4.62/2.03 YES 4.62/2.04 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.62/2.04 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.62/2.04 4.62/2.04 4.62/2.04 Termination w.r.t. Q of the given QTRS could be proven: 4.62/2.04 4.62/2.04 (0) QTRS 4.62/2.04 (1) DependencyPairsProof [EQUIVALENT, 15 ms] 4.62/2.04 (2) QDP 4.62/2.04 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.62/2.04 (4) AND 4.62/2.04 (5) QDP 4.62/2.04 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.62/2.04 (7) QDP 4.62/2.04 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.62/2.04 (9) QDP 4.62/2.04 (10) QReductionProof [EQUIVALENT, 0 ms] 4.62/2.04 (11) QDP 4.62/2.04 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.62/2.04 (13) YES 4.62/2.04 (14) QDP 4.62/2.04 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.62/2.04 (16) QDP 4.62/2.04 (17) ATransformationProof [EQUIVALENT, 2 ms] 4.62/2.04 (18) QDP 4.62/2.04 (19) QReductionProof [EQUIVALENT, 0 ms] 4.62/2.04 (20) QDP 4.62/2.04 (21) QDPOrderProof [EQUIVALENT, 19 ms] 4.62/2.04 (22) QDP 4.62/2.04 (23) PisEmptyProof [EQUIVALENT, 0 ms] 4.62/2.04 (24) YES 4.62/2.04 (25) QDP 4.62/2.04 (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.62/2.04 (27) YES 4.62/2.04 4.62/2.04 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (0) 4.62/2.04 Obligation: 4.62/2.04 Q restricted rewrite system: 4.62/2.04 The TRS R consists of the following rules: 4.62/2.04 4.62/2.04 app(f, 0) -> true 4.62/2.04 app(f, 1) -> false 4.62/2.04 app(f, app(s, x)) -> app(f, x) 4.62/2.04 app(app(app(if, true), x), y) -> x 4.62/2.04 app(app(app(if, false), x), y) -> y 4.62/2.04 app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.04 app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.04 app(app(map, fun), nil) -> nil 4.62/2.04 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.62/2.04 app(app(filter, fun), nil) -> nil 4.62/2.04 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.04 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.62/2.04 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.62/2.04 4.62/2.04 The set Q consists of the following terms: 4.62/2.04 4.62/2.04 app(f, 0) 4.62/2.04 app(f, 1) 4.62/2.04 app(f, app(s, x0)) 4.62/2.04 app(app(app(if, true), x0), x1) 4.62/2.04 app(app(app(if, false), x0), x1) 4.62/2.04 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.04 app(app(g, x0), app(c, x1)) 4.62/2.04 app(app(map, x0), nil) 4.62/2.04 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.04 app(app(filter, x0), nil) 4.62/2.04 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.04 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.04 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.04 4.62/2.04 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (1) DependencyPairsProof (EQUIVALENT) 4.62/2.04 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (2) 4.62/2.04 Obligation: 4.62/2.04 Q DP problem: 4.62/2.04 The TRS P consists of the following rules: 4.62/2.04 4.62/2.04 APP(f, app(s, x)) -> APP(f, x) 4.62/2.04 APP(app(g, app(s, x)), app(s, y)) -> APP(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.04 APP(app(g, app(s, x)), app(s, y)) -> APP(app(if, app(f, x)), app(s, x)) 4.62/2.04 APP(app(g, app(s, x)), app(s, y)) -> APP(if, app(f, x)) 4.62/2.04 APP(app(g, app(s, x)), app(s, y)) -> APP(f, x) 4.62/2.04 APP(app(g, x), app(c, y)) -> APP(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.04 APP(app(g, x), app(c, y)) -> APP(app(g, app(s, app(c, y))), y) 4.62/2.04 APP(app(g, x), app(c, y)) -> APP(g, app(s, app(c, y))) 4.62/2.04 APP(app(g, x), app(c, y)) -> APP(s, app(c, y)) 4.62/2.04 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.62/2.04 APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) 4.62/2.04 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.62/2.04 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.62/2.04 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.04 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) 4.62/2.04 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) 4.62/2.04 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) 4.62/2.04 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.62/2.04 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) 4.62/2.04 APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) 4.62/2.04 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.62/2.04 APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) 4.62/2.04 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.62/2.04 APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) 4.62/2.04 4.62/2.04 The TRS R consists of the following rules: 4.62/2.04 4.62/2.04 app(f, 0) -> true 4.62/2.04 app(f, 1) -> false 4.62/2.04 app(f, app(s, x)) -> app(f, x) 4.62/2.04 app(app(app(if, true), x), y) -> x 4.62/2.04 app(app(app(if, false), x), y) -> y 4.62/2.04 app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.04 app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.04 app(app(map, fun), nil) -> nil 4.62/2.04 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.62/2.04 app(app(filter, fun), nil) -> nil 4.62/2.04 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.04 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.62/2.04 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.62/2.04 4.62/2.04 The set Q consists of the following terms: 4.62/2.04 4.62/2.04 app(f, 0) 4.62/2.04 app(f, 1) 4.62/2.04 app(f, app(s, x0)) 4.62/2.04 app(app(app(if, true), x0), x1) 4.62/2.04 app(app(app(if, false), x0), x1) 4.62/2.04 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.04 app(app(g, x0), app(c, x1)) 4.62/2.04 app(app(map, x0), nil) 4.62/2.04 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.04 app(app(filter, x0), nil) 4.62/2.04 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.04 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.04 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.04 4.62/2.04 We have to consider all minimal (P,Q,R)-chains. 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (3) DependencyGraphProof (EQUIVALENT) 4.62/2.04 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 15 less nodes. 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (4) 4.62/2.04 Complex Obligation (AND) 4.62/2.04 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (5) 4.62/2.04 Obligation: 4.62/2.04 Q DP problem: 4.62/2.04 The TRS P consists of the following rules: 4.62/2.04 4.62/2.04 APP(f, app(s, x)) -> APP(f, x) 4.62/2.04 4.62/2.04 The TRS R consists of the following rules: 4.62/2.04 4.62/2.04 app(f, 0) -> true 4.62/2.04 app(f, 1) -> false 4.62/2.04 app(f, app(s, x)) -> app(f, x) 4.62/2.04 app(app(app(if, true), x), y) -> x 4.62/2.04 app(app(app(if, false), x), y) -> y 4.62/2.04 app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.04 app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.04 app(app(map, fun), nil) -> nil 4.62/2.04 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.62/2.04 app(app(filter, fun), nil) -> nil 4.62/2.04 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.04 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.62/2.04 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.62/2.04 4.62/2.04 The set Q consists of the following terms: 4.62/2.04 4.62/2.04 app(f, 0) 4.62/2.04 app(f, 1) 4.62/2.04 app(f, app(s, x0)) 4.62/2.04 app(app(app(if, true), x0), x1) 4.62/2.04 app(app(app(if, false), x0), x1) 4.62/2.04 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.04 app(app(g, x0), app(c, x1)) 4.62/2.04 app(app(map, x0), nil) 4.62/2.04 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.04 app(app(filter, x0), nil) 4.62/2.04 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.04 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.04 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.04 4.62/2.04 We have to consider all minimal (P,Q,R)-chains. 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (6) UsableRulesProof (EQUIVALENT) 4.62/2.04 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. 4.62/2.04 ---------------------------------------- 4.62/2.04 4.62/2.04 (7) 4.62/2.04 Obligation: 4.62/2.04 Q DP problem: 4.62/2.04 The TRS P consists of the following rules: 4.62/2.04 4.62/2.04 APP(f, app(s, x)) -> APP(f, x) 4.62/2.04 4.62/2.04 R is empty. 4.62/2.04 The set Q consists of the following terms: 4.62/2.04 4.62/2.04 app(f, 0) 4.62/2.04 app(f, 1) 4.62/2.04 app(f, app(s, x0)) 4.62/2.04 app(app(app(if, true), x0), x1) 4.62/2.04 app(app(app(if, false), x0), x1) 4.62/2.04 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.04 app(app(g, x0), app(c, x1)) 4.62/2.05 app(app(map, x0), nil) 4.62/2.05 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(filter, x0), nil) 4.62/2.05 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.05 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (8) ATransformationProof (EQUIVALENT) 4.62/2.05 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (9) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 f1(s(x)) -> f1(x) 4.62/2.05 4.62/2.05 R is empty. 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 f(0) 4.62/2.05 f(1) 4.62/2.05 f(s(x0)) 4.62/2.05 if(true, x0, x1) 4.62/2.05 if(false, x0, x1) 4.62/2.05 g(s(x0), s(x1)) 4.62/2.05 g(x0, c(x1)) 4.62/2.05 map(x0, nil) 4.62/2.05 map(x0, cons(x1, x2)) 4.62/2.05 filter(x0, nil) 4.62/2.05 filter(x0, cons(x1, x2)) 4.62/2.05 filter2(true, x0, x1, x2) 4.62/2.05 filter2(false, x0, x1, x2) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (10) QReductionProof (EQUIVALENT) 4.62/2.05 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.62/2.05 4.62/2.05 f(0) 4.62/2.05 f(1) 4.62/2.05 f(s(x0)) 4.62/2.05 if(true, x0, x1) 4.62/2.05 if(false, x0, x1) 4.62/2.05 g(s(x0), s(x1)) 4.62/2.05 g(x0, c(x1)) 4.62/2.05 map(x0, nil) 4.62/2.05 map(x0, cons(x1, x2)) 4.62/2.05 filter(x0, nil) 4.62/2.05 filter(x0, cons(x1, x2)) 4.62/2.05 filter2(true, x0, x1, x2) 4.62/2.05 filter2(false, x0, x1, x2) 4.62/2.05 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (11) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 f1(s(x)) -> f1(x) 4.62/2.05 4.62/2.05 R is empty. 4.62/2.05 Q is empty. 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (12) QDPSizeChangeProof (EQUIVALENT) 4.62/2.05 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. 4.62/2.05 4.62/2.05 From the DPs we obtained the following set of size-change graphs: 4.62/2.05 *f1(s(x)) -> f1(x) 4.62/2.05 The graph contains the following edges 1 > 1 4.62/2.05 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (13) 4.62/2.05 YES 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (14) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 APP(app(g, x), app(c, y)) -> APP(app(g, app(s, app(c, y))), y) 4.62/2.05 APP(app(g, x), app(c, y)) -> APP(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.05 4.62/2.05 The TRS R consists of the following rules: 4.62/2.05 4.62/2.05 app(f, 0) -> true 4.62/2.05 app(f, 1) -> false 4.62/2.05 app(f, app(s, x)) -> app(f, x) 4.62/2.05 app(app(app(if, true), x), y) -> x 4.62/2.05 app(app(app(if, false), x), y) -> y 4.62/2.05 app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.05 app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.05 app(app(map, fun), nil) -> nil 4.62/2.05 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.62/2.05 app(app(filter, fun), nil) -> nil 4.62/2.05 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.05 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.62/2.05 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.62/2.05 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 app(f, 0) 4.62/2.05 app(f, 1) 4.62/2.05 app(f, app(s, x0)) 4.62/2.05 app(app(app(if, true), x0), x1) 4.62/2.05 app(app(app(if, false), x0), x1) 4.62/2.05 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.05 app(app(g, x0), app(c, x1)) 4.62/2.05 app(app(map, x0), nil) 4.62/2.05 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(filter, x0), nil) 4.62/2.05 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.05 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (15) UsableRulesProof (EQUIVALENT) 4.62/2.05 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. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (16) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 APP(app(g, x), app(c, y)) -> APP(app(g, app(s, app(c, y))), y) 4.62/2.05 APP(app(g, x), app(c, y)) -> APP(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.05 4.62/2.05 The TRS R consists of the following rules: 4.62/2.05 4.62/2.05 app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.05 app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.05 app(f, 0) -> true 4.62/2.05 app(f, 1) -> false 4.62/2.05 app(f, app(s, x)) -> app(f, x) 4.62/2.05 app(app(app(if, true), x), y) -> x 4.62/2.05 app(app(app(if, false), x), y) -> y 4.62/2.05 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 app(f, 0) 4.62/2.05 app(f, 1) 4.62/2.05 app(f, app(s, x0)) 4.62/2.05 app(app(app(if, true), x0), x1) 4.62/2.05 app(app(app(if, false), x0), x1) 4.62/2.05 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.05 app(app(g, x0), app(c, x1)) 4.62/2.05 app(app(map, x0), nil) 4.62/2.05 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(filter, x0), nil) 4.62/2.05 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.05 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (17) ATransformationProof (EQUIVALENT) 4.62/2.05 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (18) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 g1(x, c(y)) -> g1(s(c(y)), y) 4.62/2.05 g1(x, c(y)) -> g1(x, g(s(c(y)), y)) 4.62/2.05 4.62/2.05 The TRS R consists of the following rules: 4.62/2.05 4.62/2.05 g(s(x), s(y)) -> if(f(x), s(x), s(y)) 4.62/2.05 g(x, c(y)) -> g(x, g(s(c(y)), y)) 4.62/2.05 f(0) -> true 4.62/2.05 f(1) -> false 4.62/2.05 f(s(x)) -> f(x) 4.62/2.05 if(true, x, y) -> x 4.62/2.05 if(false, x, y) -> y 4.62/2.05 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 f(0) 4.62/2.05 f(1) 4.62/2.05 f(s(x0)) 4.62/2.05 if(true, x0, x1) 4.62/2.05 if(false, x0, x1) 4.62/2.05 g(s(x0), s(x1)) 4.62/2.05 g(x0, c(x1)) 4.62/2.05 map(x0, nil) 4.62/2.05 map(x0, cons(x1, x2)) 4.62/2.05 filter(x0, nil) 4.62/2.05 filter(x0, cons(x1, x2)) 4.62/2.05 filter2(true, x0, x1, x2) 4.62/2.05 filter2(false, x0, x1, x2) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (19) QReductionProof (EQUIVALENT) 4.62/2.05 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.62/2.05 4.62/2.05 map(x0, nil) 4.62/2.05 map(x0, cons(x1, x2)) 4.62/2.05 filter(x0, nil) 4.62/2.05 filter(x0, cons(x1, x2)) 4.62/2.05 filter2(true, x0, x1, x2) 4.62/2.05 filter2(false, x0, x1, x2) 4.62/2.05 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (20) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 g1(x, c(y)) -> g1(s(c(y)), y) 4.62/2.05 g1(x, c(y)) -> g1(x, g(s(c(y)), y)) 4.62/2.05 4.62/2.05 The TRS R consists of the following rules: 4.62/2.05 4.62/2.05 g(s(x), s(y)) -> if(f(x), s(x), s(y)) 4.62/2.05 g(x, c(y)) -> g(x, g(s(c(y)), y)) 4.62/2.05 f(0) -> true 4.62/2.05 f(1) -> false 4.62/2.05 f(s(x)) -> f(x) 4.62/2.05 if(true, x, y) -> x 4.62/2.05 if(false, x, y) -> y 4.62/2.05 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 f(0) 4.62/2.05 f(1) 4.62/2.05 f(s(x0)) 4.62/2.05 if(true, x0, x1) 4.62/2.05 if(false, x0, x1) 4.62/2.05 g(s(x0), s(x1)) 4.62/2.05 g(x0, c(x1)) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (21) QDPOrderProof (EQUIVALENT) 4.62/2.05 We use the reduction pair processor [LPAR04,JAR06]. 4.62/2.05 4.62/2.05 4.62/2.05 The following pairs can be oriented strictly and are deleted. 4.62/2.05 4.62/2.05 g1(x, c(y)) -> g1(s(c(y)), y) 4.62/2.05 g1(x, c(y)) -> g1(x, g(s(c(y)), y)) 4.62/2.05 The remaining pairs can at least be oriented weakly. 4.62/2.05 Used ordering: Combined order from the following AFS and order. 4.62/2.05 g1(x1, x2) = x2 4.62/2.05 4.62/2.05 c(x1) = c(x1) 4.62/2.05 4.62/2.05 g(x1, x2) = g 4.62/2.05 4.62/2.05 if(x1, x2, x3) = if(x2, x3) 4.62/2.05 4.62/2.05 s(x1) = s 4.62/2.05 4.62/2.05 4.62/2.05 Knuth-Bendix order [KBO] with precedence:trivial 4.62/2.05 4.62/2.05 and weight map: 4.62/2.05 4.62/2.05 s=1 4.62/2.05 c_1=4 4.62/2.05 g=4 4.62/2.05 if_2=1 4.62/2.05 4.62/2.05 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 4.62/2.05 4.62/2.05 g(s(x), s(y)) -> if(f(x), s(x), s(y)) 4.62/2.05 g(x, c(y)) -> g(x, g(s(c(y)), y)) 4.62/2.05 if(true, x, y) -> x 4.62/2.05 if(false, x, y) -> y 4.62/2.05 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (22) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 P is empty. 4.62/2.05 The TRS R consists of the following rules: 4.62/2.05 4.62/2.05 g(s(x), s(y)) -> if(f(x), s(x), s(y)) 4.62/2.05 g(x, c(y)) -> g(x, g(s(c(y)), y)) 4.62/2.05 f(0) -> true 4.62/2.05 f(1) -> false 4.62/2.05 f(s(x)) -> f(x) 4.62/2.05 if(true, x, y) -> x 4.62/2.05 if(false, x, y) -> y 4.62/2.05 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 f(0) 4.62/2.05 f(1) 4.62/2.05 f(s(x0)) 4.62/2.05 if(true, x0, x1) 4.62/2.05 if(false, x0, x1) 4.62/2.05 g(s(x0), s(x1)) 4.62/2.05 g(x0, c(x1)) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (23) PisEmptyProof (EQUIVALENT) 4.62/2.05 The TRS P is empty. Hence, there is no (P,Q,R) chain. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (24) 4.62/2.05 YES 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (25) 4.62/2.05 Obligation: 4.62/2.05 Q DP problem: 4.62/2.05 The TRS P consists of the following rules: 4.62/2.05 4.62/2.05 APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.62/2.05 APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.62/2.05 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.05 APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.62/2.05 APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.62/2.05 APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.62/2.05 4.62/2.05 The TRS R consists of the following rules: 4.62/2.05 4.62/2.05 app(f, 0) -> true 4.62/2.05 app(f, 1) -> false 4.62/2.05 app(f, app(s, x)) -> app(f, x) 4.62/2.05 app(app(app(if, true), x), y) -> x 4.62/2.05 app(app(app(if, false), x), y) -> y 4.62/2.05 app(app(g, app(s, x)), app(s, y)) -> app(app(app(if, app(f, x)), app(s, x)), app(s, y)) 4.62/2.05 app(app(g, x), app(c, y)) -> app(app(g, x), app(app(g, app(s, app(c, y))), y)) 4.62/2.05 app(app(map, fun), nil) -> nil 4.62/2.05 app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) 4.62/2.05 app(app(filter, fun), nil) -> nil 4.62/2.05 app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.05 app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) 4.62/2.05 app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) 4.62/2.05 4.62/2.05 The set Q consists of the following terms: 4.62/2.05 4.62/2.05 app(f, 0) 4.62/2.05 app(f, 1) 4.62/2.05 app(f, app(s, x0)) 4.62/2.05 app(app(app(if, true), x0), x1) 4.62/2.05 app(app(app(if, false), x0), x1) 4.62/2.05 app(app(g, app(s, x0)), app(s, x1)) 4.62/2.05 app(app(g, x0), app(c, x1)) 4.62/2.05 app(app(map, x0), nil) 4.62/2.05 app(app(map, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(filter, x0), nil) 4.62/2.05 app(app(filter, x0), app(app(cons, x1), x2)) 4.62/2.05 app(app(app(app(filter2, true), x0), x1), x2) 4.62/2.05 app(app(app(app(filter2, false), x0), x1), x2) 4.62/2.05 4.62/2.05 We have to consider all minimal (P,Q,R)-chains. 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (26) QDPSizeChangeProof (EQUIVALENT) 4.62/2.05 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. 4.62/2.05 4.62/2.05 From the DPs we obtained the following set of size-change graphs: 4.62/2.05 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.62/2.05 The graph contains the following edges 1 > 1, 2 > 2 4.62/2.05 4.62/2.05 4.62/2.05 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) 4.62/2.05 The graph contains the following edges 1 > 1, 2 > 2 4.62/2.05 4.62/2.05 4.62/2.05 *APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) 4.62/2.05 The graph contains the following edges 1 >= 1, 2 > 2 4.62/2.05 4.62/2.05 4.62/2.05 *APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) 4.62/2.05 The graph contains the following edges 2 > 2 4.62/2.05 4.62/2.05 4.62/2.05 *APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) 4.62/2.05 The graph contains the following edges 2 >= 2 4.62/2.05 4.62/2.05 4.62/2.05 *APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) 4.62/2.05 The graph contains the following edges 2 >= 2 4.62/2.05 4.62/2.05 4.62/2.05 ---------------------------------------- 4.62/2.05 4.62/2.05 (27) 4.62/2.05 YES 4.86/2.08 EOF