4.08/1.90 YES 4.08/1.91 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.08/1.91 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.08/1.91 4.08/1.91 4.08/1.91 Termination w.r.t. Q of the given QTRS could be proven: 4.08/1.91 4.08/1.91 (0) QTRS 4.08/1.91 (1) DependencyPairsProof [EQUIVALENT, 21 ms] 4.08/1.91 (2) QDP 4.08/1.91 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.08/1.91 (4) AND 4.08/1.91 (5) QDP 4.08/1.91 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.08/1.91 (7) QDP 4.08/1.91 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.08/1.91 (9) QDP 4.08/1.91 (10) QReductionProof [EQUIVALENT, 0 ms] 4.08/1.91 (11) QDP 4.08/1.91 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.08/1.91 (13) YES 4.08/1.91 (14) QDP 4.08/1.91 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.08/1.91 (16) QDP 4.08/1.91 (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.08/1.91 (18) YES 4.08/1.91 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (0) 4.08/1.91 Obligation: 4.08/1.91 Q restricted rewrite system: 4.08/1.91 The TRS R consists of the following rules: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, y), y), app(g, x)) 4.08/1.91 app(g, app(s, x)) -> app(s, app(g, x)) 4.08/1.91 app(g, 0) -> 0 4.08/1.91 app(app(map, fun), nil) -> nil 4.08/1.91 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.08/1.91 app(app(filter, fun), nil) -> nil 4.08/1.91 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.08/1.91 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.08/1.91 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.08/1.91 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.08/1.91 app(g, app(s, x0)) 4.08/1.91 app(g, 0) 4.08/1.91 app(app(map, x0), nil) 4.08/1.91 app(app(map, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(filter, x0), nil) 4.08/1.91 app(app(filter, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(app(app(filter2, true), x0), x1), x2) 4.08/1.91 app(app(app(app(filter2, false), x0), x1), x2) 4.08/1.91 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (1) DependencyPairsProof (EQUIVALENT) 4.08/1.91 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (2) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(app(app(f, y), y), app(g, x)) 4.08/1.91 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(app(f, y), y) 4.08/1.91 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(f, y) 4.08/1.91 APP(g, app(s, x)) -> APP(s, app(g, x)) 4.08/1.91 APP(g, app(s, x)) -> APP(g, x) 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) 4.08/1.91 APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) 4.08/1.91 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) 4.08/1.91 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) 4.08/1.91 4.08/1.91 The TRS R consists of the following rules: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, y), y), app(g, x)) 4.08/1.91 app(g, app(s, x)) -> app(s, app(g, x)) 4.08/1.91 app(g, 0) -> 0 4.08/1.91 app(app(map, fun), nil) -> nil 4.08/1.91 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.08/1.91 app(app(filter, fun), nil) -> nil 4.08/1.91 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.08/1.91 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.08/1.91 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.08/1.91 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.08/1.91 app(g, app(s, x0)) 4.08/1.91 app(g, 0) 4.08/1.91 app(app(map, x0), nil) 4.08/1.91 app(app(map, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(filter, x0), nil) 4.08/1.91 app(app(filter, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(app(app(filter2, true), x0), x1), x2) 4.08/1.91 app(app(app(app(filter2, false), x0), x1), x2) 4.08/1.91 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (3) DependencyGraphProof (EQUIVALENT) 4.08/1.91 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 14 less nodes. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (4) 4.08/1.91 Complex Obligation (AND) 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (5) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 APP(g, app(s, x)) -> APP(g, x) 4.08/1.91 4.08/1.91 The TRS R consists of the following rules: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, y), y), app(g, x)) 4.08/1.91 app(g, app(s, x)) -> app(s, app(g, x)) 4.08/1.91 app(g, 0) -> 0 4.08/1.91 app(app(map, fun), nil) -> nil 4.08/1.91 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.08/1.91 app(app(filter, fun), nil) -> nil 4.08/1.91 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.08/1.91 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.08/1.91 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.08/1.91 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.08/1.91 app(g, app(s, x0)) 4.08/1.91 app(g, 0) 4.08/1.91 app(app(map, x0), nil) 4.08/1.91 app(app(map, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(filter, x0), nil) 4.08/1.91 app(app(filter, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(app(app(filter2, true), x0), x1), x2) 4.08/1.91 app(app(app(app(filter2, false), x0), x1), x2) 4.08/1.91 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (6) UsableRulesProof (EQUIVALENT) 4.08/1.91 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. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (7) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 APP(g, app(s, x)) -> APP(g, x) 4.08/1.91 4.08/1.91 R is empty. 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.08/1.91 app(g, app(s, x0)) 4.08/1.91 app(g, 0) 4.08/1.91 app(app(map, x0), nil) 4.08/1.91 app(app(map, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(filter, x0), nil) 4.08/1.91 app(app(filter, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(app(app(filter2, true), x0), x1), x2) 4.08/1.91 app(app(app(app(filter2, false), x0), x1), x2) 4.08/1.91 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (8) ATransformationProof (EQUIVALENT) 4.08/1.91 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (9) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 g1(s(x)) -> g1(x) 4.08/1.91 4.08/1.91 R is empty. 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 f(g(x0), s(0), x1) 4.08/1.91 g(s(x0)) 4.08/1.91 g(0) 4.08/1.91 map(x0, nil) 4.08/1.91 map(x0, cons(x1, x2)) 4.08/1.91 filter(x0, nil) 4.08/1.91 filter(x0, cons(x1, x2)) 4.08/1.91 filter2(true, x0, x1, x2) 4.08/1.91 filter2(false, x0, x1, x2) 4.08/1.91 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (10) QReductionProof (EQUIVALENT) 4.08/1.91 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.08/1.91 4.08/1.91 f(g(x0), s(0), x1) 4.08/1.91 g(s(x0)) 4.08/1.91 g(0) 4.08/1.91 map(x0, nil) 4.08/1.91 map(x0, cons(x1, x2)) 4.08/1.91 filter(x0, nil) 4.08/1.91 filter(x0, cons(x1, x2)) 4.08/1.91 filter2(true, x0, x1, x2) 4.08/1.91 filter2(false, x0, x1, x2) 4.08/1.91 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (11) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 g1(s(x)) -> g1(x) 4.08/1.91 4.08/1.91 R is empty. 4.08/1.91 Q is empty. 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (12) QDPSizeChangeProof (EQUIVALENT) 4.08/1.91 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. 4.08/1.91 4.08/1.91 From the DPs we obtained the following set of size-change graphs: 4.08/1.91 *g1(s(x)) -> g1(x) 4.08/1.91 The graph contains the following edges 1 > 1 4.08/1.91 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (13) 4.08/1.91 YES 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (14) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 4.08/1.91 The TRS R consists of the following rules: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, y), y), app(g, x)) 4.08/1.91 app(g, app(s, x)) -> app(s, app(g, x)) 4.08/1.91 app(g, 0) -> 0 4.08/1.91 app(app(map, fun), nil) -> nil 4.08/1.91 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.08/1.91 app(app(filter, fun), nil) -> nil 4.08/1.91 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.08/1.91 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.08/1.91 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.08/1.91 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.08/1.91 app(g, app(s, x0)) 4.08/1.91 app(g, 0) 4.08/1.91 app(app(map, x0), nil) 4.08/1.91 app(app(map, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(filter, x0), nil) 4.08/1.91 app(app(filter, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(app(app(filter2, true), x0), x1), x2) 4.08/1.91 app(app(app(app(filter2, false), x0), x1), x2) 4.08/1.91 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (15) UsableRulesProof (EQUIVALENT) 4.08/1.91 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. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (16) 4.08/1.91 Obligation: 4.08/1.91 Q DP problem: 4.08/1.91 The TRS P consists of the following rules: 4.08/1.91 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.08/1.91 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 4.08/1.91 R is empty. 4.08/1.91 The set Q consists of the following terms: 4.08/1.91 4.08/1.91 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.08/1.91 app(g, app(s, x0)) 4.08/1.91 app(g, 0) 4.08/1.91 app(app(map, x0), nil) 4.08/1.91 app(app(map, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(filter, x0), nil) 4.08/1.91 app(app(filter, x0), app(app(cons, x1), x2)) 4.08/1.91 app(app(app(app(filter2, true), x0), x1), x2) 4.08/1.91 app(app(app(app(filter2, false), x0), x1), x2) 4.08/1.91 4.08/1.91 We have to consider all minimal (P,Q,R)-chains. 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (17) QDPSizeChangeProof (EQUIVALENT) 4.08/1.91 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. 4.08/1.91 4.08/1.91 From the DPs we obtained the following set of size-change graphs: 4.08/1.91 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 The graph contains the following edges 1 > 1, 2 > 2 4.08/1.91 4.08/1.91 4.08/1.91 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.08/1.91 The graph contains the following edges 1 > 1, 2 > 2 4.08/1.91 4.08/1.91 4.08/1.91 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.08/1.91 The graph contains the following edges 1 >= 1, 2 > 2 4.08/1.91 4.08/1.91 4.08/1.91 *APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 The graph contains the following edges 2 >= 2 4.08/1.91 4.08/1.91 4.08/1.91 *APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.08/1.91 The graph contains the following edges 2 >= 2 4.08/1.91 4.08/1.91 4.08/1.91 ---------------------------------------- 4.08/1.91 4.08/1.91 (18) 4.08/1.91 YES 4.35/1.95 EOF