Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

AND(tt, X) → ACTIVATE(X)
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(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)
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(activate(X1), activate(X2))
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__take(x1, x2)  =  n__take(x1, x2)
TAKE(x1, x2)  =  TAKE(x2)
activate(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  x2
zeros  =  zeros
0  =  0
n__zeros  =  n__zeros
and(x1, x2)  =  and(x1, x2)
tt  =  tt
length(x1)  =  length
nil  =  nil
take(x1, x2)  =  take(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[zeros, nzeros] > [0, length] > [ntake2, TAKE1, nil, take2]
and2 > [ntake2, TAKE1, nil, take2]
tt > [ntake2, TAKE1, nil, take2]

Status:
ntake2: [2,1]
TAKE1: [1]
zeros: []
0: []
nzeros: []
and2: [1,2]
tt: []
length: []
nil: []
take2: [2,1]


The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

Q DP problem:
P is empty.
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(activate(X1), activate(X2))
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) 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)
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(activate(X1), activate(X2))
activate(X) → X

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