/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 150 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 41 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) MRRProof [EQUIVALENT, 27 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) AND (13) QDP (14) MRRProof [EQUIVALENT, 66 ms] (15) QDP (16) DependencyGraphProof [EQUIVALENT, 0 ms] (17) QDP (18) UsableRulesProof [EQUIVALENT, 0 ms] (19) QDP (20) QDPSizeChangeProof [EQUIVALENT, 0 ms] (21) YES (22) QDP (23) TransformationProof [EQUIVALENT, 0 ms] (24) QDP (25) DependencyGraphProof [EQUIVALENT, 0 ms] (26) QDP (27) TransformationProof [EQUIVALENT, 0 ms] (28) QDP (29) DependencyGraphProof [EQUIVALENT, 0 ms] (30) QDP (31) QDPOrderProof [EQUIVALENT, 65 ms] (32) QDP (33) QDPOrderProof [EQUIVALENT, 66 ms] (34) QDP (35) QDPOrderProof [EQUIVALENT, 79 ms] (36) QDP (37) QDPOrderProof [EQUIVALENT, 37 ms] (38) QDP (39) NonTerminationLoopProof [COMPLETE, 0 ms] (40) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(nil) -> 0 a__length(cons(N, L)) -> a__U11(tt, L) a__take(0, IL) -> nil a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(U11(x_1, x_2)) = x_1 + x_2 POL(U12(x_1, x_2)) = 2*x_1 + x_2 POL(U21(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + x_4 POL(U22(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + x_4 POL(U23(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + x_4 POL(a__U11(x_1, x_2)) = x_1 + x_2 POL(a__U12(x_1, x_2)) = 2*x_1 + x_2 POL(a__U21(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + x_4 POL(a__U22(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + x_4 POL(a__U23(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + x_4 POL(a__length(x_1)) = x_1 POL(a__take(x_1, x_2)) = 1 + x_1 + x_2 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = x_1 + x_2 POL(length(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 1 + x_1 + x_2 POL(tt) = 0 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__take(0, IL) -> nil ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(nil) -> 0 a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(U11(x_1, x_2)) = x_1 + 2*x_2 POL(U12(x_1, x_2)) = 2*x_1 + 2*x_2 POL(U21(x_1, x_2, x_3, x_4)) = x_1 + 2*x_2 + 2*x_3 + 2*x_4 POL(U22(x_1, x_2, x_3, x_4)) = x_1 + 2*x_2 + 2*x_3 + 2*x_4 POL(U23(x_1, x_2, x_3, x_4)) = x_1 + 2*x_2 + 2*x_3 + 2*x_4 POL(a__U11(x_1, x_2)) = x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = 2*x_1 + 2*x_2 POL(a__U21(x_1, x_2, x_3, x_4)) = x_1 + 2*x_2 + 2*x_3 + 2*x_4 POL(a__U22(x_1, x_2, x_3, x_4)) = x_1 + 2*x_2 + 2*x_3 + 2*x_4 POL(a__U23(x_1, x_2, x_3, x_4)) = x_1 + 2*x_2 + 2*x_3 + 2*x_4 POL(a__length(x_1)) = x_1 POL(a__take(x_1, x_2)) = x_1 + x_2 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = 2*x_1 + 2*x_2 POL(length(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nil) = 2 POL(s(x_1)) = 2*x_1 POL(take(x_1, x_2)) = x_1 + x_2 POL(tt) = 0 POL(zeros) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a__length(nil) -> 0 ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, L) -> A__LENGTH(mark(L)) A__U12(tt, L) -> MARK(L) A__U21(tt, IL, M, N) -> A__U22(tt, IL, M, N) A__U22(tt, IL, M, N) -> A__U23(tt, IL, M, N) A__U23(tt, IL, M, N) -> MARK(N) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__TAKE(s(M), cons(N, IL)) -> A__U21(tt, IL, M, N) MARK(zeros) -> A__ZEROS MARK(U11(X1, X2)) -> A__U11(mark(X1), X2) MARK(U11(X1, X2)) -> MARK(X1) MARK(U12(X1, X2)) -> A__U12(mark(X1), X2) MARK(U12(X1, X2)) -> MARK(X1) MARK(length(X)) -> A__LENGTH(mark(X)) MARK(length(X)) -> MARK(X) MARK(U21(X1, X2, X3, X4)) -> A__U21(mark(X1), X2, X3, X4) MARK(U21(X1, X2, X3, X4)) -> MARK(X1) MARK(U22(X1, X2, X3, X4)) -> A__U22(mark(X1), X2, X3, X4) MARK(U22(X1, X2, X3, X4)) -> MARK(X1) MARK(U23(X1, X2, X3, X4)) -> A__U23(mark(X1), X2, X3, X4) MARK(U23(X1, X2, X3, X4)) -> MARK(X1) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, L) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, L) -> MARK(L) MARK(U11(X1, X2)) -> A__U11(mark(X1), X2) MARK(U11(X1, X2)) -> MARK(X1) MARK(U12(X1, X2)) -> A__U12(mark(X1), X2) MARK(U12(X1, X2)) -> MARK(X1) MARK(length(X)) -> A__LENGTH(mark(X)) MARK(length(X)) -> MARK(X) MARK(U21(X1, X2, X3, X4)) -> A__U21(mark(X1), X2, X3, X4) A__U21(tt, IL, M, N) -> A__U22(tt, IL, M, N) A__U22(tt, IL, M, N) -> A__U23(tt, IL, M, N) A__U23(tt, IL, M, N) -> MARK(N) MARK(U21(X1, X2, X3, X4)) -> MARK(X1) MARK(U22(X1, X2, X3, X4)) -> A__U22(mark(X1), X2, X3, X4) MARK(U22(X1, X2, X3, X4)) -> MARK(X1) MARK(U23(X1, X2, X3, X4)) -> A__U23(mark(X1), X2, X3, X4) MARK(U23(X1, X2, X3, X4)) -> MARK(X1) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) A__TAKE(s(M), cons(N, IL)) -> A__U21(tt, IL, M, N) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) MRRProof (EQUIVALENT) 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. Strictly oriented dependency pairs: MARK(U11(X1, X2)) -> A__U11(mark(X1), X2) MARK(U11(X1, X2)) -> MARK(X1) MARK(U12(X1, X2)) -> A__U12(mark(X1), X2) MARK(U12(X1, X2)) -> MARK(X1) MARK(length(X)) -> A__LENGTH(mark(X)) MARK(length(X)) -> MARK(X) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__LENGTH(x_1)) = 2*x_1 POL(A__TAKE(x_1, x_2)) = 2*x_1 + 2*x_2 POL(A__U11(x_1, x_2)) = 2*x_1 + 2*x_2 POL(A__U12(x_1, x_2)) = x_1 + 2*x_2 POL(A__U21(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(A__U22(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(A__U23(x_1, x_2, x_3, x_4)) = 2*x_1 + x_2 + x_3 + 2*x_4 POL(MARK(x_1)) = 2*x_1 POL(U11(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(U12(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(U21(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(U22(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(U23(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(a__U11(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(a__U12(x_1, x_2)) = 1 + x_1 + 2*x_2 POL(a__U21(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(a__U22(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(a__U23(x_1, x_2, x_3, x_4)) = x_1 + x_2 + x_3 + 2*x_4 POL(a__length(x_1)) = 1 + 2*x_1 POL(a__take(x_1, x_2)) = x_1 + x_2 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(length(x_1)) = 1 + 2*x_1 POL(mark(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = x_1 + x_2 POL(tt) = 0 POL(zeros) = 0 ---------------------------------------- (10) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, L) -> A__LENGTH(mark(L)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, L) -> MARK(L) MARK(U21(X1, X2, X3, X4)) -> A__U21(mark(X1), X2, X3, X4) A__U21(tt, IL, M, N) -> A__U22(tt, IL, M, N) A__U22(tt, IL, M, N) -> A__U23(tt, IL, M, N) A__U23(tt, IL, M, N) -> MARK(N) MARK(U21(X1, X2, X3, X4)) -> MARK(X1) MARK(U22(X1, X2, X3, X4)) -> A__U22(mark(X1), X2, X3, X4) MARK(U22(X1, X2, X3, X4)) -> MARK(X1) MARK(U23(X1, X2, X3, X4)) -> A__U23(mark(X1), X2, X3, X4) MARK(U23(X1, X2, X3, X4)) -> MARK(X1) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) A__TAKE(s(M), cons(N, IL)) -> A__U21(tt, IL, M, N) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (11) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (12) Complex Obligation (AND) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: A__U21(tt, IL, M, N) -> A__U22(tt, IL, M, N) A__U22(tt, IL, M, N) -> A__U23(tt, IL, M, N) A__U23(tt, IL, M, N) -> MARK(N) MARK(U21(X1, X2, X3, X4)) -> A__U21(mark(X1), X2, X3, X4) MARK(U21(X1, X2, X3, X4)) -> MARK(X1) MARK(U22(X1, X2, X3, X4)) -> A__U22(mark(X1), X2, X3, X4) MARK(U22(X1, X2, X3, X4)) -> MARK(X1) MARK(U23(X1, X2, X3, X4)) -> A__U23(mark(X1), X2, X3, X4) MARK(U23(X1, X2, X3, X4)) -> MARK(X1) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) A__TAKE(s(M), cons(N, IL)) -> A__U21(tt, IL, M, N) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) MRRProof (EQUIVALENT) 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. Strictly oriented dependency pairs: A__U21(tt, IL, M, N) -> A__U22(tt, IL, M, N) A__U22(tt, IL, M, N) -> A__U23(tt, IL, M, N) MARK(U21(X1, X2, X3, X4)) -> MARK(X1) MARK(U22(X1, X2, X3, X4)) -> A__U22(mark(X1), X2, X3, X4) MARK(U22(X1, X2, X3, X4)) -> MARK(X1) MARK(U23(X1, X2, X3, X4)) -> A__U23(mark(X1), X2, X3, X4) MARK(U23(X1, X2, X3, X4)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X1) MARK(take(X1, X2)) -> MARK(X2) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(A__TAKE(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(A__U21(x_1, x_2, x_3, x_4)) = 2 + 2*x_1 + x_2 + x_3 + 2*x_4 POL(A__U22(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + x_3 + 2*x_4 POL(A__U23(x_1, x_2, x_3, x_4)) = 2*x_1 + x_2 + x_3 + 2*x_4 POL(MARK(x_1)) = 2*x_1 POL(U11(x_1, x_2)) = 2*x_1 + x_2 POL(U12(x_1, x_2)) = 2*x_1 + x_2 POL(U21(x_1, x_2, x_3, x_4)) = 1 + x_1 + x_2 + 2*x_3 + 2*x_4 POL(U22(x_1, x_2, x_3, x_4)) = 1 + x_1 + x_2 + 2*x_3 + 2*x_4 POL(U23(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + 2*x_3 + 2*x_4 POL(a__U11(x_1, x_2)) = 2*x_1 + x_2 POL(a__U12(x_1, x_2)) = 2*x_1 + x_2 POL(a__U21(x_1, x_2, x_3, x_4)) = 1 + x_1 + x_2 + 2*x_3 + 2*x_4 POL(a__U22(x_1, x_2, x_3, x_4)) = 1 + x_1 + x_2 + 2*x_3 + 2*x_4 POL(a__U23(x_1, x_2, x_3, x_4)) = 1 + 2*x_1 + x_2 + 2*x_3 + 2*x_4 POL(a__length(x_1)) = x_1 POL(a__take(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(a__zeros) = 0 POL(cons(x_1, x_2)) = 2*x_1 + x_2 POL(length(x_1)) = x_1 POL(mark(x_1)) = x_1 POL(nil) = 0 POL(s(x_1)) = x_1 POL(take(x_1, x_2)) = 1 + 2*x_1 + x_2 POL(tt) = 0 POL(zeros) = 0 ---------------------------------------- (15) Obligation: Q DP problem: The TRS P consists of the following rules: A__U23(tt, IL, M, N) -> MARK(N) MARK(U21(X1, X2, X3, X4)) -> A__U21(mark(X1), X2, X3, X4) MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2)) A__TAKE(s(M), cons(N, IL)) -> A__U21(tt, IL, M, N) MARK(cons(X1, X2)) -> MARK(X1) MARK(s(X)) -> MARK(X) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (17) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (18) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (19) Obligation: Q DP problem: The TRS P consists of the following rules: MARK(s(X)) -> MARK(X) MARK(cons(X1, X2)) -> MARK(X1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (20) 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: *MARK(s(X)) -> MARK(X) The graph contains the following edges 1 > 1 *MARK(cons(X1, X2)) -> MARK(X1) The graph contains the following edges 1 > 1 ---------------------------------------- (21) YES ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, L) -> A__LENGTH(mark(L)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule A__U12(tt, L) -> A__LENGTH(mark(L)) at position [0] we obtained the following new rules [LPAR04]: (A__U12(tt, zeros) -> A__LENGTH(a__zeros),A__U12(tt, zeros) -> A__LENGTH(a__zeros)) (A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1)),A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1))) (A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1)),A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1))) (A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0))),A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0)))) (A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)),A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3))) (A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)),A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3))) (A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)),A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3))) (A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))),A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1)))) (A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)),A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1))) (A__U12(tt, 0) -> A__LENGTH(0),A__U12(tt, 0) -> A__LENGTH(0)) (A__U12(tt, tt) -> A__LENGTH(tt),A__U12(tt, tt) -> A__LENGTH(tt)) (A__U12(tt, s(x0)) -> A__LENGTH(s(mark(x0))),A__U12(tt, s(x0)) -> A__LENGTH(s(mark(x0)))) (A__U12(tt, nil) -> A__LENGTH(nil),A__U12(tt, nil) -> A__LENGTH(nil)) ---------------------------------------- (24) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(a__zeros) A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1)) A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1)) A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0))) A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)) A__U12(tt, 0) -> A__LENGTH(0) A__U12(tt, tt) -> A__LENGTH(tt) A__U12(tt, s(x0)) -> A__LENGTH(s(mark(x0))) A__U12(tt, nil) -> A__LENGTH(nil) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (25) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, zeros) -> A__LENGTH(a__zeros) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1)) A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1)) A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0))) A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (27) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule A__U12(tt, zeros) -> A__LENGTH(a__zeros) at position [0] we obtained the following new rules [LPAR04]: (A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)),A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros))) (A__U12(tt, zeros) -> A__LENGTH(zeros),A__U12(tt, zeros) -> A__LENGTH(zeros)) ---------------------------------------- (28) Obligation: Q DP problem: The TRS P consists of the following rules: A__U11(tt, L) -> A__U12(tt, L) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1)) A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1)) A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0))) A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) A__U12(tt, zeros) -> A__LENGTH(zeros) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (29) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1)) A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1)) A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0))) A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (31) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1)) A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1)) A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0))) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A__LENGTH_1(x_1) ) = x_1 POL( a__U11_2(x_1, x_2) ) = max{0, -2} POL( a__U12_2(x_1, x_2) ) = max{0, -2} POL( a__U21_4(x_1, ..., x_4) ) = 1 POL( a__U22_4(x_1, ..., x_4) ) = 1 POL( a__U23_4(x_1, ..., x_4) ) = 1 POL( a__length_1(x_1) ) = max{0, -2} POL( a__take_2(x_1, x_2) ) = 1 POL( cons_2(x_1, x_2) ) = 1 POL( s_1(x_1) ) = max{0, -2} POL( mark_1(x_1) ) = 2 POL( zeros ) = 0 POL( a__zeros ) = 0 POL( U11_2(x_1, x_2) ) = 0 POL( U12_2(x_1, x_2) ) = 0 POL( length_1(x_1) ) = 0 POL( U21_4(x_1, ..., x_4) ) = 0 POL( U22_4(x_1, ..., x_4) ) = 0 POL( U23_4(x_1, ..., x_4) ) = 1 POL( take_2(x_1, x_2) ) = 0 POL( 0 ) = 0 POL( tt ) = 0 POL( nil ) = 1 POL( A__U12_2(x_1, x_2) ) = 2x_1 + 1 POL( A__U11_2(x_1, x_2) ) = 2x_1 + 1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__U11(tt, L) -> a__U12(tt, L) a__U11(X1, X2) -> U11(X1, X2) a__U12(tt, L) -> s(a__length(mark(L))) a__U12(X1, X2) -> U12(X1, X2) a__length(cons(N, L)) -> a__U11(tt, L) a__length(X) -> length(X) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) a__take(X1, X2) -> take(X1, X2) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (33) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3)) A__U12(tt, cons(x0, x1)) -> A__LENGTH(cons(mark(x0), x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A__LENGTH_1(x_1) ) = max{0, x_1 - 2} POL( a__U11_2(x_1, x_2) ) = max{0, -2} POL( a__U12_2(x_1, x_2) ) = max{0, -2} POL( a__U21_4(x_1, ..., x_4) ) = x_1 + 2x_2 + 2 POL( a__U22_4(x_1, ..., x_4) ) = 2x_2 + 2 POL( a__U23_4(x_1, ..., x_4) ) = x_1 + 2x_2 + 2 POL( a__length_1(x_1) ) = 0 POL( a__take_2(x_1, x_2) ) = x_2 POL( cons_2(x_1, x_2) ) = 2x_2 + 2 POL( s_1(x_1) ) = x_1 POL( mark_1(x_1) ) = x_1 + 2 POL( zeros ) = 0 POL( a__zeros ) = 2 POL( U11_2(x_1, x_2) ) = 0 POL( U12_2(x_1, x_2) ) = 0 POL( length_1(x_1) ) = 0 POL( U21_4(x_1, ..., x_4) ) = x_1 + 2x_2 + 2 POL( U22_4(x_1, ..., x_4) ) = 2x_2 + 2 POL( U23_4(x_1, ..., x_4) ) = x_1 + 2x_2 + 2 POL( take_2(x_1, x_2) ) = x_2 POL( 0 ) = 0 POL( tt ) = 0 POL( nil ) = 2 POL( A__U11_2(x_1, x_2) ) = 2x_1 + x_2 POL( A__U12_2(x_1, x_2) ) = 2x_1 + x_2 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) a__take(X1, X2) -> take(X1, X2) a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__length(cons(N, L)) -> a__U11(tt, L) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__zeros -> cons(0, zeros) a__zeros -> zeros ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (35) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A__LENGTH_1(x_1) ) = max{0, x_1 - 2} POL( a__U11_2(x_1, x_2) ) = 2 POL( a__U12_2(x_1, x_2) ) = 2 POL( a__U21_4(x_1, ..., x_4) ) = 2 POL( a__U22_4(x_1, ..., x_4) ) = 2 POL( a__U23_4(x_1, ..., x_4) ) = 2 POL( a__length_1(x_1) ) = 2 POL( a__take_2(x_1, x_2) ) = 2 POL( cons_2(x_1, x_2) ) = x_2 + 2 POL( s_1(x_1) ) = max{0, -2} POL( mark_1(x_1) ) = x_1 + 2 POL( zeros ) = 0 POL( a__zeros ) = 0 POL( U11_2(x_1, x_2) ) = 1 POL( U12_2(x_1, x_2) ) = 0 POL( length_1(x_1) ) = 2 POL( U21_4(x_1, ..., x_4) ) = 2 POL( U22_4(x_1, ..., x_4) ) = 0 POL( U23_4(x_1, ..., x_4) ) = 0 POL( take_2(x_1, x_2) ) = 0 POL( 0 ) = 0 POL( tt ) = 0 POL( nil ) = 2 POL( A__U11_2(x_1, x_2) ) = 2x_1 + x_2 POL( A__U12_2(x_1, x_2) ) = max{0, 2x_1 + x_2 - 1} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) a__take(X1, X2) -> take(X1, X2) a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__length(cons(N, L)) -> a__U11(tt, L) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (37) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: POL( A__LENGTH_1(x_1) ) = max{0, 2x_1 - 2} POL( a__U11_2(x_1, x_2) ) = 2 POL( a__U12_2(x_1, x_2) ) = 2 POL( a__U21_4(x_1, ..., x_4) ) = 1 POL( a__U22_4(x_1, ..., x_4) ) = 1 POL( a__U23_4(x_1, ..., x_4) ) = 1 POL( a__length_1(x_1) ) = 2 POL( a__take_2(x_1, x_2) ) = 1 POL( cons_2(x_1, x_2) ) = 2x_2 + 1 POL( s_1(x_1) ) = 2 POL( mark_1(x_1) ) = 2x_1 + 1 POL( zeros ) = 0 POL( a__zeros ) = 0 POL( U11_2(x_1, x_2) ) = 2 POL( U12_2(x_1, x_2) ) = 2 POL( length_1(x_1) ) = 2 POL( U21_4(x_1, ..., x_4) ) = 0 POL( U22_4(x_1, ..., x_4) ) = 1 POL( U23_4(x_1, ..., x_4) ) = 1 POL( take_2(x_1, x_2) ) = 0 POL( 0 ) = 0 POL( tt ) = 0 POL( nil ) = 1 POL( A__U11_2(x_1, x_2) ) = 2x_1 + 2x_2 POL( A__U12_2(x_1, x_2) ) = max{0, 2x_1 + 2x_2 - 1} The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(s(X)) -> s(mark(X)) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) a__take(X1, X2) -> take(X1, X2) a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__length(cons(N, L)) -> a__U11(tt, L) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) ---------------------------------------- (38) Obligation: Q DP problem: The TRS P consists of the following rules: A__LENGTH(cons(N, L)) -> A__U11(tt, L) A__U11(tt, L) -> A__U12(tt, L) A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1))) A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) The TRS R consists of the following rules: a__zeros -> cons(0, zeros) a__U11(tt, L) -> a__U12(tt, L) a__U12(tt, L) -> s(a__length(mark(L))) a__U21(tt, IL, M, N) -> a__U22(tt, IL, M, N) a__U22(tt, IL, M, N) -> a__U23(tt, IL, M, N) a__U23(tt, IL, M, N) -> cons(mark(N), take(M, IL)) a__length(cons(N, L)) -> a__U11(tt, L) a__take(s(M), cons(N, IL)) -> a__U21(tt, IL, M, N) mark(zeros) -> a__zeros mark(U11(X1, X2)) -> a__U11(mark(X1), X2) mark(U12(X1, X2)) -> a__U12(mark(X1), X2) mark(length(X)) -> a__length(mark(X)) mark(U21(X1, X2, X3, X4)) -> a__U21(mark(X1), X2, X3, X4) mark(U22(X1, X2, X3, X4)) -> a__U22(mark(X1), X2, X3, X4) mark(U23(X1, X2, X3, X4)) -> a__U23(mark(X1), X2, X3, X4) mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) mark(cons(X1, X2)) -> cons(mark(X1), X2) mark(0) -> 0 mark(tt) -> tt mark(s(X)) -> s(mark(X)) mark(nil) -> nil a__zeros -> zeros a__U11(X1, X2) -> U11(X1, X2) a__U12(X1, X2) -> U12(X1, X2) a__length(X) -> length(X) a__U21(X1, X2, X3, X4) -> U21(X1, X2, X3, X4) a__U22(X1, X2, X3, X4) -> U22(X1, X2, X3, X4) a__U23(X1, X2, X3, X4) -> U23(X1, X2, X3, X4) a__take(X1, X2) -> take(X1, X2) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (39) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = A__U11(tt, zeros) evaluates to t =A__U11(tt, zeros) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence A__U11(tt, zeros) -> A__U12(tt, zeros) with rule A__U11(tt, L) -> A__U12(tt, L) at position [] and matcher [L / zeros] A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) with rule A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros)) at position [] and matcher [ ] A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros) with rule A__LENGTH(cons(N, L)) -> A__U11(tt, L) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (40) NO