x)) APP(perfectp, app(s, x)) -> APP(app(f, x), app(s, 0)) APP(perfectp, app(s, x)) -> APP(f, x) APP(perfectp, app(s, x)) -> APP(s, 0) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(f, x), u), app(app(minus, z), app(s, x))) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(f, x), u) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(f, x) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(minus, z), app(s, x)) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(minus, z) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(if, app(app(le, x), y)) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(le, x), y) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(le, x) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(f, app(s, x)), app(app(minus, y), x)), z) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(f, app(s, x)), app(app(minus, y), x)) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(minus, y), x) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(minus, y) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, x), u), z), u) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(f, x), u), z) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(f, x), u) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(f, x) APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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 4 SCCs with 33 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(le, app(s, x)), app(s, y)) -> APP(app(le, x), y) The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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(le, app(s, x)), app(s, y)) -> APP(app(le, x), y) R is empty. The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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: le1(s(x), s(y)) -> le1(x, y) R is empty. The set Q consists of the following terms: minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) if(true, x0, x1) if(false, x0, x1) perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(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]. minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) if(true, x0, x1) if(false, x0, x1) perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(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) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: le1(s(x), s(y)) -> le1(x, 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: *le1(s(x), s(y)) -> le1(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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. ---------------------------------------- (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(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y) R is empty. The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: minus1(s(x), s(y)) -> minus1(x, y) R is empty. The set Q consists of the following terms: minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) if(true, x0, x1) if(false, x0, x1) perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(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. ---------------------------------------- (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]. minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) if(true, x0, x1) if(false, x0, x1) perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(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) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: minus1(s(x), s(y)) -> minus1(x, y) R is empty. Q is empty. 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: *minus1(s(x), s(y)) -> minus1(x, y) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (24) YES ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, x), u), z), u) The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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. ---------------------------------------- (26) 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. ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u) APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, x), u), z), u) The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(minus, app(s, x)), 0) -> app(s, x) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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. ---------------------------------------- (28) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: f1(s(x), s(y), z, u) -> f1(s(x), minus(y, x), z, u) f1(s(x), 0, z, u) -> f1(x, u, minus(z, s(x)), u) f1(s(x), s(y), z, u) -> f1(x, u, z, u) The TRS R consists of the following rules: minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) minus(s(x), 0) -> s(x) The set Q consists of the following terms: minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) if(true, x0, x1) if(false, x0, x1) perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(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. ---------------------------------------- (30) 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]. le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) if(true, x0, x1) if(false, x0, x1) perfectp(0) perfectp(s(x0)) f(0, x0, 0, x1) f(0, x0, s(x1), x2) f(s(x0), 0, x1, x2) f(s(x0), s(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) ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: f1(s(x), s(y), z, u) -> f1(s(x), minus(y, x), z, u) f1(s(x), 0, z, u) -> f1(x, u, minus(z, s(x)), u) f1(s(x), s(y), z, u) -> f1(x, u, z, u) The TRS R consists of the following rules: minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) minus(s(x), 0) -> s(x) The set Q consists of the following terms: minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. f1(s(x), 0, z, u) -> f1(x, u, minus(z, s(x)), u) f1(s(x), s(y), z, u) -> f1(x, u, z, u) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. f1(x1, x2, x3, x4) = x1 s(x1) = s(x1) Knuth-Bendix order [KBO] with precedence:trivial and weight map: s_1=1 dummyConstant=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (33) Obligation: Q DP problem: The TRS P consists of the following rules: f1(s(x), s(y), z, u) -> f1(s(x), minus(y, x), z, u) The TRS R consists of the following rules: minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) minus(s(x), 0) -> s(x) The set Q consists of the following terms: minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (34) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. f1(s(x), s(y), z, u) -> f1(s(x), minus(y, x), z, u) The remaining pairs can at least be oriented weakly. Used ordering: Combined order from the following AFS and order. f1(x1, x2, x3, x4) = x2 s(x1) = s(x1) minus(x1, x2) = x1 0 = 0 Knuth-Bendix order [KBO] with precedence:trivial and weight map: s_1=1 0=1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) minus(s(x), 0) -> s(x) ---------------------------------------- (35) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: minus(0, y) -> 0 minus(s(x), s(y)) -> minus(x, y) minus(s(x), 0) -> s(x) The set Q consists of the following terms: minus(0, x0) minus(s(x0), 0) minus(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (36) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (37) YES ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) The TRS R consists of the following rules: app(app(minus, 0), y) -> 0 app(app(minus, app(s, x)), 0) -> app(s, x) app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y) app(app(le, 0), y) -> true app(app(le, app(s, x)), 0) -> false app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y) app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(perfectp, 0) -> false app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x)) app(app(app(app(f, 0), y), 0), u) -> true app(app(app(app(f, 0), y), app(s, z)), u) -> false app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u) app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u)) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(minus, 0), x0) app(app(minus, app(s, x0)), 0) app(app(minus, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(perfectp, 0) app(perfectp, app(s, x0)) app(app(app(app(f, 0), x0), 0), x1) app(app(app(app(f, 0), x0), app(s, x1)), x2) app(app(app(app(f, app(s, x0)), 0), x1), x2) app(app(app(app(f, app(s, x0)), app(s, 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. ---------------------------------------- (39) 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, fun), app(app(cons, x), xs)) -> APP(fun, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) The graph contains the following edges 1 >= 1, 2 > 2 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) The graph contains the following edges 2 > 2 *APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) The graph contains the following edges 2 >= 2 *APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) The graph contains the following edges 2 >= 2 ---------------------------------------- (40) YES