4.45/2.01 YES 4.45/2.02 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.45/2.02 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.45/2.02 4.45/2.02 4.45/2.02 Termination w.r.t. Q of the given QTRS could be proven: 4.45/2.02 4.45/2.02 (0) QTRS 4.45/2.02 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 4.45/2.02 (2) QDP 4.45/2.02 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.45/2.02 (4) AND 4.45/2.02 (5) QDP 4.45/2.02 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.45/2.02 (7) QDP 4.45/2.02 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.45/2.02 (9) QDP 4.45/2.02 (10) QReductionProof [EQUIVALENT, 0 ms] 4.45/2.02 (11) QDP 4.45/2.02 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.45/2.02 (13) YES 4.45/2.02 (14) QDP 4.45/2.02 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.45/2.02 (16) QDP 4.45/2.02 (17) ATransformationProof [EQUIVALENT, 0 ms] 4.45/2.02 (18) QDP 4.45/2.02 (19) QReductionProof [EQUIVALENT, 0 ms] 4.45/2.02 (20) QDP 4.45/2.02 (21) TransformationProof [EQUIVALENT, 0 ms] 4.45/2.02 (22) QDP 4.45/2.02 (23) DependencyGraphProof [EQUIVALENT, 0 ms] 4.45/2.02 (24) TRUE 4.45/2.02 (25) QDP 4.45/2.02 (26) UsableRulesProof [EQUIVALENT, 0 ms] 4.45/2.02 (27) QDP 4.45/2.02 (28) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.45/2.02 (29) YES 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (0) 4.45/2.02 Obligation: 4.45/2.02 Q restricted rewrite system: 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 app(g, app(s, x)) -> app(s, app(g, x)) 4.45/2.02 app(g, 0) -> 0 4.45/2.02 app(app(map, fun), nil) -> nil 4.45/2.02 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.45/2.02 app(app(filter, fun), nil) -> nil 4.45/2.02 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.45/2.02 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.45/2.02 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (1) DependencyPairsProof (EQUIVALENT) 4.45/2.02 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (2) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(app(f, app(g, app(s, 0))), y) 4.45/2.02 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(f, app(g, app(s, 0))) 4.45/2.02 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(g, app(s, 0)) 4.45/2.02 APP(g, app(s, x)) -> APP(s, app(g, x)) 4.45/2.02 APP(g, app(s, x)) -> APP(g, x) 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) 4.45/2.02 APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) 4.45/2.02 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) 4.45/2.02 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 app(g, app(s, x)) -> app(s, app(g, x)) 4.45/2.02 app(g, 0) -> 0 4.45/2.02 app(app(map, fun), nil) -> nil 4.45/2.02 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.45/2.02 app(app(filter, fun), nil) -> nil 4.45/2.02 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.45/2.02 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.45/2.02 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (3) DependencyGraphProof (EQUIVALENT) 4.45/2.02 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 14 less nodes. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (4) 4.45/2.02 Complex Obligation (AND) 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (5) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(g, app(s, x)) -> APP(g, x) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 app(g, app(s, x)) -> app(s, app(g, x)) 4.45/2.02 app(g, 0) -> 0 4.45/2.02 app(app(map, fun), nil) -> nil 4.45/2.02 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.45/2.02 app(app(filter, fun), nil) -> nil 4.45/2.02 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.45/2.02 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.45/2.02 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (6) UsableRulesProof (EQUIVALENT) 4.45/2.02 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. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (7) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(g, app(s, x)) -> APP(g, x) 4.45/2.02 4.45/2.02 R is empty. 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (8) ATransformationProof (EQUIVALENT) 4.45/2.02 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (9) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 g1(s(x)) -> g1(x) 4.45/2.02 4.45/2.02 R is empty. 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 f(g(x0), s(0), x1) 4.45/2.02 g(s(x0)) 4.45/2.02 g(0) 4.45/2.02 map(x0, nil) 4.45/2.02 map(x0, cons(x1, x2)) 4.45/2.02 filter(x0, nil) 4.45/2.02 filter(x0, cons(x1, x2)) 4.45/2.02 filter2(true, x0, x1, x2) 4.45/2.02 filter2(false, x0, x1, x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (10) QReductionProof (EQUIVALENT) 4.45/2.02 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.45/2.02 4.45/2.02 f(g(x0), s(0), x1) 4.45/2.02 g(s(x0)) 4.45/2.02 g(0) 4.45/2.02 map(x0, nil) 4.45/2.02 map(x0, cons(x1, x2)) 4.45/2.02 filter(x0, nil) 4.45/2.02 filter(x0, cons(x1, x2)) 4.45/2.02 filter2(true, x0, x1, x2) 4.45/2.02 filter2(false, x0, x1, x2) 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (11) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 g1(s(x)) -> g1(x) 4.45/2.02 4.45/2.02 R is empty. 4.45/2.02 Q is empty. 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (12) QDPSizeChangeProof (EQUIVALENT) 4.45/2.02 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. 4.45/2.02 4.45/2.02 From the DPs we obtained the following set of size-change graphs: 4.45/2.02 *g1(s(x)) -> g1(x) 4.45/2.02 The graph contains the following edges 1 > 1 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (13) 4.45/2.02 YES 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (14) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 app(g, app(s, x)) -> app(s, app(g, x)) 4.45/2.02 app(g, 0) -> 0 4.45/2.02 app(app(map, fun), nil) -> nil 4.45/2.02 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.45/2.02 app(app(filter, fun), nil) -> nil 4.45/2.02 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.45/2.02 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.45/2.02 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (15) UsableRulesProof (EQUIVALENT) 4.45/2.02 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. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (16) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(app(app(f, app(g, x)), app(s, 0)), y) -> APP(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 app(g, app(s, x)) -> app(s, app(g, x)) 4.45/2.02 app(g, 0) -> 0 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (17) ATransformationProof (EQUIVALENT) 4.45/2.02 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (18) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 f1(g(x), s(0), y) -> f1(g(s(0)), y, g(x)) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 g(s(x)) -> s(g(x)) 4.45/2.02 g(0) -> 0 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 f(g(x0), s(0), x1) 4.45/2.02 g(s(x0)) 4.45/2.02 g(0) 4.45/2.02 map(x0, nil) 4.45/2.02 map(x0, cons(x1, x2)) 4.45/2.02 filter(x0, nil) 4.45/2.02 filter(x0, cons(x1, x2)) 4.45/2.02 filter2(true, x0, x1, x2) 4.45/2.02 filter2(false, x0, x1, x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (19) QReductionProof (EQUIVALENT) 4.45/2.02 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.45/2.02 4.45/2.02 f(g(x0), s(0), x1) 4.45/2.02 map(x0, nil) 4.45/2.02 map(x0, cons(x1, x2)) 4.45/2.02 filter(x0, nil) 4.45/2.02 filter(x0, cons(x1, x2)) 4.45/2.02 filter2(true, x0, x1, x2) 4.45/2.02 filter2(false, x0, x1, x2) 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (20) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 f1(g(x), s(0), y) -> f1(g(s(0)), y, g(x)) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 g(s(x)) -> s(g(x)) 4.45/2.02 g(0) -> 0 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 g(s(x0)) 4.45/2.02 g(0) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (21) TransformationProof (EQUIVALENT) 4.45/2.02 By rewriting [LPAR04] the rule f1(g(x), s(0), y) -> f1(g(s(0)), y, g(x)) at position [0] we obtained the following new rules [LPAR04]: 4.45/2.02 4.45/2.02 (f1(g(x), s(0), y) -> f1(s(g(0)), y, g(x)),f1(g(x), s(0), y) -> f1(s(g(0)), y, g(x))) 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (22) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 f1(g(x), s(0), y) -> f1(s(g(0)), y, g(x)) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 g(s(x)) -> s(g(x)) 4.45/2.02 g(0) -> 0 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 g(s(x0)) 4.45/2.02 g(0) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (23) DependencyGraphProof (EQUIVALENT) 4.45/2.02 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (24) 4.45/2.02 TRUE 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (25) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 4.45/2.02 The TRS R consists of the following rules: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x)), app(s, 0)), y) -> app(app(app(f, app(g, app(s, 0))), y), app(g, x)) 4.45/2.02 app(g, app(s, x)) -> app(s, app(g, x)) 4.45/2.02 app(g, 0) -> 0 4.45/2.02 app(app(map, fun), nil) -> nil 4.45/2.02 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.45/2.02 app(app(filter, fun), nil) -> nil 4.45/2.02 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.45/2.02 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.45/2.02 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.45/2.02 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (26) UsableRulesProof (EQUIVALENT) 4.45/2.02 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. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (27) 4.45/2.02 Obligation: 4.45/2.02 Q DP problem: 4.45/2.02 The TRS P consists of the following rules: 4.45/2.02 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.45/2.02 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 4.45/2.02 R is empty. 4.45/2.02 The set Q consists of the following terms: 4.45/2.02 4.45/2.02 app(app(app(f, app(g, x0)), app(s, 0)), x1) 4.45/2.02 app(g, app(s, x0)) 4.45/2.02 app(g, 0) 4.45/2.02 app(app(map, x0), nil) 4.45/2.02 app(app(map, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(filter, x0), nil) 4.45/2.02 app(app(filter, x0), app(app(cons, x1), x2)) 4.45/2.02 app(app(app(app(filter2, true), x0), x1), x2) 4.45/2.02 app(app(app(app(filter2, false), x0), x1), x2) 4.45/2.02 4.45/2.02 We have to consider all minimal (P,Q,R)-chains. 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (28) QDPSizeChangeProof (EQUIVALENT) 4.45/2.02 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. 4.45/2.02 4.45/2.02 From the DPs we obtained the following set of size-change graphs: 4.45/2.02 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 The graph contains the following edges 1 > 1, 2 > 2 4.45/2.02 4.45/2.02 4.45/2.02 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.45/2.02 The graph contains the following edges 1 > 1, 2 > 2 4.45/2.02 4.45/2.02 4.45/2.02 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.45/2.02 The graph contains the following edges 1 >= 1, 2 > 2 4.45/2.02 4.45/2.02 4.45/2.02 *APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 The graph contains the following edges 2 >= 2 4.45/2.02 4.45/2.02 4.45/2.02 *APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.45/2.02 The graph contains the following edges 2 >= 2 4.45/2.02 4.45/2.02 4.45/2.02 ---------------------------------------- 4.45/2.02 4.45/2.02 (29) 4.45/2.02 YES 4.57/2.08 EOF