Termination w.r.t. Q proof of /tmp/tmp3_9HFk/OvConsOS_nosorts_noand

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(nil) → 0
a__length(cons(N, L)) → a__U11(tt, L)
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(nil) → 0
a__length(cons(N, L)) → a__U11(tt, L)
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.

The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(nil) → 0
a__length(cons(N, L)) → a__U11(tt, L)
a__take(0, IL) → nil
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

a__length(nil) → 0
a__take(0, IL) → nil

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(U11(x₁, x₂)) = x₁ + x₂
POL(U12(x₁, x₂)) = x₁ + x₂
POL(U21(x₁, x₂, x₃, x₄)) = 2 + x₁ + x₂ + x₃ + x₄
POL(U22(x₁, x₂, x₃, x₄)) = 2 + x₁ + x₂ + x₃ + x₄
POL(U23(x₁, x₂, x₃, x₄)) = 2 + x₁ + x₂ + x₃ + x₄
POL(a__U11(x₁, x₂)) = x₁ + x₂
POL(a__U12(x₁, x₂)) = x₁ + x₂
POL(a__U21(x₁, x₂, x₃, x₄)) = 2 + x₁ + x₂ + x₃ + x₄
POL(a__U22(x₁, x₂, x₃, x₄)) = 2 + x₁ + x₂ + x₃ + x₄
POL(a__U23(x₁, x₂, x₃, x₄)) = 2 + x₁ + x₂ + x₃ + x₄
POL(a__length(x₁)) = x₁
POL(a__take(x₁, x₂)) = 2 + x₁ + x₂
POL(a__zeros) = 0
POL(cons(x₁, x₂)) = x₁ + x₂
POL(length(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(nil) = 1
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 2 + x₁ + x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK(U22(X1, X2, X3, X4)) → MARK(X1)
A__U22(tt, IL, M, N) → A__U23(tt, IL, M, N)
MARK(U23(X1, X2, X3, X4)) → A__U23(mark(X1), X2, X3, X4)
MARK(take(X1, X2)) → MARK(X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(zeros) → A__ZEROS
MARK(U21(X1, X2, X3, X4)) → A__U21(mark(X1), X2, X3, X4)
MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
MARK(U22(X1, X2, X3, X4)) → A__U22(mark(X1), X2, X3, X4)
MARK(length(X)) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
MARK(take(X1, X2)) → MARK(X1)
MARK(U23(X1, X2, X3, X4)) → MARK(X1)
A__U12(tt, L) → A__LENGTH(mark(L))
A__U21(tt, IL, M, N) → A__U22(tt, IL, M, N)
MARK(U12(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__U23(tt, IL, M, N) → MARK(N)
A__U12(tt, L) → MARK(L)
A__U11(tt, L) → A__U12(tt, L)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)
A__TAKE(s(M), cons(N, IL)) → A__U21(tt, IL, M, N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MARK(U22(X1, X2, X3, X4)) → MARK(X1)
A__U22(tt, IL, M, N) → A__U23(tt, IL, M, N)
MARK(U23(X1, X2, X3, X4)) → A__U23(mark(X1), X2, X3, X4)
MARK(take(X1, X2)) → MARK(X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(zeros) → A__ZEROS
MARK(U21(X1, X2, X3, X4)) → A__U21(mark(X1), X2, X3, X4)
MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
MARK(U22(X1, X2, X3, X4)) → A__U22(mark(X1), X2, X3, X4)
MARK(length(X)) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
MARK(take(X1, X2)) → MARK(X1)
MARK(U23(X1, X2, X3, X4)) → MARK(X1)
A__U12(tt, L) → A__LENGTH(mark(L))
A__U21(tt, IL, M, N) → A__U22(tt, IL, M, N)
MARK(U12(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__U23(tt, IL, M, N) → MARK(N)
A__U12(tt, L) → MARK(L)
A__U11(tt, L) → A__U12(tt, L)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)
A__TAKE(s(M), cons(N, IL)) → A__U21(tt, IL, M, N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

Q DP problem:
The TRS P consists of the following rules:

MARK(U22(X1, X2, X3, X4)) → MARK(X1)
A__U22(tt, IL, M, N) → A__U23(tt, IL, M, N)
MARK(U23(X1, X2, X3, X4)) → A__U23(mark(X1), X2, X3, X4)
MARK(take(X1, X2)) → MARK(X2)
MARK(U21(X1, X2, X3, X4)) → A__U21(mark(X1), X2, X3, X4)
MARK(U11(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
MARK(U22(X1, X2, X3, X4)) → A__U22(mark(X1), X2, X3, X4)
MARK(length(X)) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
MARK(take(X1, X2)) → MARK(X1)
MARK(U23(X1, X2, X3, X4)) → MARK(X1)
A__U12(tt, L) → A__LENGTH(mark(L))
A__U21(tt, IL, M, N) → A__U22(tt, IL, M, N)
MARK(U12(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__U23(tt, IL, M, N) → MARK(N)
A__U12(tt, L) → MARK(L)
A__U11(tt, L) → A__U12(tt, L)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)
A__TAKE(s(M), cons(N, IL)) → A__U21(tt, IL, M, N)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(U22(X1, X2, X3, X4)) → MARK(X1)
A__U22(tt, IL, M, N) → A__U23(tt, IL, M, N)
MARK(U23(X1, X2, X3, X4)) → A__U23(mark(X1), X2, X3, X4)
MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → MARK(X1)
MARK(U23(X1, X2, X3, X4)) → MARK(X1)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = x₁
POL(A__TAKE(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(A__U11(x₁, x₂)) = x₁ + x₂
POL(A__U12(x₁, x₂)) = 2·x₁ + x₂
POL(A__U21(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₂ + x₃ + 2·x₄
POL(A__U22(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + x₃ + 2·x₄
POL(A__U23(x₁, x₂, x₃, x₄)) = 2·x₁ + x₂ + x₃ + 2·x₄
POL(MARK(x₁)) = x₁
POL(U11(x₁, x₂)) = x₁ + 2·x₂
POL(U12(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U21(x₁, x₂, x₃, x₄)) = 1 + x₁ + 2·x₂ + x₃ + 2·x₄
POL(U22(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + 2·x₂ + x₃ + 2·x₄
POL(U23(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + 2·x₂ + x₃ + 2·x₄
POL(a__U11(x₁, x₂)) = x₁ + 2·x₂
POL(a__U12(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a__U21(x₁, x₂, x₃, x₄)) = 1 + x₁ + 2·x₂ + x₃ + 2·x₄
POL(a__U22(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + 2·x₂ + x₃ + 2·x₄
POL(a__U23(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + 2·x₂ + x₃ + 2·x₄
POL(a__length(x₁)) = 2·x₁
POL(a__take(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(a__zeros) = 0
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

A__U21(tt, IL, M, N) → A__U22(tt, IL, M, N)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(U12(X1, X2)) → MARK(X1)
A__U23(tt, IL, M, N) → MARK(N)
MARK(U21(X1, X2, X3, X4)) → A__U21(mark(X1), X2, X3, X4)
MARK(U11(X1, X2)) → MARK(X1)
A__U12(tt, L) → MARK(L)
MARK(s(X)) → MARK(X)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
MARK(U22(X1, X2, X3, X4)) → A__U22(mark(X1), X2, X3, X4)
MARK(length(X)) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
A__U11(tt, L) → A__U12(tt, L)
A__TAKE(s(M), cons(N, IL)) → A__U21(tt, IL, M, N)
A__U12(tt, L) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

Q DP problem:
The TRS P consists of the following rules:

MARK(length(X)) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U12(X1, X2)) → MARK(X1)
MARK(U11(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
A__U12(tt, L) → MARK(L)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U11(tt, L) → A__U12(tt, L)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → A__LENGTH(mark(X))
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)

MARK(length(X)) → MARK(X)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U12(X1, X2)) → MARK(X1)
MARK(U11(X1, X2)) → MARK(X1)
MARK(U12(X1, X2)) → A__U12(mark(X1), X2)
A__U12(tt, L) → MARK(L)
MARK(length(X)) → A__LENGTH(mark(X))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(A__LENGTH(x₁)) = 2 + 2·x₁
POL(A__U11(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(A__U12(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(MARK(x₁)) = 1 + 2·x₁
POL(U11(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(U12(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(U21(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + x₃ + x₄
POL(U22(x₁, x₂, x₃, x₄)) = x₁ + 2·x₂ + x₃ + x₄
POL(U23(x₁, x₂, x₃, x₄)) = x₁ + 2·x₂ + x₃ + x₄
POL(a__U11(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(a__U12(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(a__U21(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + x₃ + x₄
POL(a__U22(x₁, x₂, x₃, x₄)) = x₁ + 2·x₂ + x₃ + x₄
POL(a__U23(x₁, x₂, x₃, x₄)) = x₁ + 2·x₂ + x₃ + x₄
POL(a__length(x₁)) = 2 + 2·x₁
POL(a__take(x₁, x₂)) = x₁ + 2·x₂
POL(a__zeros) = 0
POL(cons(x₁, x₂)) = x₁ + x₂
POL(length(x₁)) = 2 + 2·x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

MARK(s(X)) → MARK(X)
A__U11(tt, L) → A__U12(tt, L)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
MARK(cons(X1, X2)) → MARK(X1)
A__U12(tt, L) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ Narrowing

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, L) → A__LENGTH(mark(L))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__U12(tt, L) → A__LENGTH(mark(L)) at position [0] we obtained the following new rules:

A__U12(tt, zeros) → A__LENGTH(a__zeros)
A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__U12(tt, tt) → A__LENGTH(tt)
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, cons(x0, x1)) → A__LENGTH(cons(mark(x0), x1))
A__U12(tt, 0) → A__LENGTH(0)
A__U12(tt, U11(x0, x1)) → A__LENGTH(a__U11(mark(x0), x1))
A__U12(tt, U22(x0, x1, x2, x3)) → A__LENGTH(a__U22(mark(x0), x1, x2, x3))
A__U12(tt, s(x0)) → A__LENGTH(s(mark(x0)))
A__U12(tt, nil) → A__LENGTH(nil)
A__U12(tt, U23(x0, x1, x2, x3)) → A__LENGTH(a__U23(mark(x0), x1, x2, x3))
A__U12(tt, U21(x0, x1, x2, x3)) → A__LENGTH(a__U21(mark(x0), x1, x2, x3))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

A__U12(tt, zeros) → A__LENGTH(a__zeros)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__U12(tt, tt) → A__LENGTH(tt)
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, cons(x0, x1)) → A__LENGTH(cons(mark(x0), x1))
A__U12(tt, 0) → A__LENGTH(0)
A__U12(tt, U11(x0, x1)) → A__LENGTH(a__U11(mark(x0), x1))
A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, U22(x0, x1, x2, x3)) → A__LENGTH(a__U22(mark(x0), x1, x2, x3))
A__U12(tt, nil) → A__LENGTH(nil)
A__U12(tt, s(x0)) → A__LENGTH(s(mark(x0)))
A__U12(tt, U23(x0, x1, x2, x3)) → A__LENGTH(a__U23(mark(x0), x1, x2, x3))
A__U12(tt, U21(x0, x1, x2, x3)) → A__LENGTH(a__U21(mark(x0), x1, x2, x3))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Narrowing

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

A__U12(tt, cons(x0, x1)) → A__LENGTH(cons(mark(x0), x1))
A__U12(tt, U11(x0, x1)) → A__LENGTH(a__U11(mark(x0), x1))
A__U12(tt, zeros) → A__LENGTH(a__zeros)
A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__U11(tt, L) → A__U12(tt, L)
A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U12(tt, U22(x0, x1, x2, x3)) → A__LENGTH(a__U22(mark(x0), x1, x2, x3))
A__U12(tt, U12(x0, x1)) → A__LENGTH(a__U12(mark(x0), x1))
A__U12(tt, U23(x0, x1, x2, x3)) → A__LENGTH(a__U23(mark(x0), x1, x2, x3))
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))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A__U12(tt, zeros) → A__LENGTH(a__zeros) at position [0] we obtained the following new rules:

A__U12(tt, zeros) → A__LENGTH(zeros)
A__U12(tt, zeros) → A__LENGTH(cons(0, zeros))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(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, cons(x0, x1)) → A__LENGTH(cons(mark(x0), x1))
A__U12(tt, U11(x0, x1)) → A__LENGTH(a__U11(mark(x0), x1))
A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, U22(x0, x1, x2, x3)) → A__LENGTH(a__U22(mark(x0), x1, x2, x3))
A__U12(tt, zeros) → A__LENGTH(zeros)
A__U12(tt, zeros) → A__LENGTH(cons(0, zeros))
A__U12(tt, U23(x0, x1, x2, x3)) → A__LENGTH(a__U23(mark(x0), x1, x2, x3))
A__U12(tt, U21(x0, x1, x2, x3)) → A__LENGTH(a__U21(mark(x0), x1, x2, x3))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

A__U12(tt, cons(x0, x1)) → A__LENGTH(cons(mark(x0), x1))
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, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, zeros) → A__LENGTH(cons(0, zeros))
A__U12(tt, U22(x0, x1, x2, x3)) → A__LENGTH(a__U22(mark(x0), x1, x2, x3))
A__U12(tt, U12(x0, x1)) → A__LENGTH(a__U12(mark(x0), x1))
A__U12(tt, U23(x0, x1, x2, x3)) → A__LENGTH(a__U23(mark(x0), x1, x2, x3))
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))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

A__U12(tt, cons(x0, x1)) → A__LENGTH(cons(mark(x0), x1))
A__U12(tt, U11(x0, x1)) → A__LENGTH(a__U11(mark(x0), x1))
A__U12(tt, U22(x0, x1, x2, x3)) → A__LENGTH(a__U22(mark(x0), x1, x2, x3))
A__U12(tt, U12(x0, x1)) → A__LENGTH(a__U12(mark(x0), x1))
A__U12(tt, U23(x0, x1, x2, x3)) → A__LENGTH(a__U23(mark(x0), x1, x2, x3))
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))

The remaining pairs can at least be oriented weakly.

A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, zeros) → A__LENGTH(cons(0, zeros))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 1
POL(A__LENGTH(x₁)) = x₁
POL(A__U11(x₁, x₂)) = 1 + x₂
POL(A__U12(x₁, x₂)) = x₁ + x₂
POL(U11(x₁, x₂)) = 1
POL(U12(x₁, x₂)) = 1
POL(U21(x₁, x₂, x₃, x₄)) = 1
POL(U22(x₁, x₂, x₃, x₄)) = 1
POL(U23(x₁, x₂, x₃, x₄)) = 1
POL(a__U11(x₁, x₂)) = 1
POL(a__U12(x₁, x₂)) = 1
POL(a__U21(x₁, x₂, x₃, x₄)) = 1
POL(a__U22(x₁, x₂, x₃, x₄)) = 1
POL(a__U23(x₁, x₂, x₃, x₄)) = 1
POL(a__length(x₁)) = 1
POL(a__take(x₁, x₂)) = 1
POL(a__zeros) = 1
POL(cons(x₁, x₂)) = 1 + x₂
POL(length(x₁)) = 1
POL(mark(x₁)) = 1 + x₁
POL(nil) = 1
POL(s(x₁)) = 1
POL(take(x₁, x₂)) = 0
POL(tt) = 1
POL(zeros) = 0

The following usable rules [17] were oriented:

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(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U11(X1, X2) → U11(X1, X2)
mark(s(X)) → s(mark(X))
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__length(X) → length(X)
a__U12(X1, X2) → U12(X1, X2)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(length(X)) → a__length(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ NonTerminationProof

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(N, L)) → A__U11(tt, L)
A__U12(tt, zeros) → A__LENGTH(cons(0, zeros))

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

A__U11(tt, L) → A__U12(tt, L)
A__U12(tt, take(x0, x1)) → A__LENGTH(a__take(mark(x0), mark(x1)))
A__LENGTH(cons(N, L)) → A__U11(tt, L)
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)

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:

Semiunifier: [ ]
Matcher: [ ]

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(N, L')) → A__U11(tt, L') at position [] and matcher [L' / zeros, N / 0]

A__U11(tt, zeros) → A__U12(tt, zeros)
with rule A__U11(tt, L) → A__U12(tt, L)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ RuleRemovalProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__zeros → cons(0, zeros)
a__U11(tt, L) → a__U12(tt, L)
a__U12(tt, L) → s(a__length(mark(L)))
a__U21(tt, IL, M, N) → a__U22(tt, IL, M, N)
a__U22(tt, IL, M, N) → a__U23(tt, IL, M, N)
a__U23(tt, IL, M, N) → cons(mark(N), take(M, IL))
a__length(cons(N, L)) → a__U11(tt, L)
a__take(s(M), cons(N, IL)) → a__U21(tt, IL, M, N)
mark(zeros) → a__zeros
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U12(X1, X2)) → a__U12(mark(X1), X2)
mark(length(X)) → a__length(mark(X))
mark(U21(X1, X2, X3, X4)) → a__U21(mark(X1), X2, X3, X4)
mark(U22(X1, X2, X3, X4)) → a__U22(mark(X1), X2, X3, X4)
mark(U23(X1, X2, X3, X4)) → a__U23(mark(X1), X2, X3, X4)
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(nil) → nil
a__zeros → zeros
a__U11(X1, X2) → U11(X1, X2)
a__U12(X1, X2) → U12(X1, X2)
a__length(X) → length(X)
a__U21(X1, X2, X3, X4) → U21(X1, X2, X3, X4)
a__U22(X1, X2, X3, X4) → U22(X1, X2, X3, X4)
a__U23(X1, X2, X3, X4) → U23(X1, X2, X3, X4)
a__take(X1, X2) → take(X1, X2)

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