YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 79 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) ATransformationProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) ATransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPOrderProof [EQUIVALENT, 22 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) TRUE (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) ATransformationProof [EQUIVALENT, 0 ms] (29) QDP (30) QReductionProof [EQUIVALENT, 0 ms] (31) QDP (32) QDPSizeChangeProof [EQUIVALENT, 0 ms] (33) YES (34) QDP (35) UsableRulesProof [EQUIVALENT, 0 ms] (36) QDP (37) ATransformationProof [EQUIVALENT, 0 ms] (38) QDP (39) QReductionProof [EQUIVALENT, 0 ms] (40) QDP (41) QDPSizeChangeProof [EQUIVALENT, 0 ms] (42) YES (43) QDP (44) UsableRulesProof [EQUIVALENT, 0 ms] (45) QDP (46) ATransformationProof [EQUIVALENT, 0 ms] (47) QDP (48) QReductionProof [EQUIVALENT, 0 ms] (49) QDP (50) QDPOrderProof [EQUIVALENT, 69 ms] (51) QDP (52) PisEmptyProof [EQUIVALENT, 0 ms] (53) YES (54) QDP (55) QDPSizeChangeProof [EQUIVALENT, 0 ms] (56) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(eq, app(s, n)), app(s, m)) -> APP(app(eq, n), m) APP(app(eq, app(s, n)), app(s, m)) -> APP(eq, n) APP(app(le, app(s, n)), app(s, m)) -> APP(app(le, n), m) APP(app(le, app(s, n)), app(s, m)) -> APP(le, n) APP(min, app(app(cons, n), app(app(cons, m), x))) -> APP(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) APP(min, app(app(cons, n), app(app(cons, m), x))) -> APP(if_min, app(app(le, n), m)) APP(min, app(app(cons, n), app(app(cons, m), x))) -> APP(app(le, n), m) APP(min, app(app(cons, n), app(app(cons, m), x))) -> APP(le, n) APP(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> APP(min, app(app(cons, n), x)) APP(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> APP(app(cons, n), x) APP(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> APP(min, app(app(cons, m), x)) APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(app(app(if_replace, app(app(eq, n), k)), n), m) APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(app(if_replace, app(app(eq, n), k)), n) APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(if_replace, app(app(eq, n), k)) APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(app(eq, n), k) APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(eq, n) APP(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> APP(app(cons, m), x) APP(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> APP(cons, m) APP(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> APP(app(cons, k), app(app(app(replace, n), m), x)) APP(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> APP(app(app(replace, n), m), x) APP(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> APP(app(replace, n), m) APP(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> APP(replace, n) APP(sort, app(app(cons, n), x)) -> APP(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) APP(sort, app(app(cons, n), x)) -> APP(cons, app(min, app(app(cons, n), x))) APP(sort, app(app(cons, n), x)) -> APP(min, app(app(cons, n), x)) APP(sort, app(app(cons, n), x)) -> APP(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x)) APP(sort, app(app(cons, n), x)) -> APP(app(app(replace, app(min, app(app(cons, n), x))), n), x) APP(sort, app(app(cons, n), x)) -> APP(app(replace, app(min, app(app(cons, n), x))), n) APP(sort, app(app(cons, n), x)) -> APP(replace, app(min, app(app(cons, n), x))) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 31 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(le, app(s, n)), app(s, m)) -> APP(app(le, n), m) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (6) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(le, app(s, n)), app(s, m)) -> APP(app(le, n), m) R is empty. The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: le1(s(n), s(m)) -> le1(n, m) R is empty. The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: le1(s(n), s(m)) -> le1(n, m) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *le1(s(n), s(m)) -> le1(n, m) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (13) YES ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: APP(min, app(app(cons, n), app(app(cons, m), x))) -> APP(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) APP(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> APP(min, app(app(cons, n), x)) APP(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> APP(min, app(app(cons, m), x)) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: APP(min, app(app(cons, n), app(app(cons, m), x))) -> APP(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) APP(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> APP(min, app(app(cons, n), x)) APP(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> APP(min, app(app(cons, m), x)) The TRS R consists of the following rules: app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: min1(cons(n, cons(m, x))) -> if_min1(le(n, m), cons(n, cons(m, x))) if_min1(true, cons(n, cons(m, x))) -> min1(cons(n, x)) if_min1(false, cons(n, cons(m, x))) -> min1(cons(m, x)) The TRS R consists of the following rules: le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: min1(cons(n, cons(m, x))) -> if_min1(le(n, m), cons(n, cons(m, x))) if_min1(true, cons(n, cons(m, x))) -> min1(cons(n, x)) if_min1(false, cons(n, cons(m, x))) -> min1(cons(m, x)) The TRS R consists of the following rules: le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. if_min1(true, cons(n, cons(m, x))) -> min1(cons(n, x)) if_min1(false, cons(n, cons(m, x))) -> min1(cons(m, x)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( if_min1_2(x_1, x_2) ) = 2x_2 + 2 POL( le_2(x_1, x_2) ) = 0 POL( 0 ) = 0 POL( true ) = 2 POL( s_1(x_1) ) = 2x_1 + 2 POL( false ) = 2 POL( min1_1(x_1) ) = 2x_1 + 2 POL( cons_2(x_1, x_2) ) = 2x_2 + 2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: none ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: min1(cons(n, cons(m, x))) -> if_min1(le(n, m), cons(n, cons(m, x))) The TRS R consists of the following rules: le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) The set Q consists of the following terms: le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (24) TRUE ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(eq, app(s, n)), app(s, m)) -> APP(app(eq, n), m) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(eq, app(s, n)), app(s, m)) -> APP(app(eq, n), m) R is empty. The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (29) Obligation: Q DP problem: The TRS P consists of the following rules: eq1(s(n), s(m)) -> eq1(n, m) R is empty. The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (30) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) ---------------------------------------- (31) Obligation: Q DP problem: The TRS P consists of the following rules: eq1(s(n), s(m)) -> eq1(n, m) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (32) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *eq1(s(n), s(m)) -> eq1(n, m) The graph contains the following edges 1 > 1, 2 > 2 ---------------------------------------- (33) YES ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) APP(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> APP(app(app(replace, n), m), x) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(replace, n), m), app(app(cons, k), x)) -> APP(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) APP(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> APP(app(app(replace, n), m), x) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: replace1(n, m, cons(k, x)) -> if_replace1(eq(n, k), n, m, cons(k, x)) if_replace1(false, n, m, cons(k, x)) -> replace1(n, m, x) The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) ---------------------------------------- (40) Obligation: Q DP problem: The TRS P consists of the following rules: replace1(n, m, cons(k, x)) -> if_replace1(eq(n, k), n, m, cons(k, x)) if_replace1(false, n, m, cons(k, x)) -> replace1(n, m, x) The TRS R consists of the following rules: eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (41) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *if_replace1(false, n, m, cons(k, x)) -> replace1(n, m, x) The graph contains the following edges 2 >= 1, 3 >= 2, 4 > 3 *replace1(n, m, cons(k, x)) -> if_replace1(eq(n, k), n, m, cons(k, x)) The graph contains the following edges 1 >= 2, 2 >= 3, 3 >= 4 ---------------------------------------- (42) YES ---------------------------------------- (43) Obligation: Q DP problem: The TRS P consists of the following rules: APP(sort, app(app(cons, n), x)) -> APP(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x)) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (44) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (45) Obligation: Q DP problem: The TRS P consists of the following rules: APP(sort, app(app(cons, n), x)) -> APP(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x)) The TRS R consists of the following rules: app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (46) ATransformationProof (EQUIVALENT) We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. ---------------------------------------- (47) Obligation: Q DP problem: The TRS P consists of the following rules: sort1(cons(n, x)) -> sort1(replace(min(cons(n, x)), n, x)) The TRS R consists of the following rules: min(cons(0, nil)) -> 0 min(cons(s(n), nil)) -> s(n) min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x))) if_min(true, cons(n, cons(m, x))) -> min(cons(n, x)) if_min(false, cons(n, cons(m, x))) -> min(cons(m, x)) replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (48) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. sort(nil) sort(cons(x0, x1)) map(x0, nil) map(x0, cons(x1, x2)) filter(x0, nil) filter(x0, cons(x1, x2)) filter2(true, x0, x1, x2) filter2(false, x0, x1, x2) ---------------------------------------- (49) Obligation: Q DP problem: The TRS P consists of the following rules: sort1(cons(n, x)) -> sort1(replace(min(cons(n, x)), n, x)) The TRS R consists of the following rules: min(cons(0, nil)) -> 0 min(cons(s(n), nil)) -> s(n) min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x))) if_min(true, cons(n, cons(m, x))) -> min(cons(n, x)) if_min(false, cons(n, cons(m, x))) -> min(cons(m, x)) replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (50) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. sort1(cons(n, x)) -> sort1(replace(min(cons(n, x)), n, x)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( replace_3(x_1, ..., x_3) ) = max{0, 2x_3 - 2} POL( sort1_1(x_1) ) = max{0, x_1 - 1} POL( min_1(x_1) ) = max{0, -2} POL( cons_2(x_1, x_2) ) = 2x_2 + 2 POL( 0 ) = 2 POL( nil ) = 0 POL( s_1(x_1) ) = 2 POL( if_min_2(x_1, x_2) ) = 2x_1 POL( le_2(x_1, x_2) ) = 0 POL( true ) = 1 POL( false ) = 0 POL( if_replace_4(x_1, ..., x_4) ) = max{0, 2x_4 - 2} POL( eq_2(x_1, x_2) ) = 0 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) ---------------------------------------- (51) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: min(cons(0, nil)) -> 0 min(cons(s(n), nil)) -> s(n) min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x))) if_min(true, cons(n, cons(m, x))) -> min(cons(n, x)) if_min(false, cons(n, cons(m, x))) -> min(cons(m, x)) replace(n, m, nil) -> nil replace(n, m, cons(k, x)) -> if_replace(eq(n, k), n, m, cons(k, x)) eq(0, 0) -> true eq(0, s(m)) -> false eq(s(n), 0) -> false eq(s(n), s(m)) -> eq(n, m) if_replace(true, n, m, cons(k, x)) -> cons(m, x) if_replace(false, n, m, cons(k, x)) -> cons(k, replace(n, m, x)) le(0, m) -> true le(s(n), 0) -> false le(s(n), s(m)) -> le(n, m) The set Q consists of the following terms: eq(0, 0) eq(0, s(x0)) eq(s(x0), 0) eq(s(x0), s(x1)) le(0, x0) le(s(x0), 0) le(s(x0), s(x1)) min(cons(0, nil)) min(cons(s(x0), nil)) min(cons(x0, cons(x1, x2))) if_min(true, cons(x0, cons(x1, x2))) if_min(false, cons(x0, cons(x1, x2))) replace(x0, x1, nil) replace(x0, x1, cons(x2, x3)) if_replace(true, x0, x1, cons(x2, x3)) if_replace(false, x0, x1, cons(x2, x3)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (52) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (53) YES ---------------------------------------- (54) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(eq, 0), 0) -> true app(app(eq, 0), app(s, m)) -> false app(app(eq, app(s, n)), 0) -> false app(app(eq, app(s, n)), app(s, m)) -> app(app(eq, n), m) app(app(le, 0), m) -> true app(app(le, app(s, n)), 0) -> false app(app(le, app(s, n)), app(s, m)) -> app(app(le, n), m) app(min, app(app(cons, 0), nil)) -> 0 app(min, app(app(cons, app(s, n)), nil)) -> app(s, n) app(min, app(app(cons, n), app(app(cons, m), x))) -> app(app(if_min, app(app(le, n), m)), app(app(cons, n), app(app(cons, m), x))) app(app(if_min, true), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, n), x)) app(app(if_min, false), app(app(cons, n), app(app(cons, m), x))) -> app(min, app(app(cons, m), x)) app(app(app(replace, n), m), nil) -> nil app(app(app(replace, n), m), app(app(cons, k), x)) -> app(app(app(app(if_replace, app(app(eq, n), k)), n), m), app(app(cons, k), x)) app(app(app(app(if_replace, true), n), m), app(app(cons, k), x)) -> app(app(cons, m), x) app(app(app(app(if_replace, false), n), m), app(app(cons, k), x)) -> app(app(cons, k), app(app(app(replace, n), m), x)) app(sort, nil) -> nil app(sort, app(app(cons, n), x)) -> app(app(cons, app(min, app(app(cons, n), x))), app(sort, app(app(app(replace, app(min, app(app(cons, n), x))), n), x))) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) The set Q consists of the following terms: app(app(eq, 0), 0) app(app(eq, 0), app(s, x0)) app(app(eq, app(s, x0)), 0) app(app(eq, app(s, x0)), app(s, x1)) app(app(le, 0), x0) app(app(le, app(s, x0)), 0) app(app(le, app(s, x0)), app(s, x1)) app(min, app(app(cons, 0), nil)) app(min, app(app(cons, app(s, x0)), nil)) app(min, app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, true), app(app(cons, x0), app(app(cons, x1), x2))) app(app(if_min, false), app(app(cons, x0), app(app(cons, x1), x2))) app(app(app(replace, x0), x1), nil) app(app(app(replace, x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, true), x0), x1), app(app(cons, x2), x3)) app(app(app(app(if_replace, false), x0), x1), app(app(cons, x2), x3)) app(sort, nil) app(sort, app(app(cons, x0), x1)) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (55) QDPSizeChangeProof (EQUIVALENT) 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. From the DPs we obtained the following set of size-change graphs: *APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) The graph contains the following edges 1 > 1, 2 > 2 *APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) The graph contains the following edges 1 >= 1, 2 > 2 *APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) The graph contains the following edges 2 > 2 *APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 *APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) The graph contains the following edges 2 >= 2 ---------------------------------------- (56) YES