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(X1, 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(X1, 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(X1, 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)) = x1 + x2   
POL(U12(x1, x2)) = x1 + x2   
POL(U21(x1, x2, x3, x4)) = 1 + x1 + 2·x2 + 2·x3 + 2·x4   
POL(U22(x1, x2, x3, x4)) = 1 + 2·x1 + 2·x2 + 2·x3 + 2·x4   
POL(U23(x1, x2, x3, x4)) = 1 + x1 + 2·x2 + 2·x3 + 2·x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(length(x1)) = x1   
POL(n__take(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(n__zeros) = 0   
POL(nil) = 1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + 2·x1 + 2·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(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(X1, 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(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(X1, 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)) = x1 + 2·x2   
POL(U12(x1, x2)) = x1 + 2·x2   
POL(U21(x1, x2, x3, x4)) = 1 + x1 + 2·x2 + 2·x3 + 2·x4   
POL(U22(x1, x2, x3, x4)) = 1 + 2·x1 + 2·x2 + 2·x3 + 2·x4   
POL(U23(x1, x2, x3, x4)) = 1 + 2·x1 + 2·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 + 2·x2   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + 2·x1 + 2·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(X1, 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))
U221(tt, IL, M, N) → ACTIVATE(N)
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))
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)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
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(X1, 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))
U221(tt, IL, M, N) → ACTIVATE(N)
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))
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)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
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(X1, 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))
U221(tt, IL, M, N) → ACTIVATE(N)
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)
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)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
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(X1, 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))
U221(tt, IL, M, N) → ACTIVATE(N)
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)
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)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
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(X1, 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:

TAKE(s(M), cons(N, IL)) → 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)
Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(TAKE(x1, x2)) = 2·x1 + x2   
POL(U21(x1, x2, x3, x4)) = x1 + 2·x2 + 2·x3 + 2·x4   
POL(U211(x1, x2, x3, x4)) = 2·x1 + x2 + 2·x3 + x4   
POL(U22(x1, x2, x3, x4)) = 2·x1 + 2·x2 + 2·x3 + 2·x4   
POL(U221(x1, x2, x3, x4)) = x1 + x2 + x3 + x4   
POL(U23(x1, x2, x3, x4)) = 2·x1 + 2·x2 + 2·x3 + 2·x4   
POL(U231(x1, x2, x3, x4)) = x1 + x2 + x3 + x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(n__take(x1, x2)) = 2·x1 + 2·x2   
POL(n__zeros) = 0   
POL(s(x1)) = 2 + x1   
POL(take(x1, x2)) = 2·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)
U221(tt, IL, M, N) → ACTIVATE(M)
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(IL)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
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)
U211(tt, IL, M, N) → ACTIVATE(N)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, 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))
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 0 SCCs with 12 less nodes.

↳ 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(X1, 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
                  ↳ 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(X1, 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.

No dependency pairs are removed.

The following rules are removed from R:

take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(LENGTH(x1)) = 2·x1   
POL(U111(x1, x2)) = 2·x1 + 2·x2   
POL(U121(x1, x2)) = 2·x1 + 2·x2   
POL(U21(x1, x2, x3, x4)) = 1 + x1 + x2 + 2·x3 + 2·x4   
POL(U22(x1, x2, x3, x4)) = 1 + x1 + x2 + 2·x3 + 2·x4   
POL(U23(x1, x2, x3, x4)) = 2·x1 + x2 + 2·x3 + 2·x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(n__take(x1, x2)) = 2·x1 + x2   
POL(n__zeros) = 0   
POL(s(x1)) = 1 + 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(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
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)))
zeros → cons(0, n__zeros)
zeros → n__zeros

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
                  ↳ 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(X1, 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(x0, 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(x0, 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(X1, 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__take(x0, x1)) → LENGTH(take(x0, x1)) at position [0] we obtained the following new rules:

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



↳ 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__take(x0, x1)) → LENGTH(n__take(x0, 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(X1, 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
                                            ↳ Narrowing

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

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(X1, 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
                                          ↳ 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__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(X1, 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
                                            ↳ Narrowing
                                              ↳ QDP
                                                ↳ DependencyGraphProof
                                                  ↳ 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(X1, 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(X1, 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:



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.