Termination w.r.t. Q proof of /tmp/tmpPp5-iU/OvConsOS_nosorts_noand

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

  ↳ 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(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.

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

↳ 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(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.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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(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.
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.

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

The remaining pairs can at least be oriented weakly.

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(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(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(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__U23(tt, IL, M, N) → MARK(N)
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)

Used ordering: Polynomial interpretation [25,35]:

POL(U23(x₁, x₂, x₃, x₄)) = (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(a__U11(x₁, x₂)) = 1/4 + (3/2)x_1 + (4)x_2
POL(a__zeros) = 0
POL(U22(x₁, x₂, x₃, x₄)) = (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(mark(x₁)) = x_1
POL(U12(x₁, x₂)) = 1/4 + (4)x_1 + (4)x_2
POL(take(x₁, x₂)) = (4)x_1 + x_2
POL(A__U21(x₁, x₂, x₃, x₄)) = (1/4)x_2 + x_4
POL(a__length(x₁)) = 1/4 + (4)x_1
POL(tt) = 0
POL(zeros) = 0
POL(s(x₁)) = x_1
POL(nil) = 0
POL(A__U22(x₁, x₂, x₃, x₄)) = (4)x_1 + (1/4)x_2 + x_4
POL(a__U22(x₁, x₂, x₃, x₄)) = (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(A__LENGTH(x₁)) = 1/4 + (2)x_1
POL(A__U12(x₁, x₂)) = 1/4 + (2)x_2
POL(A__U11(x₁, x₂)) = 1/4 + (2)x_2
POL(A__U23(x₁, x₂, x₃, x₄)) = (4)x_1 + (1/4)x_2 + x_4
POL(U11(x₁, x₂)) = 1/4 + (3/2)x_1 + (4)x_2
POL(U21(x₁, x₂, x₃, x₄)) = (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(a__take(x₁, x₂)) = (4)x_1 + x_2
POL(0) = 0
POL(A__TAKE(x₁, x₂)) = (1/4)x_2
POL(cons(x₁, x₂)) = (4)x_1 + x_2
POL(MARK(x₁)) = x_1
POL(a__U12(x₁, x₂)) = 1/4 + (4)x_1 + (4)x_2
POL(a__U23(x₁, x₂, x₃, x₄)) = (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(a__U21(x₁, x₂, x₃, x₄)) = (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(length(x₁)) = 1/4 + (4)x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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(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)) → A__TAKE(mark(X1), mark(X2))
A__U23(tt, IL, M, N) → MARK(N)
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(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)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X1)
A__TAKE(s(M), cons(N, IL)) → A__U21(tt, IL, M, N)
MARK(U23(X1, X2, X3, X4)) → 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(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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

A__U21(tt, IL, M, N) → A__U22(tt, IL, M, N)
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)) → A__TAKE(mark(X1), mark(X2))
MARK(U21(X1, X2, X3, X4)) → A__U21(mark(X1), X2, X3, X4)
A__U23(tt, IL, M, N) → MARK(N)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(U22(X1, X2, X3, X4)) → A__U22(mark(X1), X2, X3, X4)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X1)
MARK(U23(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(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.
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.

MARK(U22(X1, X2, X3, X4)) → MARK(X1)
MARK(U23(X1, X2, X3, X4)) → A__U23(mark(X1), X2, X3, X4)
MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(U21(X1, X2, X3, X4)) → A__U21(mark(X1), X2, X3, X4)
MARK(cons(X1, X2)) → MARK(X1)
MARK(U22(X1, X2, X3, X4)) → A__U22(mark(X1), X2, X3, X4)
MARK(U21(X1, X2, X3, X4)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X1)
MARK(U23(X1, X2, X3, X4)) → MARK(X1)
A__TAKE(s(M), cons(N, IL)) → A__U21(tt, IL, M, N)

The remaining pairs can at least be oriented weakly.

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(s(X)) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(U23(x₁, x₂, x₃, x₄)) = 3/2 + (4)x_1 + x_2 + x_3 + (4)x_4
POL(a__U11(x₁, x₂)) = 0
POL(a__zeros) = 4
POL(U22(x₁, x₂, x₃, x₄)) = 5/2 + x_1 + (1/4)x_2 + (1/2)x_3 + (2)x_4
POL(mark(x₁)) = 1/2 + (4)x_1
POL(U12(x₁, x₂)) = 0
POL(take(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(A__U21(x₁, x₂, x₃, x₄)) = 7/4 + (1/4)x_4
POL(a__length(x₁)) = 0
POL(tt) = 0
POL(zeros) = 4
POL(s(x₁)) = x_1
POL(A__U22(x₁, x₂, x₃, x₄)) = 7/4 + (1/4)x_4
POL(nil) = 7/4
POL(a__U22(x₁, x₂, x₃, x₄)) = 4 + x_1 + x_2 + (2)x_3 + (4)x_4
POL(A__U23(x₁, x₂, x₃, x₄)) = 7/4 + (1/4)x_4
POL(U11(x₁, x₂)) = 0
POL(U21(x₁, x₂, x₃, x₄)) = 3/2 + (4)x_1 + (1/4)x_2 + (3/2)x_3 + (4)x_4
POL(0) = 0
POL(a__take(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(A__TAKE(x₁, x₂)) = 5/2 + (1/4)x_2
POL(cons(x₁, x₂)) = 5/2 + x_1 + (1/4)x_2
POL(MARK(x₁)) = 7/4 + (1/4)x_1
POL(a__U21(x₁, x₂, x₃, x₄)) = 4 + (4)x_1 + x_2 + (4)x_3 + (4)x_4
POL(a__U12(x₁, x₂)) = 0
POL(a__U23(x₁, x₂, x₃, x₄)) = 4 + (4)x_1 + x_2 + x_3 + (4)x_4
POL(length(x₁)) = 0

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ 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)
A__U22(tt, IL, M, N) → A__U23(tt, IL, M, N)
A__U23(tt, IL, M, N) → MARK(N)
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(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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

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(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.
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.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x₁)) = (2)x_1
POL(s(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.