Termination w.r.t. Q proof of /tmp/tmpNDkaEw/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:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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:

U21¹(tt, IL, M, N) → ACTIVATE(M)
U12¹(tt, L) → ACTIVATE(L)
U22¹(tt, IL, M, N) → ACTIVATE(IL)
LENGTH(cons(N, L)) → ACTIVATE(L)
U22¹(tt, IL, M, N) → U23¹(tt, activate(IL), activate(M), activate(N))
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
U22¹(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
U21¹(tt, IL, M, N) → U22¹(tt, activate(IL), activate(M), activate(N))
U23¹(tt, IL, M, N) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(M)
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
U21¹(tt, IL, M, N) → ACTIVATE(N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(N)
U22¹(tt, IL, M, N) → ACTIVATE(M)
U21¹(tt, IL, M, N) → ACTIVATE(IL)
U11¹(tt, L) → ACTIVATE(L)
TAKE(s(M), cons(N, IL)) → U21¹(tt, activate(IL), M, N)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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:

U21¹(tt, IL, M, N) → ACTIVATE(M)
U12¹(tt, L) → ACTIVATE(L)
U22¹(tt, IL, M, N) → ACTIVATE(IL)
LENGTH(cons(N, L)) → ACTIVATE(L)
U22¹(tt, IL, M, N) → U23¹(tt, activate(IL), activate(M), activate(N))
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
U22¹(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
U21¹(tt, IL, M, N) → U22¹(tt, activate(IL), activate(M), activate(N))
U23¹(tt, IL, M, N) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(M)
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
U21¹(tt, IL, M, N) → ACTIVATE(N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(N)
U22¹(tt, IL, M, N) → ACTIVATE(M)
U21¹(tt, IL, M, N) → ACTIVATE(IL)
U11¹(tt, L) → ACTIVATE(L)
TAKE(s(M), cons(N, IL)) → U21¹(tt, activate(IL), M, N)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

U21¹(tt, IL, M, N) → ACTIVATE(M)
U22¹(tt, IL, M, N) → ACTIVATE(IL)
U22¹(tt, IL, M, N) → U23¹(tt, activate(IL), activate(M), activate(N))
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
U22¹(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
U21¹(tt, IL, M, N) → U22¹(tt, activate(IL), activate(M), activate(N))
U23¹(tt, IL, M, N) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(M)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
U21¹(tt, IL, M, N) → ACTIVATE(N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(N)
U22¹(tt, IL, M, N) → ACTIVATE(M)
U21¹(tt, IL, M, N) → ACTIVATE(IL)
TAKE(s(M), cons(N, IL)) → U21¹(tt, activate(IL), M, N)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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.

TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
TAKE(s(M), cons(N, IL)) → U21¹(tt, activate(IL), M, N)

The remaining pairs can at least be oriented weakly.

U21¹(tt, IL, M, N) → ACTIVATE(M)
U22¹(tt, IL, M, N) → ACTIVATE(IL)
U22¹(tt, IL, M, N) → U23¹(tt, activate(IL), activate(M), activate(N))
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
U22¹(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
U21¹(tt, IL, M, N) → U22¹(tt, activate(IL), activate(M), activate(N))
U23¹(tt, IL, M, N) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(M)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
U21¹(tt, IL, M, N) → ACTIVATE(N)
U23¹(tt, IL, M, N) → ACTIVATE(N)
U22¹(tt, IL, M, N) → ACTIVATE(M)
U21¹(tt, IL, M, N) → ACTIVATE(IL)

Used ordering: Polynomial interpretation [25,35]:

POL(U23(x₁, x₂, x₃, x₄)) = (4)x_2 + (4)x_3 + (2)x_4
POL(U22(x₁, x₂, x₃, x₄)) = (2)x_1 + (4)x_2 + (4)x_3 + (3)x_4
POL(n__zeros) = 0
POL(U21¹(x₁, x₂, x₃, x₄)) = (4)x_2 + (4)x_3 + (4)x_4
POL(activate(x₁)) = x_1
POL(U21(x₁, x₂, x₃, x₄)) = (2)x_1 + (4)x_2 + (4)x_3 + (3)x_4
POL(take(x₁, x₂)) = (2)x_1 + (2)x_2
POL(0) = 0
POL(TAKE(x₁, x₂)) = (2)x_1 + (2)x_2
POL(cons(x₁, x₂)) = (2)x_1 + (2)x_2
POL(tt) = 4
POL(U22¹(x₁, x₂, x₃, x₄)) = (4)x_2 + (4)x_3 + (4)x_4
POL(zeros) = 0
POL(U23¹(x₁, x₂, x₃, x₄)) = (4)x_2 + (4)x_3 + (4)x_4
POL(n__take(x₁, x₂)) = (2)x_1 + (2)x_2
POL(s(x₁)) = 4 + (2)x_1
POL(ACTIVATE(x₁)) = (4)x_1
POL(nil) = 0

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

zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
activate(X) → X
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
zeros → cons(0, n__zeros)
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

U21¹(tt, IL, M, N) → ACTIVATE(M)
U22¹(tt, IL, M, N) → ACTIVATE(IL)
U22¹(tt, IL, M, N) → U23¹(tt, activate(IL), activate(M), activate(N))
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
U22¹(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
U21¹(tt, IL, M, N) → U22¹(tt, activate(IL), activate(M), activate(N))
U23¹(tt, IL, M, N) → ACTIVATE(IL)
U23¹(tt, IL, M, N) → ACTIVATE(M)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
U21¹(tt, IL, M, N) → ACTIVATE(N)
U23¹(tt, IL, M, N) → ACTIVATE(N)
U22¹(tt, IL, M, N) → ACTIVATE(M)
U21¹(tt, IL, M, N) → ACTIVATE(IL)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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 12 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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.

ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

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

POL(n__take(x₁, x₂)) = 1 + (4)x_1 + (4)x_2
POL(ACTIVATE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

U12¹(tt, L) → LENGTH(activate(L))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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