4.44/1.99 YES 4.44/2.00 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.44/2.00 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.44/2.00 4.44/2.00 4.44/2.00 Termination w.r.t. Q of the given QTRS could be proven: 4.44/2.00 4.44/2.00 (0) QTRS 4.44/2.00 (1) DependencyPairsProof [EQUIVALENT, 32 ms] 4.44/2.00 (2) QDP 4.44/2.00 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.44/2.00 (4) AND 4.44/2.00 (5) QDP 4.44/2.00 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.44/2.00 (7) QDP 4.44/2.00 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.44/2.00 (9) QDP 4.44/2.00 (10) QReductionProof [EQUIVALENT, 0 ms] 4.44/2.00 (11) QDP 4.44/2.00 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.44/2.00 (13) YES 4.44/2.00 (14) QDP 4.44/2.00 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.44/2.00 (16) QDP 4.44/2.00 (17) ATransformationProof [EQUIVALENT, 0 ms] 4.44/2.00 (18) QDP 4.44/2.00 (19) QReductionProof [EQUIVALENT, 0 ms] 4.44/2.00 (20) QDP 4.44/2.00 (21) MRRProof [EQUIVALENT, 12 ms] 4.44/2.00 (22) QDP 4.44/2.00 (23) DependencyGraphProof [EQUIVALENT, 0 ms] 4.44/2.00 (24) TRUE 4.44/2.00 (25) QDP 4.44/2.00 (26) UsableRulesProof [EQUIVALENT, 0 ms] 4.44/2.00 (27) QDP 4.44/2.00 (28) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.44/2.00 (29) YES 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (0) 4.44/2.00 Obligation: 4.44/2.00 Q restricted rewrite system: 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 app(half, 0) -> 0 4.44/2.00 app(half, app(s, 0)) -> 0 4.44/2.00 app(half, app(s, app(s, x))) -> app(s, app(half, x)) 4.44/2.00 app(bits, 0) -> 0 4.44/2.00 app(bits, app(s, x)) -> app(s, app(bits, app(half, app(s, x)))) 4.44/2.00 app(app(map, f), nil) -> nil 4.44/2.00 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.44/2.00 app(app(filter, f), nil) -> nil 4.44/2.00 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.44/2.00 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.44/2.00 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (1) DependencyPairsProof (EQUIVALENT) 4.44/2.00 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (2) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(half, app(s, app(s, x))) -> APP(s, app(half, x)) 4.44/2.00 APP(half, app(s, app(s, x))) -> APP(half, x) 4.44/2.00 APP(bits, app(s, x)) -> APP(s, app(bits, app(half, app(s, x)))) 4.44/2.00 APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x))) 4.44/2.00 APP(bits, app(s, x)) -> APP(half, app(s, x)) 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) 4.44/2.00 APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) 4.44/2.00 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) 4.44/2.00 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 app(half, 0) -> 0 4.44/2.00 app(half, app(s, 0)) -> 0 4.44/2.00 app(half, app(s, app(s, x))) -> app(s, app(half, x)) 4.44/2.00 app(bits, 0) -> 0 4.44/2.00 app(bits, app(s, x)) -> app(s, app(bits, app(half, app(s, x)))) 4.44/2.00 app(app(map, f), nil) -> nil 4.44/2.00 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.44/2.00 app(app(filter, f), nil) -> nil 4.44/2.00 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.44/2.00 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.44/2.00 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (3) DependencyGraphProof (EQUIVALENT) 4.44/2.00 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 13 less nodes. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (4) 4.44/2.00 Complex Obligation (AND) 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (5) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(half, app(s, app(s, x))) -> APP(half, x) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 app(half, 0) -> 0 4.44/2.00 app(half, app(s, 0)) -> 0 4.44/2.00 app(half, app(s, app(s, x))) -> app(s, app(half, x)) 4.44/2.00 app(bits, 0) -> 0 4.44/2.00 app(bits, app(s, x)) -> app(s, app(bits, app(half, app(s, x)))) 4.44/2.00 app(app(map, f), nil) -> nil 4.44/2.00 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.44/2.00 app(app(filter, f), nil) -> nil 4.44/2.00 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.44/2.00 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.44/2.00 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (6) UsableRulesProof (EQUIVALENT) 4.44/2.00 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. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (7) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(half, app(s, app(s, x))) -> APP(half, x) 4.44/2.00 4.44/2.00 R is empty. 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (8) ATransformationProof (EQUIVALENT) 4.44/2.00 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (9) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 half1(s(s(x))) -> half1(x) 4.44/2.00 4.44/2.00 R is empty. 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 half(0) 4.44/2.00 half(s(0)) 4.44/2.00 half(s(s(x0))) 4.44/2.00 bits(0) 4.44/2.00 bits(s(x0)) 4.44/2.00 map(x0, nil) 4.44/2.00 map(x0, cons(x1, x2)) 4.44/2.00 filter(x0, nil) 4.44/2.00 filter(x0, cons(x1, x2)) 4.44/2.00 filter2(true, x0, x1, x2) 4.44/2.00 filter2(false, x0, x1, x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (10) QReductionProof (EQUIVALENT) 4.44/2.00 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.44/2.00 4.44/2.00 half(0) 4.44/2.00 half(s(0)) 4.44/2.00 half(s(s(x0))) 4.44/2.00 bits(0) 4.44/2.00 bits(s(x0)) 4.44/2.00 map(x0, nil) 4.44/2.00 map(x0, cons(x1, x2)) 4.44/2.00 filter(x0, nil) 4.44/2.00 filter(x0, cons(x1, x2)) 4.44/2.00 filter2(true, x0, x1, x2) 4.44/2.00 filter2(false, x0, x1, x2) 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (11) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 half1(s(s(x))) -> half1(x) 4.44/2.00 4.44/2.00 R is empty. 4.44/2.00 Q is empty. 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (12) QDPSizeChangeProof (EQUIVALENT) 4.44/2.00 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. 4.44/2.00 4.44/2.00 From the DPs we obtained the following set of size-change graphs: 4.44/2.00 *half1(s(s(x))) -> half1(x) 4.44/2.00 The graph contains the following edges 1 > 1 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (13) 4.44/2.00 YES 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (14) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x))) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 app(half, 0) -> 0 4.44/2.00 app(half, app(s, 0)) -> 0 4.44/2.00 app(half, app(s, app(s, x))) -> app(s, app(half, x)) 4.44/2.00 app(bits, 0) -> 0 4.44/2.00 app(bits, app(s, x)) -> app(s, app(bits, app(half, app(s, x)))) 4.44/2.00 app(app(map, f), nil) -> nil 4.44/2.00 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.44/2.00 app(app(filter, f), nil) -> nil 4.44/2.00 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.44/2.00 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.44/2.00 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (15) UsableRulesProof (EQUIVALENT) 4.44/2.00 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. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (16) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(bits, app(s, x)) -> APP(bits, app(half, app(s, x))) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 app(half, app(s, 0)) -> 0 4.44/2.00 app(half, app(s, app(s, x))) -> app(s, app(half, x)) 4.44/2.00 app(half, 0) -> 0 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (17) ATransformationProof (EQUIVALENT) 4.44/2.00 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (18) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 bits1(s(x)) -> bits1(half(s(x))) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 half(s(0)) -> 0 4.44/2.00 half(s(s(x))) -> s(half(x)) 4.44/2.00 half(0) -> 0 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 half(0) 4.44/2.00 half(s(0)) 4.44/2.00 half(s(s(x0))) 4.44/2.00 bits(0) 4.44/2.00 bits(s(x0)) 4.44/2.00 map(x0, nil) 4.44/2.00 map(x0, cons(x1, x2)) 4.44/2.00 filter(x0, nil) 4.44/2.00 filter(x0, cons(x1, x2)) 4.44/2.00 filter2(true, x0, x1, x2) 4.44/2.00 filter2(false, x0, x1, x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (19) QReductionProof (EQUIVALENT) 4.44/2.00 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.44/2.00 4.44/2.00 bits(0) 4.44/2.00 bits(s(x0)) 4.44/2.00 map(x0, nil) 4.44/2.00 map(x0, cons(x1, x2)) 4.44/2.00 filter(x0, nil) 4.44/2.00 filter(x0, cons(x1, x2)) 4.44/2.00 filter2(true, x0, x1, x2) 4.44/2.00 filter2(false, x0, x1, x2) 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (20) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 bits1(s(x)) -> bits1(half(s(x))) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 half(s(0)) -> 0 4.44/2.00 half(s(s(x))) -> s(half(x)) 4.44/2.00 half(0) -> 0 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 half(0) 4.44/2.00 half(s(0)) 4.44/2.00 half(s(s(x0))) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (21) MRRProof (EQUIVALENT) 4.44/2.00 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 4.44/2.00 4.44/2.00 4.44/2.00 Strictly oriented rules of the TRS R: 4.44/2.00 4.44/2.00 half(s(0)) -> 0 4.44/2.00 half(s(s(x))) -> s(half(x)) 4.44/2.00 4.44/2.00 Used ordering: Polynomial interpretation [POLO]: 4.44/2.00 4.44/2.00 POL(0) = 0 4.44/2.00 POL(bits1(x_1)) = 2*x_1 4.44/2.00 POL(half(x_1)) = x_1 4.44/2.00 POL(s(x_1)) = 2 + x_1 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (22) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 bits1(s(x)) -> bits1(half(s(x))) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 half(0) -> 0 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 half(0) 4.44/2.00 half(s(0)) 4.44/2.00 half(s(s(x0))) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (23) DependencyGraphProof (EQUIVALENT) 4.44/2.00 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (24) 4.44/2.00 TRUE 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (25) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 4.44/2.00 The TRS R consists of the following rules: 4.44/2.00 4.44/2.00 app(half, 0) -> 0 4.44/2.00 app(half, app(s, 0)) -> 0 4.44/2.00 app(half, app(s, app(s, x))) -> app(s, app(half, x)) 4.44/2.00 app(bits, 0) -> 0 4.44/2.00 app(bits, app(s, x)) -> app(s, app(bits, app(half, app(s, x)))) 4.44/2.00 app(app(map, f), nil) -> nil 4.44/2.00 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.44/2.00 app(app(filter, f), nil) -> nil 4.44/2.00 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.44/2.00 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.44/2.00 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.44/2.00 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (26) UsableRulesProof (EQUIVALENT) 4.44/2.00 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. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (27) 4.44/2.00 Obligation: 4.44/2.00 Q DP problem: 4.44/2.00 The TRS P consists of the following rules: 4.44/2.00 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.44/2.00 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 4.44/2.00 R is empty. 4.44/2.00 The set Q consists of the following terms: 4.44/2.00 4.44/2.00 app(half, 0) 4.44/2.00 app(half, app(s, 0)) 4.44/2.00 app(half, app(s, app(s, x0))) 4.44/2.00 app(bits, 0) 4.44/2.00 app(bits, app(s, x0)) 4.44/2.00 app(app(map, x0), nil) 4.44/2.00 app(app(map, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(filter, x0), nil) 4.44/2.00 app(app(filter, x0), app(app(cons, x1), x2)) 4.44/2.00 app(app(app(app(filter2, true), x0), x1), x2) 4.44/2.00 app(app(app(app(filter2, false), x0), x1), x2) 4.44/2.00 4.44/2.00 We have to consider all minimal (P,Q,R)-chains. 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (28) QDPSizeChangeProof (EQUIVALENT) 4.44/2.00 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. 4.44/2.00 4.44/2.00 From the DPs we obtained the following set of size-change graphs: 4.44/2.00 *APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 The graph contains the following edges 1 > 1, 2 > 2 4.44/2.00 4.44/2.00 4.44/2.00 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.44/2.00 The graph contains the following edges 1 > 1, 2 > 2 4.44/2.00 4.44/2.00 4.44/2.00 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.44/2.00 The graph contains the following edges 1 >= 1, 2 > 2 4.44/2.00 4.44/2.00 4.44/2.00 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 The graph contains the following edges 2 >= 2 4.44/2.00 4.44/2.00 4.44/2.00 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.44/2.00 The graph contains the following edges 2 >= 2 4.44/2.00 4.44/2.00 4.44/2.00 ---------------------------------------- 4.44/2.00 4.44/2.00 (29) 4.44/2.00 YES 4.44/2.03 EOF