Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 QTRSRRRProof (⇔, 175 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 32 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 UsableRulesReductionPairsProof (⇔, 44 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 TRUE
    ↳14 QDP
        ↳15 NonLoopProof (⇔, 1918 ms)
        ↳16 NO

(0) Obligation:

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.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(U11(x1, x2)) = x1 + x2   
POL(U12(x1, x2)) = x1 + x2   
POL(U21(x1, x2, x3, x4)) = 2 + x1 + x2 + 2·x3 + x4   
POL(U22(x1, x2, x3, x4)) = 2 + x1 + x2 + 2·x3 + x4   
POL(U23(x1, x2, x3, x4)) = 2 + 2·x1 + x2 + 2·x3 + x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = x1   
POL(n__take(x1, x2)) = 2 + 2·x1 + x2   
POL(n__zeros) = 0   
POL(nil) = 1   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 2 + 2·x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

length(nil) → 0
take(0, IL) → nil


(2) Obligation:

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.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

(8) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

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

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.

(10) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] 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)) → U211(tt, activate(IL), M, N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
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 [POLO]:

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

(11) Obligation:

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

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

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.

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 12 less nodes.

(13) TRUE

(14) Obligation:

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

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

(15) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x0 / 0] on the rule
LENGTH(cons(0, n__zeros))[ ]n[ ] → LENGTH(cons(0, n__zeros))[ ]n[x0 / 0]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu
LENGTH(cons(x0, n__zeros))[ ]n[ ] → LENGTH(cons(0, n__zeros))[ ]n[ ]
    by Rewrite t
        LENGTH(cons(x0, n__zeros))[ ]n[ ] → LENGTH(activate(activate(zeros)))[ ]n[ ]
            by Narrowing at position: [0,0,0]
                intermediate steps: Instantiation
                LENGTH(cons(x0, x1))[ ]n[ ] → LENGTH(activate(activate(activate(x1))))[ ]n[ ]
                    by Narrowing at position: []
                        intermediate steps: Instantiation
                        LENGTH(cons(N, L))[ ]n[ ] → U111(tt, activate(L))[ ]n[ ]
                            by OriginalRule from TRS P

                        intermediate steps: Instantiation - Instantiation
                        U111(tt, x0)[ ]n[ ] → LENGTH(activate(activate(x0)))[ ]n[ ]
                            by Narrowing at position: []
                                intermediate steps: Instantiation
                                U111(tt, L)[ ]n[ ] → U121(tt, activate(L))[ ]n[ ]
                                    by OriginalRule from TRS P

                                intermediate steps: Instantiation - Instantiation
                                U121(tt, L)[ ]n[ ] → LENGTH(activate(L))[ ]n[ ]
                                    by OriginalRule from TRS P

                activate(n__zeros)[ ]n[ ] → zeros[ ]n[ ]
                    by OriginalRule from TRS R

(16) NO