4.42/2.00 YES 4.42/2.01 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 4.42/2.01 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.42/2.01 4.42/2.01 4.42/2.01 Termination w.r.t. Q of the given QTRS could be proven: 4.42/2.01 4.42/2.01 (0) QTRS 4.42/2.01 (1) DependencyPairsProof [EQUIVALENT, 50 ms] 4.42/2.01 (2) QDP 4.42/2.01 (3) DependencyGraphProof [EQUIVALENT, 0 ms] 4.42/2.01 (4) AND 4.42/2.01 (5) QDP 4.42/2.01 (6) UsableRulesProof [EQUIVALENT, 0 ms] 4.42/2.01 (7) QDP 4.42/2.01 (8) ATransformationProof [EQUIVALENT, 0 ms] 4.42/2.01 (9) QDP 4.42/2.01 (10) QReductionProof [EQUIVALENT, 0 ms] 4.42/2.01 (11) QDP 4.42/2.01 (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.42/2.01 (13) YES 4.42/2.01 (14) QDP 4.42/2.01 (15) UsableRulesProof [EQUIVALENT, 0 ms] 4.42/2.01 (16) QDP 4.42/2.01 (17) QDPSizeChangeProof [EQUIVALENT, 0 ms] 4.42/2.01 (18) YES 4.42/2.01 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (0) 4.42/2.01 Obligation: 4.42/2.01 Q restricted rewrite system: 4.42/2.01 The TRS R consists of the following rules: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, y)), app(s, z)) -> 0 4.42/2.01 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) 4.42/2.01 app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 4.42/2.01 app(app(map, f), nil) -> nil 4.42/2.01 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.42/2.01 app(app(filter, f), nil) -> nil 4.42/2.01 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.42/2.01 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.42/2.01 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.42/2.01 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, x0)), app(s, x1)) 4.42/2.01 app(app(app(quot, app(s, x0)), app(s, x1)), x2) 4.42/2.01 app(app(app(quot, x0), 0), app(s, x1)) 4.42/2.01 app(app(map, x0), nil) 4.42/2.01 app(app(map, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(filter, x0), nil) 4.42/2.01 app(app(filter, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(app(app(filter2, true), x0), x1), x2) 4.42/2.01 app(app(app(app(filter2, false), x0), x1), x2) 4.42/2.01 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (1) DependencyPairsProof (EQUIVALENT) 4.42/2.01 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (2) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(app(quot, x), y), z) 4.42/2.01 APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(quot, x), y) 4.42/2.01 APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(quot, x) 4.42/2.01 APP(app(app(quot, x), 0), app(s, z)) -> APP(s, app(app(app(quot, x), app(s, z)), app(s, z))) 4.42/2.01 APP(app(app(quot, x), 0), app(s, z)) -> APP(app(app(quot, x), app(s, z)), app(s, z)) 4.42/2.01 APP(app(app(quot, x), 0), app(s, z)) -> APP(app(quot, x), app(s, z)) 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) 4.42/2.01 APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) 4.42/2.01 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) 4.42/2.01 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) 4.42/2.01 4.42/2.01 The TRS R consists of the following rules: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, y)), app(s, z)) -> 0 4.42/2.01 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) 4.42/2.01 app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 4.42/2.01 app(app(map, f), nil) -> nil 4.42/2.01 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.42/2.01 app(app(filter, f), nil) -> nil 4.42/2.01 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.42/2.01 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.42/2.01 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.42/2.01 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, x0)), app(s, x1)) 4.42/2.01 app(app(app(quot, app(s, x0)), app(s, x1)), x2) 4.42/2.01 app(app(app(quot, x0), 0), app(s, x1)) 4.42/2.01 app(app(map, x0), nil) 4.42/2.01 app(app(map, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(filter, x0), nil) 4.42/2.01 app(app(filter, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(app(app(filter2, true), x0), x1), x2) 4.42/2.01 app(app(app(app(filter2, false), x0), x1), x2) 4.42/2.01 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (3) DependencyGraphProof (EQUIVALENT) 4.42/2.01 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 14 less nodes. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (4) 4.42/2.01 Complex Obligation (AND) 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (5) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 APP(app(app(quot, x), 0), app(s, z)) -> APP(app(app(quot, x), app(s, z)), app(s, z)) 4.42/2.01 APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(app(quot, x), y), z) 4.42/2.01 4.42/2.01 The TRS R consists of the following rules: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, y)), app(s, z)) -> 0 4.42/2.01 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) 4.42/2.01 app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 4.42/2.01 app(app(map, f), nil) -> nil 4.42/2.01 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.42/2.01 app(app(filter, f), nil) -> nil 4.42/2.01 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.42/2.01 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.42/2.01 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.42/2.01 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, x0)), app(s, x1)) 4.42/2.01 app(app(app(quot, app(s, x0)), app(s, x1)), x2) 4.42/2.01 app(app(app(quot, x0), 0), app(s, x1)) 4.42/2.01 app(app(map, x0), nil) 4.42/2.01 app(app(map, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(filter, x0), nil) 4.42/2.01 app(app(filter, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(app(app(filter2, true), x0), x1), x2) 4.42/2.01 app(app(app(app(filter2, false), x0), x1), x2) 4.42/2.01 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (6) UsableRulesProof (EQUIVALENT) 4.42/2.01 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. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (7) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 APP(app(app(quot, x), 0), app(s, z)) -> APP(app(app(quot, x), app(s, z)), app(s, z)) 4.42/2.01 APP(app(app(quot, app(s, x)), app(s, y)), z) -> APP(app(app(quot, x), y), z) 4.42/2.01 4.42/2.01 R is empty. 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, x0)), app(s, x1)) 4.42/2.01 app(app(app(quot, app(s, x0)), app(s, x1)), x2) 4.42/2.01 app(app(app(quot, x0), 0), app(s, x1)) 4.42/2.01 app(app(map, x0), nil) 4.42/2.01 app(app(map, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(filter, x0), nil) 4.42/2.01 app(app(filter, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(app(app(filter2, true), x0), x1), x2) 4.42/2.01 app(app(app(app(filter2, false), x0), x1), x2) 4.42/2.01 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (8) ATransformationProof (EQUIVALENT) 4.42/2.01 We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (9) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 quot1(x, 0, s(z)) -> quot1(x, s(z), s(z)) 4.42/2.01 quot1(s(x), s(y), z) -> quot1(x, y, z) 4.42/2.01 4.42/2.01 R is empty. 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 quot(0, s(x0), s(x1)) 4.42/2.01 quot(s(x0), s(x1), x2) 4.42/2.01 quot(x0, 0, s(x1)) 4.42/2.01 map(x0, nil) 4.42/2.01 map(x0, cons(x1, x2)) 4.42/2.01 filter(x0, nil) 4.42/2.01 filter(x0, cons(x1, x2)) 4.42/2.01 filter2(true, x0, x1, x2) 4.42/2.01 filter2(false, x0, x1, x2) 4.42/2.01 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (10) QReductionProof (EQUIVALENT) 4.42/2.01 We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. 4.42/2.01 4.42/2.01 quot(0, s(x0), s(x1)) 4.42/2.01 quot(s(x0), s(x1), x2) 4.42/2.01 quot(x0, 0, s(x1)) 4.42/2.01 map(x0, nil) 4.42/2.01 map(x0, cons(x1, x2)) 4.42/2.01 filter(x0, nil) 4.42/2.01 filter(x0, cons(x1, x2)) 4.42/2.01 filter2(true, x0, x1, x2) 4.42/2.01 filter2(false, x0, x1, x2) 4.42/2.01 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (11) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 quot1(x, 0, s(z)) -> quot1(x, s(z), s(z)) 4.42/2.01 quot1(s(x), s(y), z) -> quot1(x, y, z) 4.42/2.01 4.42/2.01 R is empty. 4.42/2.01 Q is empty. 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (12) QDPSizeChangeProof (EQUIVALENT) 4.42/2.01 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. 4.42/2.01 4.42/2.01 From the DPs we obtained the following set of size-change graphs: 4.42/2.01 *quot1(s(x), s(y), z) -> quot1(x, y, z) 4.42/2.01 The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3 4.42/2.01 4.42/2.01 4.42/2.01 *quot1(x, 0, s(z)) -> quot1(x, s(z), s(z)) 4.42/2.01 The graph contains the following edges 1 >= 1, 3 >= 2, 3 >= 3 4.42/2.01 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (13) 4.42/2.01 YES 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (14) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 4.42/2.01 The TRS R consists of the following rules: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, y)), app(s, z)) -> 0 4.42/2.01 app(app(app(quot, app(s, x)), app(s, y)), z) -> app(app(app(quot, x), y), z) 4.42/2.01 app(app(app(quot, x), 0), app(s, z)) -> app(s, app(app(app(quot, x), app(s, z)), app(s, z))) 4.42/2.01 app(app(map, f), nil) -> nil 4.42/2.01 app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) 4.42/2.01 app(app(filter, f), nil) -> nil 4.42/2.01 app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) 4.42/2.01 app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) 4.42/2.01 app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) 4.42/2.01 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, x0)), app(s, x1)) 4.42/2.01 app(app(app(quot, app(s, x0)), app(s, x1)), x2) 4.42/2.01 app(app(app(quot, x0), 0), app(s, x1)) 4.42/2.01 app(app(map, x0), nil) 4.42/2.01 app(app(map, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(filter, x0), nil) 4.42/2.01 app(app(filter, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(app(app(filter2, true), x0), x1), x2) 4.42/2.01 app(app(app(app(filter2, false), x0), x1), x2) 4.42/2.01 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (15) UsableRulesProof (EQUIVALENT) 4.42/2.01 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. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (16) 4.42/2.01 Obligation: 4.42/2.01 Q DP problem: 4.42/2.01 The TRS P consists of the following rules: 4.42/2.01 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.42/2.01 APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 4.42/2.01 R is empty. 4.42/2.01 The set Q consists of the following terms: 4.42/2.01 4.42/2.01 app(app(app(quot, 0), app(s, x0)), app(s, x1)) 4.42/2.01 app(app(app(quot, app(s, x0)), app(s, x1)), x2) 4.42/2.01 app(app(app(quot, x0), 0), app(s, x1)) 4.42/2.01 app(app(map, x0), nil) 4.42/2.01 app(app(map, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(filter, x0), nil) 4.42/2.01 app(app(filter, x0), app(app(cons, x1), x2)) 4.42/2.01 app(app(app(app(filter2, true), x0), x1), x2) 4.42/2.01 app(app(app(app(filter2, false), x0), x1), x2) 4.42/2.01 4.42/2.01 We have to consider all minimal (P,Q,R)-chains. 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (17) QDPSizeChangeProof (EQUIVALENT) 4.42/2.01 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. 4.42/2.01 4.42/2.01 From the DPs we obtained the following set of size-change graphs: 4.42/2.01 *APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 The graph contains the following edges 1 > 1, 2 > 2 4.42/2.01 4.42/2.01 4.42/2.01 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) 4.42/2.01 The graph contains the following edges 1 > 1, 2 > 2 4.42/2.01 4.42/2.01 4.42/2.01 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) 4.42/2.01 The graph contains the following edges 1 >= 1, 2 > 2 4.42/2.01 4.42/2.01 4.42/2.01 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 The graph contains the following edges 2 >= 2 4.42/2.01 4.42/2.01 4.42/2.01 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) 4.42/2.01 The graph contains the following edges 2 >= 2 4.42/2.01 4.42/2.01 4.42/2.01 ---------------------------------------- 4.42/2.01 4.42/2.01 (18) 4.42/2.01 YES 4.42/2.04 EOF