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

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

  ↳ RRRPoloQTRSProof

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.

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

take(0, IL) → nil

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(U11(x₁, x₂)) = 2·x₁ + 2·x₂
POL(U12(x₁, x₂)) = x₁ + 2·x₂
POL(U21(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₂ + 2·x₃ + 2·x₄
POL(U22(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + 2·x₃ + 2·x₄
POL(U23(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + 2·x₃ + 2·x₄
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(length(x₁)) = 2·x₁
POL(n__take(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(take(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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(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.

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

length(nil) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
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 + 2·x₃ + x₄
POL(U22(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + 2·x₃ + x₄
POL(U23(x₁, x₂, x₃, x₄)) = x₁ + 2·x₂ + 2·x₃ + x₄
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + 2·x₂
POL(length(x₁)) = x₁
POL(n__take(x₁, x₂)) = x₁ + x₂
POL(n__zeros) = 0
POL(nil) = 2
POL(s(x₁)) = 2·x₁
POL(take(x₁, x₂)) = x₁ + x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ 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(cons(N, L)) → U11(tt, activate(L))
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(cons(N, L)) → U11(tt, activate(L))
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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ 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(cons(N, L)) → U11(tt, activate(L))
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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                  ↳ 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(cons(N, L)) → U11(tt, activate(L))
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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                  ↳ 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:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

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))
U22¹(tt, IL, M, N) → ACTIVATE(N)
U21¹(tt, IL, M, N) → ACTIVATE(N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
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 following rules are removed from R:

take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVATE(x₁)) = x₁
POL(TAKE(x₁, x₂)) = x₁ + 2·x₂
POL(U21(x₁, x₂, x₃, x₄)) = 1 + x₁ + 2·x₂ + 2·x₃ + 2·x₄
POL(U21¹(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + x₃ + 2·x₄
POL(U22(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + 2·x₃ + 2·x₄
POL(U22¹(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + x₃ + x₄
POL(U23(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + x₃ + 2·x₄
POL(U23¹(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(n__take(x₁, x₂)) = x₁ + 2·x₂
POL(n__zeros) = 0
POL(s(x₁)) = 1 + 2·x₁
POL(take(x₁, x₂)) = x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                  ↳ QDP

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

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

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
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)))
zeros → cons(0, n__zeros)
zeros → n__zeros

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 5 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                  ↳ 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:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
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)))
zeros → cons(0, n__zeros)
zeros → n__zeros

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QDPSizeChangeProof

                  ↳ 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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
The graph contains the following edges 1 > 1
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

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(cons(N, L)) → U11(tt, activate(L))
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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

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:

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

take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTH(x₁)) = x₁
POL(U11¹(x₁, x₂)) = 2·x₁ + x₂
POL(U12¹(x₁, x₂)) = 2·x₁ + x₂
POL(U21(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + 2·x₃ + x₄
POL(U22(x₁, x₂, x₃, x₄)) = 1 + 2·x₁ + x₂ + 2·x₃ + x₄
POL(U23(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₂ + 2·x₃ + x₄
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(n__take(x₁, x₂)) = 2·x₁ + x₂
POL(n__zeros) = 0
POL(s(x₁)) = 2 + x₁
POL(take(x₁, x₂)) = 2·x₁ + x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ UsableRulesProof

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:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
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))
zeros → cons(0, n__zeros)
zeros → n__zeros

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ Narrowing

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:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, L) → LENGTH(activate(L)) at position [0] we obtained the following new rules:

U12¹(tt, n__take(x0, x1)) → LENGTH(take(activate(x0), activate(x1)))
U12¹(tt, n__zeros) → LENGTH(zeros)
U12¹(tt, x0) → LENGTH(x0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ Narrowing

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

U12¹(tt, n__take(x0, x1)) → LENGTH(take(activate(x0), activate(x1)))
U11¹(tt, L) → U12¹(tt, activate(L))
U12¹(tt, n__zeros) → LENGTH(zeros)
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U12¹(tt, n__zeros) → LENGTH(zeros) at position [0] we obtained the following new rules:

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U12¹(tt, n__zeros) → LENGTH(n__zeros)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

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

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U12¹(tt, n__take(x0, x1)) → LENGTH(take(activate(x0), activate(x1)))
U12¹(tt, n__zeros) → LENGTH(n__zeros)
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

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

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U12¹(tt, n__take(x0, x1)) → LENGTH(take(activate(x0), activate(x1)))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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:

U12¹(tt, n__take(x0, x1)) → LENGTH(take(activate(x0), activate(x1)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(LENGTH(x₁)) = x₁
POL(U11¹(x₁, x₂)) = x₁ + 2·x₂
POL(U12¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(n__take(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(n__zeros) = 0
POL(take(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(tt) = 0
POL(zeros) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ NonTerminationProof

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

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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:

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U11¹(tt, L) → U12¹(tt, activate(L))
LENGTH(cons(N, L)) → U11¹(tt, activate(L))
U12¹(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

s = LENGTH(cons(N, n__zeros)) evaluates to t =LENGTH(cons(0, n__zeros))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [N / 0]

Rewriting sequence

LENGTH(cons(N, n__zeros)) → U11¹(tt, activate(n__zeros))
with rule LENGTH(cons(N', L)) → U11¹(tt, activate(L)) at position [] and matcher [L / n__zeros, N' / N]

U11¹(tt, activate(n__zeros)) → U11¹(tt, n__zeros)
with rule activate(X) → X at position [1] and matcher [X / n__zeros]

U11¹(tt, n__zeros) → U12¹(tt, activate(n__zeros))
with rule U11¹(tt, L) → U12¹(tt, activate(L)) at position [] and matcher [L / n__zeros]

U12¹(tt, activate(n__zeros)) → U12¹(tt, n__zeros)
with rule activate(X) → X at position [1] and matcher [X / n__zeros]

U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))
with rule U12¹(tt, n__zeros) → LENGTH(cons(0, n__zeros))

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

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