YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) ATransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES (23) QDP (24) QDPSizeChangeProof [EQUIVALENT, 0 ms] (25) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) app(app(app(f, app(c, x)), x), y) -> app(c, y) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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) ---------------------------------------- (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(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), y), app(app(app(f, y), x), y)) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(f, y), y) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(f, y) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(f, y), x) APP(app(app(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) APP(app(app(f, app(s, x)), y), z) -> APP(app(f, x), app(s, app(c, y))) APP(app(app(f, app(s, x)), y), z) -> APP(f, x) APP(app(app(f, app(s, x)), y), z) -> APP(s, app(c, y)) APP(app(app(f, app(s, x)), y), z) -> APP(c, y) APP(app(app(f, app(s, x)), y), z) -> APP(c, z) APP(app(app(f, app(c, x)), x), y) -> APP(c, y) 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(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) app(app(app(f, app(c, x)), x), y) -> app(c, y) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 19 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) The TRS R consists of the following rules: app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) app(app(app(f, app(c, x)), x), y) -> app(c, y) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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. ---------------------------------------- (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(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) R is empty. The set Q consists of the following terms: app(app(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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) 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: f1(s(x), y, z) -> f1(x, s(c(y)), c(z)) R is empty. The set Q consists of the following terms: f(x0, c(x0), c(x1)) f(s(x0), x1, x2) f(c(x0), x0, x1) g(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. ---------------------------------------- (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]. f(x0, c(x0), c(x1)) f(s(x0), x1, x2) f(c(x0), x0, x1) g(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) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: f1(s(x), y, z) -> f1(x, s(c(y)), c(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: *f1(s(x), y, z) -> f1(x, s(c(y)), c(z)) The graph contains the following edges 1 > 1 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) The TRS R consists of the following rules: app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) app(app(app(f, app(c, x)), x), y) -> app(c, y) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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. ---------------------------------------- (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(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) R is empty. The set Q consists of the following terms: app(app(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: f1(x, c(x), c(y)) -> f1(y, x, y) R is empty. The set Q consists of the following terms: f(x0, c(x0), c(x1)) f(s(x0), x1, x2) f(c(x0), x0, x1) g(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. ---------------------------------------- (19) 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]. f(x0, c(x0), c(x1)) f(s(x0), x1, x2) f(c(x0), x0, x1) g(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) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: f1(x, c(x), c(y)) -> f1(y, x, y) R is empty. Q is empty. 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: *f1(x, c(x), c(y)) -> f1(y, x, y) The graph contains the following edges 3 > 1, 1 >= 2, 2 > 2, 3 > 3 ---------------------------------------- (22) YES ---------------------------------------- (23) 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(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) app(app(app(f, app(c, x)), x), y) -> app(c, y) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(app(f, x0), app(c, x0)), app(c, x1)) app(app(app(f, app(s, x0)), x1), x2) app(app(app(f, app(c, x0)), x0), x1) app(app(g, 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. ---------------------------------------- (24) 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 ---------------------------------------- (25) YES