4.21/1.98 YES 4.21/1.99 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 4.21/1.99 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.21/1.99 4.21/1.99 4.21/1.99 Termination w.r.t. Q of the given QTRS could be proven: 4.21/1.99 4.21/1.99 (0) QTRS 4.21/1.99 (1) DependencyPairsProof [EQUIVALENT, 78 ms] 4.21/1.99 (2) QDP 4.21/1.99 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.21/1.99 (4) AND 4.21/1.99 (5) QDP 4.21/1.99 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.21/1.99 (7) QDP 4.21/1.99 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.21/1.99 (9) QDP 4.21/1.99 (10) QReductionProof [EQUIVALENT, 0 ms] 4.21/1.99 (11) QDP 4.21/1.99 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.21/1.99 (13) YES 4.21/1.99 (14) QDP 4.21/1.99 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.21/1.99 (16) QDP 4.21/1.99 (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.21/1.99 (18) YES 4.21/1.99 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (0) 4.21/1.99 Obligation: 4.21/1.99 Q restricted rewrite system: 4.21/1.99 The TRS R consists of the following rules: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> app(app(app(app(f, app(app(g, x), y)), app(app(g, x), y)), app(app(g, x), y)), app(h, x)) 4.21/1.99 app(app(g, 0), 1) -> 0 4.21/1.99 app(app(g, 0), 1) -> 1 4.21/1.99 app(h, app(app(g, x), y)) -> app(h, x) 4.21/1.99 app(app(map, fun), nil) -> nil 4.21/1.99 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.21/1.99 app(app(filter, fun), nil) -> nil 4.21/1.99 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.21/1.99 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.21/1.99 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.21/1.99 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x0), x1)), x2) 4.21/1.99 app(app(g, 0), 1) 4.21/1.99 app(h, app(app(g, x0), x1)) 4.21/1.99 app(app(map, x0), nil) 4.21/1.99 app(app(map, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(filter, x0), nil) 4.21/1.99 app(app(filter, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(app(app(filter2, true), x0), x1), x2) 4.21/1.99 app(app(app(app(filter2, false), x0), x1), x2) 4.21/1.99 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (1) DependencyPairsProof (EQUIVALENT) 4.21/1.99 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (2) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 APP(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> APP(app(app(app(f, app(app(g, x), y)), app(app(g, x), y)), app(app(g, x), y)), app(h, x)) 4.21/1.99 APP(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> APP(app(app(f, app(app(g, x), y)), app(app(g, x), y)), app(app(g, x), y)) 4.21/1.99 APP(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> APP(app(f, app(app(g, x), y)), app(app(g, x), y)) 4.21/1.99 APP(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> APP(f, app(app(g, x), y)) 4.21/1.99 APP(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> APP(h, x) 4.21/1.99 APP(h, app(app(g, x), y)) -> APP(h, x) 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) 4.21/1.99 APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) 4.21/1.99 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) 4.21/1.99 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) 4.21/1.99 4.21/1.99 The TRS R consists of the following rules: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> app(app(app(app(f, app(app(g, x), y)), app(app(g, x), y)), app(app(g, x), y)), app(h, x)) 4.21/1.99 app(app(g, 0), 1) -> 0 4.21/1.99 app(app(g, 0), 1) -> 1 4.21/1.99 app(h, app(app(g, x), y)) -> app(h, x) 4.21/1.99 app(app(map, fun), nil) -> nil 4.21/1.99 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.21/1.99 app(app(filter, fun), nil) -> nil 4.21/1.99 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.21/1.99 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.21/1.99 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.21/1.99 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x0), x1)), x2) 4.21/1.99 app(app(g, 0), 1) 4.21/1.99 app(h, app(app(g, x0), x1)) 4.21/1.99 app(app(map, x0), nil) 4.21/1.99 app(app(map, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(filter, x0), nil) 4.21/1.99 app(app(filter, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(app(app(filter2, true), x0), x1), x2) 4.21/1.99 app(app(app(app(filter2, false), x0), x1), x2) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (3) DependencyGraphProof (EQUIVALENT) 4.21/1.99 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 15 less nodes. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (4) 4.21/1.99 Complex Obligation (AND) 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (5) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 APP(h, app(app(g, x), y)) -> APP(h, x) 4.21/1.99 4.21/1.99 The TRS R consists of the following rules: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> app(app(app(app(f, app(app(g, x), y)), app(app(g, x), y)), app(app(g, x), y)), app(h, x)) 4.21/1.99 app(app(g, 0), 1) -> 0 4.21/1.99 app(app(g, 0), 1) -> 1 4.21/1.99 app(h, app(app(g, x), y)) -> app(h, x) 4.21/1.99 app(app(map, fun), nil) -> nil 4.21/1.99 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.21/1.99 app(app(filter, fun), nil) -> nil 4.21/1.99 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.21/1.99 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.21/1.99 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.21/1.99 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x0), x1)), x2) 4.21/1.99 app(app(g, 0), 1) 4.21/1.99 app(h, app(app(g, x0), x1)) 4.21/1.99 app(app(map, x0), nil) 4.21/1.99 app(app(map, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(filter, x0), nil) 4.21/1.99 app(app(filter, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(app(app(filter2, true), x0), x1), x2) 4.21/1.99 app(app(app(app(filter2, false), x0), x1), x2) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (6) UsableRulesProof (EQUIVALENT) 4.21/1.99 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. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (7) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 APP(h, app(app(g, x), y)) -> APP(h, x) 4.21/1.99 4.21/1.99 R is empty. 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x0), x1)), x2) 4.21/1.99 app(app(g, 0), 1) 4.21/1.99 app(h, app(app(g, x0), x1)) 4.21/1.99 app(app(map, x0), nil) 4.21/1.99 app(app(map, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(filter, x0), nil) 4.21/1.99 app(app(filter, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(app(app(filter2, true), x0), x1), x2) 4.21/1.99 app(app(app(app(filter2, false), x0), x1), x2) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (8) ATransformationProof (EQUIVALENT) 4.21/1.99 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (9) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 h1(g(x, y)) -> h1(x) 4.21/1.99 4.21/1.99 R is empty. 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 f(0, 1, g(x0, x1), x2) 4.21/1.99 g(0, 1) 4.21/1.99 h(g(x0, x1)) 4.21/1.99 map(x0, nil) 4.21/1.99 map(x0, cons(x1, x2)) 4.21/1.99 filter(x0, nil) 4.21/1.99 filter(x0, cons(x1, x2)) 4.21/1.99 filter2(true, x0, x1, x2) 4.21/1.99 filter2(false, x0, x1, x2) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (10) QReductionProof (EQUIVALENT) 4.21/1.99 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.21/1.99 4.21/1.99 f(0, 1, g(x0, x1), x2) 4.21/1.99 h(g(x0, x1)) 4.21/1.99 map(x0, nil) 4.21/1.99 map(x0, cons(x1, x2)) 4.21/1.99 filter(x0, nil) 4.21/1.99 filter(x0, cons(x1, x2)) 4.21/1.99 filter2(true, x0, x1, x2) 4.21/1.99 filter2(false, x0, x1, x2) 4.21/1.99 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (11) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 h1(g(x, y)) -> h1(x) 4.21/1.99 4.21/1.99 R is empty. 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 g(0, 1) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (12) QDPSizeChangeProof (EQUIVALENT) 4.21/1.99 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. 4.21/1.99 4.21/1.99 From the DPs we obtained the following set of size-change graphs: 4.21/1.99 *h1(g(x, y)) -> h1(x) 4.21/1.99 The graph contains the following edges 1 > 1 4.21/1.99 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (13) 4.21/1.99 YES 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (14) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 4.21/1.99 The TRS R consists of the following rules: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x), y)), z) -> app(app(app(app(f, app(app(g, x), y)), app(app(g, x), y)), app(app(g, x), y)), app(h, x)) 4.21/1.99 app(app(g, 0), 1) -> 0 4.21/1.99 app(app(g, 0), 1) -> 1 4.21/1.99 app(h, app(app(g, x), y)) -> app(h, x) 4.21/1.99 app(app(map, fun), nil) -> nil 4.21/1.99 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.21/1.99 app(app(filter, fun), nil) -> nil 4.21/1.99 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.21/1.99 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.21/1.99 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.21/1.99 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x0), x1)), x2) 4.21/1.99 app(app(g, 0), 1) 4.21/1.99 app(h, app(app(g, x0), x1)) 4.21/1.99 app(app(map, x0), nil) 4.21/1.99 app(app(map, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(filter, x0), nil) 4.21/1.99 app(app(filter, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(app(app(filter2, true), x0), x1), x2) 4.21/1.99 app(app(app(app(filter2, false), x0), x1), x2) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (15) UsableRulesProof (EQUIVALENT) 4.21/1.99 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. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (16) 4.21/1.99 Obligation: 4.21/1.99 Q DP problem: 4.21/1.99 The TRS P consists of the following rules: 4.21/1.99 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.21/1.99 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 4.21/1.99 R is empty. 4.21/1.99 The set Q consists of the following terms: 4.21/1.99 4.21/1.99 app(app(app(app(f, 0), 1), app(app(g, x0), x1)), x2) 4.21/1.99 app(app(g, 0), 1) 4.21/1.99 app(h, app(app(g, x0), x1)) 4.21/1.99 app(app(map, x0), nil) 4.21/1.99 app(app(map, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(filter, x0), nil) 4.21/1.99 app(app(filter, x0), app(app(cons, x1), x2)) 4.21/1.99 app(app(app(app(filter2, true), x0), x1), x2) 4.21/1.99 app(app(app(app(filter2, false), x0), x1), x2) 4.21/1.99 4.21/1.99 We have to consider all minimal (P,Q,R)-chains. 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (17) QDPSizeChangeProof (EQUIVALENT) 4.21/1.99 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. 4.21/1.99 4.21/1.99 From the DPs we obtained the following set of size-change graphs: 4.21/1.99 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 The graph contains the following edges 1 > 1, 2 > 2 4.21/1.99 4.21/1.99 4.21/1.99 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.21/1.99 The graph contains the following edges 1 > 1, 2 > 2 4.21/1.99 4.21/1.99 4.21/1.99 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.21/1.99 The graph contains the following edges 1 >= 1, 2 > 2 4.21/1.99 4.21/1.99 4.21/1.99 *APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 The graph contains the following edges 2 >= 2 4.21/1.99 4.21/1.99 4.21/1.99 *APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.21/1.99 The graph contains the following edges 2 >= 2 4.21/1.99 4.21/1.99 4.21/1.99 ---------------------------------------- 4.21/1.99 4.21/1.99 (18) 4.21/1.99 YES 4.41/2.01 EOF