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)
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.

(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)) = x₁ + x₂ + x₃ + x₄   
POL(U22(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(U23(x₁, x₂, x₃, x₄)) = x₁ + x₂ + x₃ + x₄   
POL(activate(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(length(x₁)) = 1 + x₁   
POL(n__take(x₁, x₂)) = 1 + x₁ + x₂   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x₁)) = x₁   
POL(take(x₁, x₂)) = 1 + x₁ + x₂   
POL(tt) = 1   
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(activate(X1), activate(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(activate(X1), activate(X2))
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(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.

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

(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 + 2·x₂   
POL(U21(x₁, x₂, x₃, x₄)) = 2·x₁ + 2·x₂ + 2·x₃ + 2·x₄   
POL(U21¹(x₁, x₂, x₃, x₄)) = 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)) = 2·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)) = x₁ + x₂   
POL(n__take(x₁, x₂)) = x₁ + 2·x₂   
POL(n__zeros) = 0   
POL(s(x₁)) = 2·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(activate(X1), activate(X2))
U21¹(tt, IL, M, N) → U22¹(tt, activate(IL), activate(M), activate(N))
U22¹(tt, IL, M, N) → ACTIVATE(IL)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
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(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))
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 1 SCC with 12 less nodes.

(13) Obligation:

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

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

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))
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.

(14) 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.

(15) Obligation:

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

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

R is empty.
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:

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

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

(17) TRUE

(18) 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(activate(X1), activate(X2))
activate(X) → X

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

(19) 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.

(20) 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:

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.

(21) 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.

No dependency pairs are removed.

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))
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)))

Used ordering: POLO with Polynomial interpretation [POLO]:

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

(22) 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:

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.

(23) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U12¹(tt, L) → LENGTH(activate(L)) at position [0] we obtained the following new rules [LPAR04]:

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

(24) Obligation:

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

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

(25) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U12¹(tt, n__zeros) → LENGTH(zeros) at position [0] we obtained the following new rules [LPAR04]:

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

(26) Obligation:

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

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

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.

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(28) Obligation:

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

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

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.

(29) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, 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: Polynomial interpretation [POLO]:

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

(30) Obligation:

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

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

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.

(31) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = U12¹(tt, activate(activate(n__zeros))) evaluates to t =U12¹(tt, activate(activate(n__zeros)))

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

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

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

LENGTH(cons(0, n__zeros)) → U11¹(tt, activate(n__zeros))
with rule LENGTH(cons(N, L')) → U11¹(tt, activate(L')) at position [] and matcher [N / 0, L' / n__zeros]

U11¹(tt, activate(n__zeros)) → U12¹(tt, activate(activate(n__zeros)))
with rule U11¹(tt, L) → U12¹(tt, activate(L))

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.

(32) FALSE