NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 51 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 4 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 33 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) QDP
(9) MRRProof [EQUIVALENT, 0 ms]
(10) QDP
(11) MRRProof [EQUIVALENT, 16 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) AND
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QReductionProof [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) AND
            (31) QDP
                (32) UsableRulesProof [EQUIVALENT, 0 ms]
                (33) QDP
                (34) QReductionProof [EQUIVALENT, 0 ms]
                (35) QDP
                (36) TransformationProof [EQUIVALENT, 0 ms]
                (37) QDP
                (38) NonTerminationLoopProof [COMPLETE, 0 ms]
                (39) NO
            (40) QDP
                (41) MRRProof [EQUIVALENT, 0 ms]
                (42) QDP


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__and(tt, X) -> mark(X)
   a__length(nil) -> 0
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(0, IL) -> nil
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)


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

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(a__and(x_1, x_2)) = x_1 + 2*x_2
   POL(a__length(x_1)) = x_1
   POL(a__take(x_1, x_2)) = x_1 + 2*x_2
   POL(a__zeros) = 0
   POL(and(x_1, x_2)) = x_1 + 2*x_2
   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)) = x_1 + 2*x_2
   POL(tt) = 1
   POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a__and(tt, X) -> mark(X)




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

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(nil) -> 0
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(0, IL) -> nil
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)


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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(a__and(x_1, x_2)) = x_1 + 2*x_2
   POL(a__length(x_1)) = 2*x_1
   POL(a__take(x_1, x_2)) = 2 + x_1 + x_2
   POL(a__zeros) = 0
   POL(and(x_1, x_2)) = x_1 + 2*x_2
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(length(x_1)) = 2*x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 1
   POL(s(x_1)) = x_1
   POL(take(x_1, x_2)) = 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
   a__take(0, IL) -> nil




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   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__LENGTH(cons(N, L)) -> A__LENGTH(mark(L))
   A__LENGTH(cons(N, L)) -> MARK(L)
   A__TAKE(s(M), cons(N, IL)) -> MARK(N)
   MARK(zeros) -> A__ZEROS
   MARK(and(X1, X2)) -> A__AND(mark(X1), X2)
   MARK(and(X1, X2)) -> MARK(X1)
   MARK(length(X)) -> A__LENGTH(mark(X))
   MARK(length(X)) -> MARK(X)
   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__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   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 2 less nodes.
----------------------------------------

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(N, L)) -> MARK(L)
   MARK(and(X1, X2)) -> MARK(X1)
   MARK(length(X)) -> A__LENGTH(mark(X))
   A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L))
   MARK(length(X)) -> MARK(X)
   MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2))
   A__TAKE(s(M), cons(N, IL)) -> MARK(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__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   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(and(X1, X2)) -> MARK(X1)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(A__LENGTH(x_1)) = x_1
   POL(A__TAKE(x_1, x_2)) = x_1 + 2*x_2
   POL(MARK(x_1)) = x_1
   POL(a__and(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(a__length(x_1)) = x_1
   POL(a__take(x_1, x_2)) = x_1 + 2*x_2
   POL(a__zeros) = 0
   POL(and(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(cons(x_1, x_2)) = x_1 + 2*x_2
   POL(length(x_1)) = x_1
   POL(mark(x_1)) = x_1
   POL(nil) = 0
   POL(s(x_1)) = 2*x_1
   POL(take(x_1, x_2)) = x_1 + 2*x_2
   POL(tt) = 0
   POL(zeros) = 0


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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(N, L)) -> MARK(L)
   MARK(length(X)) -> A__LENGTH(mark(X))
   A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L))
   MARK(length(X)) -> MARK(X)
   MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2))
   A__TAKE(s(M), cons(N, IL)) -> MARK(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__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(11) 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__LENGTH(cons(N, L)) -> MARK(L)
   MARK(length(X)) -> A__LENGTH(mark(X))
   MARK(length(X)) -> MARK(X)
   MARK(take(X1, X2)) -> A__TAKE(mark(X1), mark(X2))
   A__TAKE(s(M), cons(N, IL)) -> MARK(N)
   MARK(take(X1, X2)) -> MARK(X1)
   MARK(take(X1, X2)) -> MARK(X2)


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(A__LENGTH(x_1)) = 2 + 2*x_1
   POL(A__TAKE(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(MARK(x_1)) = 2*x_1
   POL(a__and(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(a__length(x_1)) = 2 + x_1
   POL(a__take(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(a__zeros) = 0
   POL(and(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(cons(x_1, x_2)) = x_1 + x_2
   POL(length(x_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)) = 1 + x_1 + 2*x_2
   POL(tt) = 0
   POL(zeros) = 0


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

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L))
   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__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

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

(14)
Complex Obligation (AND)

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

(15)
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__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(16) 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.
----------------------------------------

(17)
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(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(18) 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(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)


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

(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__LENGTH(mark(L))

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(23) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule A__LENGTH(cons(N, L)) -> A__LENGTH(mark(L)) at position [0] we obtained the following new rules [LPAR04]:

   (A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros),A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros))
   (A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)),A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1)))
   (A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))),A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0))))
   (A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))),A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1))))
   (A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)),A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1)))
   (A__LENGTH(cons(y0, 0)) -> A__LENGTH(0),A__LENGTH(cons(y0, 0)) -> A__LENGTH(0))
   (A__LENGTH(cons(y0, tt)) -> A__LENGTH(tt),A__LENGTH(cons(y0, tt)) -> A__LENGTH(tt))
   (A__LENGTH(cons(y0, nil)) -> A__LENGTH(nil),A__LENGTH(cons(y0, nil)) -> A__LENGTH(nil))
   (A__LENGTH(cons(y0, s(x0))) -> A__LENGTH(s(mark(x0))),A__LENGTH(cons(y0, s(x0))) -> A__LENGTH(s(mark(x0))))


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros)
   A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1))
   A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0)))
   A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1)))
   A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1))
   A__LENGTH(cons(y0, 0)) -> A__LENGTH(0)
   A__LENGTH(cons(y0, tt)) -> A__LENGTH(tt)
   A__LENGTH(cons(y0, nil)) -> A__LENGTH(nil)
   A__LENGTH(cons(y0, s(x0))) -> A__LENGTH(s(mark(x0)))

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros)
   A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1))
   A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1))
   A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0)))
   A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1)))

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(27) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule A__LENGTH(cons(y0, zeros)) -> A__LENGTH(a__zeros) at position [0] we obtained the following new rules [LPAR04]:

   (A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)),A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)))
   (A__LENGTH(cons(y0, zeros)) -> A__LENGTH(zeros),A__LENGTH(cons(y0, zeros)) -> A__LENGTH(zeros))


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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1))
   A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1))
   A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0)))
   A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1)))
   A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros))
   A__LENGTH(cons(y0, zeros)) -> A__LENGTH(zeros)

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(29) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(30)
Complex Obligation (AND)

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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros))

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(32) 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.
----------------------------------------

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros))

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

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(34) 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(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)


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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros))

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

(36) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule A__LENGTH(cons(y0, zeros)) -> A__LENGTH(cons(0, zeros)) we obtained the following new rules [LPAR04]:

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


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

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(38) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = A__LENGTH(cons(0, zeros)) evaluates to  t =A__LENGTH(cons(0, zeros))

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

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

The DP semiunifies directly so there is only one rewrite step from A__LENGTH(cons(0, zeros)) to A__LENGTH(cons(0, zeros)).




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

(39)
NO

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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1))
   A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0)))
   A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1)))
   A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1))

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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

(41) 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__LENGTH(cons(y0, and(x0, x1))) -> A__LENGTH(a__and(mark(x0), x1))


Used ordering: Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(A__LENGTH(x_1)) = x_1
   POL(a__and(x_1, x_2)) = 2 + 2*x_1 + x_2
   POL(a__length(x_1)) = x_1
   POL(a__take(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(a__zeros) = 0
   POL(and(x_1, x_2)) = 2 + 2*x_1 + x_2
   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) = 0
   POL(s(x_1)) = 2*x_1
   POL(take(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(tt) = 0
   POL(zeros) = 0


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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A__LENGTH(cons(y0, length(x0))) -> A__LENGTH(a__length(mark(x0)))
   A__LENGTH(cons(y0, take(x0, x1))) -> A__LENGTH(a__take(mark(x0), mark(x1)))
   A__LENGTH(cons(y0, cons(x0, x1))) -> A__LENGTH(cons(mark(x0), x1))

The TRS R consists of the following rules:

   a__zeros -> cons(0, zeros)
   a__length(cons(N, L)) -> s(a__length(mark(L)))
   a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL))
   mark(zeros) -> a__zeros
   mark(and(X1, X2)) -> a__and(mark(X1), X2)
   mark(length(X)) -> a__length(mark(X))
   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(nil) -> nil
   mark(s(X)) -> s(mark(X))
   a__zeros -> zeros
   a__and(X1, X2) -> and(X1, X2)
   a__length(X) -> length(X)
   a__take(X1, X2) -> take(X1, X2)

The set Q consists of the following terms:

   a__zeros
   mark(zeros)
   mark(and(x0, x1))
   mark(length(x0))
   mark(take(x0, x1))
   mark(cons(x0, x1))
   mark(0)
   mark(tt)
   mark(nil)
   mark(s(x0))
   a__and(x0, x1)
   a__length(x0)
   a__take(x0, x1)

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