NO
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be disproven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 146 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 36 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 14 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) MRRProof [EQUIVALENT, 30 ms]
(10) QDP
(11) DependencyGraphProof [EQUIVALENT, 0 ms]
(12) AND
    (13) QDP
        (14) MRRProof [EQUIVALENT, 11 ms]
        (15) QDP
        (16) DependencyGraphProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) UsableRulesProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QReductionProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES
    (24) QDP
        (25) TransformationProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) DependencyGraphProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) TransformationProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) DependencyGraphProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) DependencyGraphProof [EQUIVALENT, 0 ms]
        (38) AND
            (39) QDP
                (40) UsableRulesProof [EQUIVALENT, 0 ms]
                (41) QDP
                (42) QReductionProof [EQUIVALENT, 0 ms]
                (43) QDP
                (44) TransformationProof [EQUIVALENT, 0 ms]
                (45) QDP
                (46) NonTerminationLoopProof [COMPLETE, 0 ms]
                (47) NO
            (48) QDP


----------------------------------------

(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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)


----------------------------------------

(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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)


----------------------------------------

(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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)


----------------------------------------

(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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(18) 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.
----------------------------------------

(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.
The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(20) 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].

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)


----------------------------------------

(21)
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.
----------------------------------------

(22) 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


----------------------------------------

(23)
YES

----------------------------------------

(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, 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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(25) 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))


----------------------------------------

(26)
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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(27) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
----------------------------------------

(28)
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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(29) 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))


----------------------------------------

(30)
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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(31) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

(32)
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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(33) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule A__U11(tt, L) -> A__U12(tt, L) we obtained the following new rules [LPAR04]:

   (A__U11(tt, U11(y_0, y_1)) -> A__U12(tt, U11(y_0, y_1)),A__U11(tt, U11(y_0, y_1)) -> A__U12(tt, U11(y_0, y_1)))
   (A__U11(tt, U12(y_0, y_1)) -> A__U12(tt, U12(y_0, y_1)),A__U11(tt, U12(y_0, y_1)) -> A__U12(tt, U12(y_0, y_1)))
   (A__U11(tt, length(y_0)) -> A__U12(tt, length(y_0)),A__U11(tt, length(y_0)) -> A__U12(tt, length(y_0)))
   (A__U11(tt, U21(y_0, y_1, y_2, y_3)) -> A__U12(tt, U21(y_0, y_1, y_2, y_3)),A__U11(tt, U21(y_0, y_1, y_2, y_3)) -> A__U12(tt, U21(y_0, y_1, y_2, y_3)))
   (A__U11(tt, U22(y_0, y_1, y_2, y_3)) -> A__U12(tt, U22(y_0, y_1, y_2, y_3)),A__U11(tt, U22(y_0, y_1, y_2, y_3)) -> A__U12(tt, U22(y_0, y_1, y_2, y_3)))
   (A__U11(tt, U23(y_0, y_1, y_2, y_3)) -> A__U12(tt, U23(y_0, y_1, y_2, y_3)),A__U11(tt, U23(y_0, y_1, y_2, y_3)) -> A__U12(tt, U23(y_0, y_1, y_2, y_3)))
   (A__U11(tt, take(y_0, y_1)) -> A__U12(tt, take(y_0, y_1)),A__U11(tt, take(y_0, y_1)) -> A__U12(tt, take(y_0, y_1)))
   (A__U11(tt, cons(y_0, y_1)) -> A__U12(tt, cons(y_0, y_1)),A__U11(tt, cons(y_0, y_1)) -> A__U12(tt, cons(y_0, y_1)))
   (A__U11(tt, zeros) -> A__U12(tt, zeros),A__U11(tt, zeros) -> A__U12(tt, zeros))


----------------------------------------

(34)
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__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__U11(tt, U11(y_0, y_1)) -> A__U12(tt, U11(y_0, y_1))
   A__U11(tt, U12(y_0, y_1)) -> A__U12(tt, U12(y_0, y_1))
   A__U11(tt, length(y_0)) -> A__U12(tt, length(y_0))
   A__U11(tt, U21(y_0, y_1, y_2, y_3)) -> A__U12(tt, U21(y_0, y_1, y_2, y_3))
   A__U11(tt, U22(y_0, y_1, y_2, y_3)) -> A__U12(tt, U22(y_0, y_1, y_2, y_3))
   A__U11(tt, U23(y_0, y_1, y_2, y_3)) -> A__U12(tt, U23(y_0, y_1, y_2, y_3))
   A__U11(tt, take(y_0, y_1)) -> A__U12(tt, take(y_0, y_1))
   A__U11(tt, cons(y_0, y_1)) -> A__U12(tt, cons(y_0, y_1))
   A__U11(tt, zeros) -> A__U12(tt, 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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(35) TransformationProof (EQUIVALENT)
By forward instantiating [JAR06] the rule A__LENGTH(cons(N, L)) -> A__U11(tt, L) we obtained the following new rules [LPAR04]:

   (A__LENGTH(cons(x0, U11(y_0, y_1))) -> A__U11(tt, U11(y_0, y_1)),A__LENGTH(cons(x0, U11(y_0, y_1))) -> A__U11(tt, U11(y_0, y_1)))
   (A__LENGTH(cons(x0, U12(y_0, y_1))) -> A__U11(tt, U12(y_0, y_1)),A__LENGTH(cons(x0, U12(y_0, y_1))) -> A__U11(tt, U12(y_0, y_1)))
   (A__LENGTH(cons(x0, length(y_0))) -> A__U11(tt, length(y_0)),A__LENGTH(cons(x0, length(y_0))) -> A__U11(tt, length(y_0)))
   (A__LENGTH(cons(x0, U21(y_0, y_1, y_2, y_3))) -> A__U11(tt, U21(y_0, y_1, y_2, y_3)),A__LENGTH(cons(x0, U21(y_0, y_1, y_2, y_3))) -> A__U11(tt, U21(y_0, y_1, y_2, y_3)))
   (A__LENGTH(cons(x0, U22(y_0, y_1, y_2, y_3))) -> A__U11(tt, U22(y_0, y_1, y_2, y_3)),A__LENGTH(cons(x0, U22(y_0, y_1, y_2, y_3))) -> A__U11(tt, U22(y_0, y_1, y_2, y_3)))
   (A__LENGTH(cons(x0, U23(y_0, y_1, y_2, y_3))) -> A__U11(tt, U23(y_0, y_1, y_2, y_3)),A__LENGTH(cons(x0, U23(y_0, y_1, y_2, y_3))) -> A__U11(tt, U23(y_0, y_1, y_2, y_3)))
   (A__LENGTH(cons(x0, take(y_0, y_1))) -> A__U11(tt, take(y_0, y_1)),A__LENGTH(cons(x0, take(y_0, y_1))) -> A__U11(tt, take(y_0, y_1)))
   (A__LENGTH(cons(x0, cons(y_0, y_1))) -> A__U11(tt, cons(y_0, y_1)),A__LENGTH(cons(x0, cons(y_0, y_1))) -> A__U11(tt, cons(y_0, y_1)))
   (A__LENGTH(cons(x0, zeros)) -> A__U11(tt, zeros),A__LENGTH(cons(x0, zeros)) -> A__U11(tt, zeros))


----------------------------------------

(36)
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__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__U11(tt, U11(y_0, y_1)) -> A__U12(tt, U11(y_0, y_1))
   A__U11(tt, U12(y_0, y_1)) -> A__U12(tt, U12(y_0, y_1))
   A__U11(tt, length(y_0)) -> A__U12(tt, length(y_0))
   A__U11(tt, U21(y_0, y_1, y_2, y_3)) -> A__U12(tt, U21(y_0, y_1, y_2, y_3))
   A__U11(tt, U22(y_0, y_1, y_2, y_3)) -> A__U12(tt, U22(y_0, y_1, y_2, y_3))
   A__U11(tt, U23(y_0, y_1, y_2, y_3)) -> A__U12(tt, U23(y_0, y_1, y_2, y_3))
   A__U11(tt, take(y_0, y_1)) -> A__U12(tt, take(y_0, y_1))
   A__U11(tt, cons(y_0, y_1)) -> A__U12(tt, cons(y_0, y_1))
   A__U11(tt, zeros) -> A__U12(tt, zeros)
   A__LENGTH(cons(x0, U11(y_0, y_1))) -> A__U11(tt, U11(y_0, y_1))
   A__LENGTH(cons(x0, U12(y_0, y_1))) -> A__U11(tt, U12(y_0, y_1))
   A__LENGTH(cons(x0, length(y_0))) -> A__U11(tt, length(y_0))
   A__LENGTH(cons(x0, U21(y_0, y_1, y_2, y_3))) -> A__U11(tt, U21(y_0, y_1, y_2, y_3))
   A__LENGTH(cons(x0, U22(y_0, y_1, y_2, y_3))) -> A__U11(tt, U22(y_0, y_1, y_2, y_3))
   A__LENGTH(cons(x0, U23(y_0, y_1, y_2, y_3))) -> A__U11(tt, U23(y_0, y_1, y_2, y_3))
   A__LENGTH(cons(x0, take(y_0, y_1))) -> A__U11(tt, take(y_0, y_1))
   A__LENGTH(cons(x0, cons(y_0, y_1))) -> A__U11(tt, cons(y_0, y_1))
   A__LENGTH(cons(x0, zeros)) -> A__U11(tt, 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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(37) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
----------------------------------------

(38)
Complex Obligation (AND)

----------------------------------------

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(x0, zeros)) -> A__U11(tt, zeros)
   A__U11(tt, zeros) -> A__U12(tt, zeros)
   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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(40) 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.
----------------------------------------

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(x0, zeros)) -> A__U11(tt, zeros)
   A__U11(tt, zeros) -> A__U12(tt, zeros)
   A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros))

R is empty.
The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(42) 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].

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)


----------------------------------------

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(x0, zeros)) -> A__U11(tt, zeros)
   A__U11(tt, zeros) -> A__U12(tt, zeros)
   A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(44) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule A__LENGTH(cons(x0, zeros)) -> A__U11(tt, zeros) we obtained the following new rules [LPAR04]:

   (A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros),A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros))


----------------------------------------

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__U11(tt, zeros) -> A__U12(tt, zeros)
   A__U12(tt, zeros) -> A__LENGTH(cons(0, zeros))
   A__LENGTH(cons(0, zeros)) -> A__U11(tt, zeros)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(46) 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__U12(tt, zeros) evaluates to  t =A__U12(tt, zeros)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [ ]

--------------------------------------------------------------------------------
Rewriting sequence

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(0, zeros)) -> A__U11(tt, zeros) at position [] and matcher [ ]

A__U11(tt, zeros) -> A__U12(tt, zeros)
with rule A__U11(tt, zeros) -> A__U12(tt, zeros)

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.




----------------------------------------

(47)
NO

----------------------------------------

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(x0, U11(y_0, y_1))) -> A__U11(tt, U11(y_0, y_1))
   A__U11(tt, U11(y_0, y_1)) -> A__U12(tt, U11(y_0, y_1))
   A__U12(tt, U11(x0, x1)) -> A__LENGTH(a__U11(mark(x0), x1))
   A__LENGTH(cons(x0, U12(y_0, y_1))) -> A__U11(tt, U12(y_0, y_1))
   A__U11(tt, U12(y_0, y_1)) -> A__U12(tt, U12(y_0, y_1))
   A__U12(tt, U12(x0, x1)) -> A__LENGTH(a__U12(mark(x0), x1))
   A__LENGTH(cons(x0, length(y_0))) -> A__U11(tt, length(y_0))
   A__U11(tt, length(y_0)) -> A__U12(tt, length(y_0))
   A__U12(tt, length(x0)) -> A__LENGTH(a__length(mark(x0)))
   A__LENGTH(cons(x0, U21(y_0, y_1, y_2, y_3))) -> A__U11(tt, U21(y_0, y_1, y_2, y_3))
   A__U11(tt, U21(y_0, y_1, y_2, y_3)) -> A__U12(tt, U21(y_0, y_1, y_2, y_3))
   A__U12(tt, U21(x0, x1, x2, x3)) -> A__LENGTH(a__U21(mark(x0), x1, x2, x3))
   A__LENGTH(cons(x0, U22(y_0, y_1, y_2, y_3))) -> A__U11(tt, U22(y_0, y_1, y_2, y_3))
   A__U11(tt, U22(y_0, y_1, y_2, y_3)) -> A__U12(tt, U22(y_0, y_1, y_2, y_3))
   A__U12(tt, U22(x0, x1, x2, x3)) -> A__LENGTH(a__U22(mark(x0), x1, x2, x3))
   A__LENGTH(cons(x0, U23(y_0, y_1, y_2, y_3))) -> A__U11(tt, U23(y_0, y_1, y_2, y_3))
   A__U11(tt, U23(y_0, y_1, y_2, y_3)) -> A__U12(tt, U23(y_0, y_1, y_2, y_3))
   A__U12(tt, U23(x0, x1, x2, x3)) -> A__LENGTH(a__U23(mark(x0), x1, x2, x3))
   A__LENGTH(cons(x0, take(y_0, y_1))) -> A__U11(tt, take(y_0, y_1))
   A__U11(tt, take(y_0, y_1)) -> A__U12(tt, take(y_0, y_1))
   A__U12(tt, take(x0, x1)) -> A__LENGTH(a__take(mark(x0), mark(x1)))
   A__LENGTH(cons(x0, cons(y_0, y_1))) -> A__U11(tt, cons(y_0, y_1))
   A__U11(tt, cons(y_0, y_1)) -> A__U12(tt, cons(y_0, y_1))
   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)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(U11(x0, x1))
   mark(U12(x0, x1))
   mark(length(x0))
   mark(U21(x0, x1, x2, x3))
   mark(U22(x0, x1, x2, x3))
   mark(U23(x0, x1, x2, x3))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(s(x0))
   mark(nil)
   a__U11(x0, x1)
   a__U12(x0, x1)
   a__length(x0)
   a__U21(x0, x1, x2, x3)
   a__U22(x0, x1, x2, x3)
   a__U23(x0, x1, x2, x3)
   a__take(x0, x1)

We have to consider all minimal (P,Q,R)-chains.