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

(0) Obligation:

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

zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
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(activate(x₁)) = x₁   
POL(and(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(cons(x₁, x₂)) = 2·x₁ + x₂   
POL(length(x₁)) = x₁   
POL(n__take(x₁, x₂)) = x₁ + 2·x₂   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = x₁ + 2·x₂   
POL(tt) = 2   
POL(zeros) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

and(tt, X) → activate(X)

(2) Obligation:

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

zeros → cons(0, n__zeros)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x₁)) = x₁   
POL(cons(x₁, x₂)) = 2·x₁ + x₂   
POL(length(x₁)) = 1 + 2·x₁   
POL(n__take(x₁, x₂)) = 2·x₁ + x₂   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = 2·x₁ + x₂   
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

(4) Obligation:

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

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
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.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(length(x₁)) = 2·x₁   
POL(n__take(x₁, x₂)) = 1 + x₁ + x₂   
POL(n__zeros) = 0   
POL(nil) = 2   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = 2 + 2·x₁ + 2·x₂   
POL(zeros) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

take(X1, X2) → n__take(X1, X2)

(6) Obligation:

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

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x₁)) = 2·x₁   
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(length(x₁)) = x₁   
POL(n__take(x₁, x₂)) = 1 + x₁ + x₂   
POL(n__zeros) = 0   
POL(nil) = 1   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = 2 + 2·x₁ + 2·x₂   
POL(zeros) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

take(0, IL) → nil

(8) Obligation:

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

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

Q is empty.

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x₁)) = 1 + 2·x₁   
POL(cons(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(length(x₁)) = x₁   
POL(n__take(x₁, x₂)) = x₁ + x₂   
POL(n__zeros) = 0   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(zeros) = 1   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

zeros → n__zeros
activate(X) → X

(10) Obligation:

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

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)

Q is empty.

(11) DependencyPairsProof (EQUIVALENT transformation)

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

(12) Obligation:

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

LENGTH(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
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)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) Complex Obligation (AND)

(15) Obligation:

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

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)

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

(16) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

• TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
  The graph contains the following edges 2 > 1

• ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
  The graph contains the following edges 1 > 1, 1 > 2

(17) YES

(18) Obligation:

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

LENGTH(cons(N, L)) → LENGTH(activate(L))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)

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

(19) 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 Narrowing at position: [0]
        LENGTH(cons(x0, n__zeros))[ ]ⁿ[ ] → LENGTH(zeros)[ ]ⁿ[ ]
            by Narrowing at position: [0]
                intermediate steps: Instantiation - Instantiation
                LENGTH(cons(N, L))[ ]ⁿ[ ] → LENGTH(activate(L))[ ]ⁿ[ ]
                    by OriginalRule from TRS P

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

        zeros[ ]ⁿ[ ] → cons(0, n__zeros)[ ]ⁿ[ ]
            by OriginalRule from TRS R

(20) NO