4.48/1.93 YES 4.55/1.95 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.55/1.95 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.55/1.95 4.55/1.95 4.55/1.95 Termination w.r.t. Q of the given QTRS could be proven: 4.55/1.95 4.55/1.95 (0) QTRS 4.55/1.95 (1) DependencyPairsProof [EQUIVALENT, 7 ms] 4.55/1.95 (2) QDP 4.55/1.95 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.55/1.95 (4) AND 4.55/1.95 (5) QDP 4.55/1.95 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.55/1.95 (7) QDP 4.55/1.95 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.55/1.95 (9) QDP 4.55/1.95 (10) QReductionProof [EQUIVALENT, 0 ms] 4.55/1.95 (11) QDP 4.55/1.95 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.55/1.95 (13) YES 4.55/1.95 (14) QDP 4.55/1.95 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.55/1.95 (16) QDP 4.55/1.95 (17) ATransformationProof [EQUIVALENT, 0 ms] 4.55/1.95 (18) QDP 4.55/1.95 (19) QReductionProof [EQUIVALENT, 0 ms] 4.55/1.95 (20) QDP 4.55/1.95 (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.55/1.95 (22) YES 4.55/1.95 (23) QDP 4.55/1.95 (24) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.55/1.95 (25) YES 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (0) 4.55/1.95 Obligation: 4.55/1.95 Q restricted rewrite system: 4.55/1.95 The TRS R consists of the following rules: 4.55/1.95 4.55/1.95 app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) 4.55/1.95 app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 app(app(app(f, app(c, x)), x), y) -> app(c, y) 4.55/1.95 app(app(g, x), y) -> x 4.55/1.95 app(app(g, x), y) -> y 4.55/1.95 app(app(map, fun), nil) -> nil 4.55/1.95 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.55/1.95 app(app(filter, fun), nil) -> nil 4.55/1.95 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.55/1.95 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.55/1.95 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (1) DependencyPairsProof (EQUIVALENT) 4.55/1.95 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (2) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), y), app(app(app(f, y), x), y)) 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(f, y), y) 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(f, y) 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(f, y), x) 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(app(f, x), app(s, app(c, y))) 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(f, x) 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(s, app(c, y)) 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(c, y) 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(c, z) 4.55/1.95 APP(app(app(f, app(c, x)), x), y) -> APP(c, y) 4.55/1.95 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.55/1.95 APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) 4.55/1.95 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.55/1.95 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.55/1.95 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) 4.55/1.95 APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) 4.55/1.95 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.55/1.95 APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) 4.55/1.95 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.55/1.95 APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) 4.55/1.95 4.55/1.95 The TRS R consists of the following rules: 4.55/1.95 4.55/1.95 app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) 4.55/1.95 app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 app(app(app(f, app(c, x)), x), y) -> app(c, y) 4.55/1.95 app(app(g, x), y) -> x 4.55/1.95 app(app(g, x), y) -> y 4.55/1.95 app(app(map, fun), nil) -> nil 4.55/1.95 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.55/1.95 app(app(filter, fun), nil) -> nil 4.55/1.95 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.55/1.95 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.55/1.95 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (3) DependencyGraphProof (EQUIVALENT) 4.55/1.95 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 19 less nodes. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (4) 4.55/1.95 Complex Obligation (AND) 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (5) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 4.55/1.95 The TRS R consists of the following rules: 4.55/1.95 4.55/1.95 app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) 4.55/1.95 app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 app(app(app(f, app(c, x)), x), y) -> app(c, y) 4.55/1.95 app(app(g, x), y) -> x 4.55/1.95 app(app(g, x), y) -> y 4.55/1.95 app(app(map, fun), nil) -> nil 4.55/1.95 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.55/1.95 app(app(filter, fun), nil) -> nil 4.55/1.95 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.55/1.95 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.55/1.95 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (6) UsableRulesProof (EQUIVALENT) 4.55/1.95 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. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (7) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 APP(app(app(f, app(s, x)), y), z) -> APP(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 4.55/1.95 R is empty. 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (8) ATransformationProof (EQUIVALENT) 4.55/1.95 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (9) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 f1(s(x), y, z) -> f1(x, s(c(y)), c(z)) 4.55/1.95 4.55/1.95 R is empty. 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 f(x0, c(x0), c(x1)) 4.55/1.95 f(s(x0), x1, x2) 4.55/1.95 f(c(x0), x0, x1) 4.55/1.95 g(x0, x1) 4.55/1.95 map(x0, nil) 4.55/1.95 map(x0, cons(x1, x2)) 4.55/1.95 filter(x0, nil) 4.55/1.95 filter(x0, cons(x1, x2)) 4.55/1.95 filter2(true, x0, x1, x2) 4.55/1.95 filter2(false, x0, x1, x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (10) QReductionProof (EQUIVALENT) 4.55/1.95 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.55/1.95 4.55/1.95 f(x0, c(x0), c(x1)) 4.55/1.95 f(s(x0), x1, x2) 4.55/1.95 f(c(x0), x0, x1) 4.55/1.95 g(x0, x1) 4.55/1.95 map(x0, nil) 4.55/1.95 map(x0, cons(x1, x2)) 4.55/1.95 filter(x0, nil) 4.55/1.95 filter(x0, cons(x1, x2)) 4.55/1.95 filter2(true, x0, x1, x2) 4.55/1.95 filter2(false, x0, x1, x2) 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (11) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 f1(s(x), y, z) -> f1(x, s(c(y)), c(z)) 4.55/1.95 4.55/1.95 R is empty. 4.55/1.95 Q is empty. 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (12) QDPSizeChangeProof (EQUIVALENT) 4.55/1.95 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. 4.55/1.95 4.55/1.95 From the DPs we obtained the following set of size-change graphs: 4.55/1.95 *f1(s(x), y, z) -> f1(x, s(c(y)), c(z)) 4.55/1.95 The graph contains the following edges 1 > 1 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (13) 4.55/1.95 YES 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (14) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) 4.55/1.95 4.55/1.95 The TRS R consists of the following rules: 4.55/1.95 4.55/1.95 app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) 4.55/1.95 app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 app(app(app(f, app(c, x)), x), y) -> app(c, y) 4.55/1.95 app(app(g, x), y) -> x 4.55/1.95 app(app(g, x), y) -> y 4.55/1.95 app(app(map, fun), nil) -> nil 4.55/1.95 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.55/1.95 app(app(filter, fun), nil) -> nil 4.55/1.95 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.55/1.95 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.55/1.95 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (15) UsableRulesProof (EQUIVALENT) 4.55/1.95 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. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (16) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 APP(app(app(f, x), app(c, x)), app(c, y)) -> APP(app(app(f, y), x), y) 4.55/1.95 4.55/1.95 R is empty. 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (17) ATransformationProof (EQUIVALENT) 4.55/1.95 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (18) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 f1(x, c(x), c(y)) -> f1(y, x, y) 4.55/1.95 4.55/1.95 R is empty. 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 f(x0, c(x0), c(x1)) 4.55/1.95 f(s(x0), x1, x2) 4.55/1.95 f(c(x0), x0, x1) 4.55/1.95 g(x0, x1) 4.55/1.95 map(x0, nil) 4.55/1.95 map(x0, cons(x1, x2)) 4.55/1.95 filter(x0, nil) 4.55/1.95 filter(x0, cons(x1, x2)) 4.55/1.95 filter2(true, x0, x1, x2) 4.55/1.95 filter2(false, x0, x1, x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (19) QReductionProof (EQUIVALENT) 4.55/1.95 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.55/1.95 4.55/1.95 f(x0, c(x0), c(x1)) 4.55/1.95 f(s(x0), x1, x2) 4.55/1.95 f(c(x0), x0, x1) 4.55/1.95 g(x0, x1) 4.55/1.95 map(x0, nil) 4.55/1.95 map(x0, cons(x1, x2)) 4.55/1.95 filter(x0, nil) 4.55/1.95 filter(x0, cons(x1, x2)) 4.55/1.95 filter2(true, x0, x1, x2) 4.55/1.95 filter2(false, x0, x1, x2) 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (20) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 f1(x, c(x), c(y)) -> f1(y, x, y) 4.55/1.95 4.55/1.95 R is empty. 4.55/1.95 Q is empty. 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (21) QDPSizeChangeProof (EQUIVALENT) 4.55/1.95 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. 4.55/1.95 4.55/1.95 From the DPs we obtained the following set of size-change graphs: 4.55/1.95 *f1(x, c(x), c(y)) -> f1(y, x, y) 4.55/1.95 The graph contains the following edges 3 > 1, 1 >= 2, 2 > 2, 3 > 3 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (22) 4.55/1.95 YES 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (23) 4.55/1.95 Obligation: 4.55/1.95 Q DP problem: 4.55/1.95 The TRS P consists of the following rules: 4.55/1.95 4.55/1.95 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.55/1.95 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.55/1.95 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.55/1.95 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.55/1.95 4.55/1.95 The TRS R consists of the following rules: 4.55/1.95 4.55/1.95 app(app(app(f, x), app(c, x)), app(c, y)) -> app(app(app(f, y), y), app(app(app(f, y), x), y)) 4.55/1.95 app(app(app(f, app(s, x)), y), z) -> app(app(app(f, x), app(s, app(c, y))), app(c, z)) 4.55/1.95 app(app(app(f, app(c, x)), x), y) -> app(c, y) 4.55/1.95 app(app(g, x), y) -> x 4.55/1.95 app(app(g, x), y) -> y 4.55/1.95 app(app(map, fun), nil) -> nil 4.55/1.95 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.55/1.95 app(app(filter, fun), nil) -> nil 4.55/1.95 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.55/1.95 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.55/1.95 4.55/1.95 The set Q consists of the following terms: 4.55/1.95 4.55/1.95 app(app(app(f, x0), app(c, x0)), app(c, x1)) 4.55/1.95 app(app(app(f, app(s, x0)), x1), x2) 4.55/1.95 app(app(app(f, app(c, x0)), x0), x1) 4.55/1.95 app(app(g, x0), x1) 4.55/1.95 app(app(map, x0), nil) 4.55/1.95 app(app(map, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(filter, x0), nil) 4.55/1.95 app(app(filter, x0), app(app(cons, x1), x2)) 4.55/1.95 app(app(app(app(filter2, true), x0), x1), x2) 4.55/1.95 app(app(app(app(filter2, false), x0), x1), x2) 4.55/1.95 4.55/1.95 We have to consider all minimal (P,Q,R)-chains. 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (24) QDPSizeChangeProof (EQUIVALENT) 4.55/1.95 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. 4.55/1.95 4.55/1.95 From the DPs we obtained the following set of size-change graphs: 4.55/1.95 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.55/1.95 The graph contains the following edges 1 > 1, 2 > 2 4.55/1.95 4.55/1.95 4.55/1.95 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.55/1.95 The graph contains the following edges 1 > 1, 2 > 2 4.55/1.95 4.55/1.95 4.55/1.95 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.55/1.95 The graph contains the following edges 1 >= 1, 2 > 2 4.55/1.95 4.55/1.95 4.55/1.95 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.55/1.95 The graph contains the following edges 2 > 2 4.55/1.95 4.55/1.95 4.55/1.95 *APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.55/1.95 The graph contains the following edges 2 >= 2 4.55/1.95 4.55/1.95 4.55/1.95 *APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.55/1.95 The graph contains the following edges 2 >= 2 4.55/1.95 4.55/1.95 4.55/1.95 ---------------------------------------- 4.55/1.95 4.55/1.95 (25) 4.55/1.95 YES 4.55/1.98 EOF