4.37/1.96 YES 4.37/1.97 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.37/1.97 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.37/1.97 4.37/1.97 4.37/1.97 Termination w.r.t. Q of the given QTRS could be proven: 4.37/1.97 4.37/1.97 (0) QTRS 4.37/1.97 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 4.37/1.97 (2) QDP 4.37/1.97 (3) DependencyGraphProof [EQUIVALENT, 6 ms] 4.37/1.97 (4) AND 4.37/1.97 (5) QDP 4.37/1.97 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.37/1.97 (7) QDP 4.37/1.97 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.37/1.97 (9) QDP 4.37/1.97 (10) QReductionProof [EQUIVALENT, 0 ms] 4.37/1.97 (11) QDP 4.37/1.97 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.37/1.97 (13) YES 4.37/1.97 (14) QDP 4.37/1.97 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.37/1.97 (16) QDP 4.37/1.97 (17) ATransformationProof [EQUIVALENT, 0 ms] 4.37/1.97 (18) QDP 4.37/1.97 (19) QReductionProof [EQUIVALENT, 0 ms] 4.37/1.97 (20) QDP 4.37/1.97 (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.37/1.97 (22) YES 4.37/1.97 (23) QDP 4.37/1.97 (24) UsableRulesProof [EQUIVALENT, 0 ms] 4.37/1.97 (25) QDP 4.37/1.97 (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.37/1.97 (27) YES 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (0) 4.37/1.97 Obligation: 4.37/1.97 Q restricted rewrite system: 4.37/1.97 The TRS R consists of the following rules: 4.37/1.97 4.37/1.97 app(intlist, nil) -> nil 4.37/1.97 app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y)) 4.37/1.97 app(app(int, 0), 0) -> app(app(cons, 0), nil) 4.37/1.97 app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y))) 4.37/1.97 app(app(int, app(s, x)), 0) -> nil 4.37/1.97 app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y)) 4.37/1.97 app(app(map, f), nil) -> nil 4.37/1.97 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.37/1.97 app(app(filter, f), nil) -> nil 4.37/1.97 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.37/1.97 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.37/1.97 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.37/1.97 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (1) DependencyPairsProof (EQUIVALENT) 4.37/1.97 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (2) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(intlist, app(app(cons, x), y)) -> APP(app(cons, app(s, x)), app(intlist, y)) 4.37/1.97 APP(intlist, app(app(cons, x), y)) -> APP(cons, app(s, x)) 4.37/1.97 APP(intlist, app(app(cons, x), y)) -> APP(s, x) 4.37/1.97 APP(intlist, app(app(cons, x), y)) -> APP(intlist, y) 4.37/1.97 APP(app(int, 0), 0) -> APP(app(cons, 0), nil) 4.37/1.97 APP(app(int, 0), 0) -> APP(cons, 0) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(app(cons, 0), app(app(int, app(s, 0)), app(s, y))) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(cons, 0) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(app(int, app(s, 0)), app(s, y)) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(int, app(s, 0)) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(s, 0) 4.37/1.97 APP(app(int, app(s, x)), app(s, y)) -> APP(intlist, app(app(int, x), y)) 4.37/1.97 APP(app(int, app(s, x)), app(s, y)) -> APP(app(int, x), y) 4.37/1.97 APP(app(int, app(s, x)), app(s, y)) -> APP(int, x) 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) 4.37/1.97 APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) 4.37/1.97 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) 4.37/1.97 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) 4.37/1.97 4.37/1.97 The TRS R consists of the following rules: 4.37/1.97 4.37/1.97 app(intlist, nil) -> nil 4.37/1.97 app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y)) 4.37/1.97 app(app(int, 0), 0) -> app(app(cons, 0), nil) 4.37/1.97 app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y))) 4.37/1.97 app(app(int, app(s, x)), 0) -> nil 4.37/1.97 app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y)) 4.37/1.97 app(app(map, f), nil) -> nil 4.37/1.97 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.37/1.97 app(app(filter, f), nil) -> nil 4.37/1.97 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.37/1.97 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.37/1.97 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.37/1.97 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (3) DependencyGraphProof (EQUIVALENT) 4.37/1.97 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 21 less nodes. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (4) 4.37/1.97 Complex Obligation (AND) 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (5) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(intlist, app(app(cons, x), y)) -> APP(intlist, y) 4.37/1.97 4.37/1.97 The TRS R consists of the following rules: 4.37/1.97 4.37/1.97 app(intlist, nil) -> nil 4.37/1.97 app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y)) 4.37/1.97 app(app(int, 0), 0) -> app(app(cons, 0), nil) 4.37/1.97 app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y))) 4.37/1.97 app(app(int, app(s, x)), 0) -> nil 4.37/1.97 app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y)) 4.37/1.97 app(app(map, f), nil) -> nil 4.37/1.97 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.37/1.97 app(app(filter, f), nil) -> nil 4.37/1.97 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.37/1.97 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.37/1.97 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.37/1.97 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (6) UsableRulesProof (EQUIVALENT) 4.37/1.97 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. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (7) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(intlist, app(app(cons, x), y)) -> APP(intlist, y) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (8) ATransformationProof (EQUIVALENT) 4.37/1.97 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (9) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 intlist1(cons(x, y)) -> intlist1(y) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 intlist(nil) 4.37/1.97 intlist(cons(x0, x1)) 4.37/1.97 int(0, 0) 4.37/1.97 int(0, s(x0)) 4.37/1.97 int(s(x0), 0) 4.37/1.97 int(s(x0), s(x1)) 4.37/1.97 map(x0, nil) 4.37/1.97 map(x0, cons(x1, x2)) 4.37/1.97 filter(x0, nil) 4.37/1.97 filter(x0, cons(x1, x2)) 4.37/1.97 filter2(true, x0, x1, x2) 4.37/1.97 filter2(false, x0, x1, x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (10) QReductionProof (EQUIVALENT) 4.37/1.97 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.37/1.97 4.37/1.97 intlist(nil) 4.37/1.97 intlist(cons(x0, x1)) 4.37/1.97 int(0, 0) 4.37/1.97 int(0, s(x0)) 4.37/1.97 int(s(x0), 0) 4.37/1.97 int(s(x0), s(x1)) 4.37/1.97 map(x0, nil) 4.37/1.97 map(x0, cons(x1, x2)) 4.37/1.97 filter(x0, nil) 4.37/1.97 filter(x0, cons(x1, x2)) 4.37/1.97 filter2(true, x0, x1, x2) 4.37/1.97 filter2(false, x0, x1, x2) 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (11) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 intlist1(cons(x, y)) -> intlist1(y) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 Q is empty. 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (12) QDPSizeChangeProof (EQUIVALENT) 4.37/1.97 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. 4.37/1.97 4.37/1.97 From the DPs we obtained the following set of size-change graphs: 4.37/1.97 *intlist1(cons(x, y)) -> intlist1(y) 4.37/1.97 The graph contains the following edges 1 > 1 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (13) 4.37/1.97 YES 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (14) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(app(int, app(s, x)), app(s, y)) -> APP(app(int, x), y) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(app(int, app(s, 0)), app(s, y)) 4.37/1.97 4.37/1.97 The TRS R consists of the following rules: 4.37/1.97 4.37/1.97 app(intlist, nil) -> nil 4.37/1.97 app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y)) 4.37/1.97 app(app(int, 0), 0) -> app(app(cons, 0), nil) 4.37/1.97 app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y))) 4.37/1.97 app(app(int, app(s, x)), 0) -> nil 4.37/1.97 app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y)) 4.37/1.97 app(app(map, f), nil) -> nil 4.37/1.97 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.37/1.97 app(app(filter, f), nil) -> nil 4.37/1.97 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.37/1.97 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.37/1.97 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.37/1.97 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (15) UsableRulesProof (EQUIVALENT) 4.37/1.97 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. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (16) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(app(int, app(s, x)), app(s, y)) -> APP(app(int, x), y) 4.37/1.97 APP(app(int, 0), app(s, y)) -> APP(app(int, app(s, 0)), app(s, y)) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (17) ATransformationProof (EQUIVALENT) 4.37/1.97 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (18) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 int1(s(x), s(y)) -> int1(x, y) 4.37/1.97 int1(0, s(y)) -> int1(s(0), s(y)) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 intlist(nil) 4.37/1.97 intlist(cons(x0, x1)) 4.37/1.97 int(0, 0) 4.37/1.97 int(0, s(x0)) 4.37/1.97 int(s(x0), 0) 4.37/1.97 int(s(x0), s(x1)) 4.37/1.97 map(x0, nil) 4.37/1.97 map(x0, cons(x1, x2)) 4.37/1.97 filter(x0, nil) 4.37/1.97 filter(x0, cons(x1, x2)) 4.37/1.97 filter2(true, x0, x1, x2) 4.37/1.97 filter2(false, x0, x1, x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (19) QReductionProof (EQUIVALENT) 4.37/1.97 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.37/1.97 4.37/1.97 intlist(nil) 4.37/1.97 intlist(cons(x0, x1)) 4.37/1.97 int(0, 0) 4.37/1.97 int(0, s(x0)) 4.37/1.97 int(s(x0), 0) 4.37/1.97 int(s(x0), s(x1)) 4.37/1.97 map(x0, nil) 4.37/1.97 map(x0, cons(x1, x2)) 4.37/1.97 filter(x0, nil) 4.37/1.97 filter(x0, cons(x1, x2)) 4.37/1.97 filter2(true, x0, x1, x2) 4.37/1.97 filter2(false, x0, x1, x2) 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (20) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 int1(s(x), s(y)) -> int1(x, y) 4.37/1.97 int1(0, s(y)) -> int1(s(0), s(y)) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 Q is empty. 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (21) QDPSizeChangeProof (EQUIVALENT) 4.37/1.97 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. 4.37/1.97 4.37/1.97 From the DPs we obtained the following set of size-change graphs: 4.37/1.97 *int1(s(x), s(y)) -> int1(x, y) 4.37/1.97 The graph contains the following edges 1 > 1, 2 > 2 4.37/1.97 4.37/1.97 4.37/1.97 *int1(0, s(y)) -> int1(s(0), s(y)) 4.37/1.97 The graph contains the following edges 2 >= 2 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (22) 4.37/1.97 YES 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (23) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 4.37/1.97 The TRS R consists of the following rules: 4.37/1.97 4.37/1.97 app(intlist, nil) -> nil 4.37/1.97 app(intlist, app(app(cons, x), y)) -> app(app(cons, app(s, x)), app(intlist, y)) 4.37/1.97 app(app(int, 0), 0) -> app(app(cons, 0), nil) 4.37/1.97 app(app(int, 0), app(s, y)) -> app(app(cons, 0), app(app(int, app(s, 0)), app(s, y))) 4.37/1.97 app(app(int, app(s, x)), 0) -> nil 4.37/1.97 app(app(int, app(s, x)), app(s, y)) -> app(intlist, app(app(int, x), y)) 4.37/1.97 app(app(map, f), nil) -> nil 4.37/1.97 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.37/1.97 app(app(filter, f), nil) -> nil 4.37/1.97 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.37/1.97 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.37/1.97 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.37/1.97 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (24) UsableRulesProof (EQUIVALENT) 4.37/1.97 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. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (25) 4.37/1.97 Obligation: 4.37/1.97 Q DP problem: 4.37/1.97 The TRS P consists of the following rules: 4.37/1.97 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.37/1.97 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 4.37/1.97 R is empty. 4.37/1.97 The set Q consists of the following terms: 4.37/1.97 4.37/1.97 app(intlist, nil) 4.37/1.97 app(intlist, app(app(cons, x0), x1)) 4.37/1.97 app(app(int, 0), 0) 4.37/1.97 app(app(int, 0), app(s, x0)) 4.37/1.97 app(app(int, app(s, x0)), 0) 4.37/1.97 app(app(int, app(s, x0)), app(s, x1)) 4.37/1.97 app(app(map, x0), nil) 4.37/1.97 app(app(map, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(filter, x0), nil) 4.37/1.97 app(app(filter, x0), app(app(cons, x1), x2)) 4.37/1.97 app(app(app(app(filter2, true), x0), x1), x2) 4.37/1.97 app(app(app(app(filter2, false), x0), x1), x2) 4.37/1.97 4.37/1.97 We have to consider all minimal (P,Q,R)-chains. 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (26) QDPSizeChangeProof (EQUIVALENT) 4.37/1.97 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. 4.37/1.97 4.37/1.97 From the DPs we obtained the following set of size-change graphs: 4.37/1.97 *APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 The graph contains the following edges 1 > 1, 2 > 2 4.37/1.97 4.37/1.97 4.37/1.97 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.37/1.97 The graph contains the following edges 1 > 1, 2 > 2 4.37/1.97 4.37/1.97 4.37/1.97 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.37/1.97 The graph contains the following edges 1 >= 1, 2 > 2 4.37/1.97 4.37/1.97 4.37/1.97 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 The graph contains the following edges 2 >= 2 4.37/1.97 4.37/1.97 4.37/1.97 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.37/1.97 The graph contains the following edges 2 >= 2 4.37/1.97 4.37/1.97 4.37/1.97 ---------------------------------------- 4.37/1.97 4.37/1.97 (27) 4.37/1.97 YES 4.54/2.00 EOF