Termination w.r.t. Q proof of /tmp/tmpMtVZ8a/OvConsOS_nosorts

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 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 PisEmptyProof (⇔)

    ↳28 TRUE

  ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 PisEmptyProof (⇔)

    ↳35 TRUE

  ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 QDPOrderProof (⇔)

    ↳40 QDP

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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:

ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(and(tt, X)) → MARK(X)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
ACTIVE(length(cons(N, L))) → S(length(L))
ACTIVE(length(cons(N, L))) → LENGTH(L)
ACTIVE(take(0, IL)) → MARK(nil)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
ACTIVE(take(s(M), cons(N, IL))) → CONS(N, take(M, IL))
ACTIVE(take(s(M), cons(N, IL))) → TAKE(M, IL)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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 6 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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.

TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = TAKE(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = and(x2)
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [mark₁, 0, take₂] > TAKE₁ > [nil, s]
zeros > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
zeros > [mark₁, 0, take₂] > tt > active₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > TAKE₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > tt > active₁ > [nil, s]
length > [mark₁, 0, take₂] > TAKE₁ > [nil, s]
length > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
length > [mark₁, 0, take₂] > tt > active₁ > [nil, s]

Status:

TAKE₁: [1]
mark₁: [1]
s: []
length: []
take₂: [2,1]
0: []
active₁: [1]
cons: []
tt: []
zeros: []
and₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(7) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = and(x1, x2)
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [mark₁, 0, and₂] > tt
zeros > [mark₁, 0, and₂] > nil > active₁ > [cons, s]
length > [mark₁, 0, and₂] > tt
length > [mark₁, 0, and₂] > nil > active₁ > [cons, s]
take > [mark₁, 0, and₂] > tt
take > [mark₁, 0, and₂] > nil > active₁ > [cons, s]

Status:

active₁: [1]
cons: []
tt: []
zeros: []
mark₁: [1]
and₂: [2,1]
s: []
length: []
0: []
nil: []
take: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)
S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
active(x1) = active(x1)
mark(x1) = mark(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = x2
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [active₁, mark₁] > 0 > [cons, nil]
zeros > [active₁, mark₁] > tt > [cons, nil]
zeros > [active₁, mark₁] > take > [cons, nil]
[length, s] > [active₁, mark₁] > 0 > [cons, nil]
[length, s] > [active₁, mark₁] > tt > [cons, nil]
[length, s] > [active₁, mark₁] > take > [cons, nil]

Status:

active₁: [1]
cons: []
tt: []
zeros: []
mark₁: [1]
s: []
length: []
0: []
nil: []
take: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = x1
active(x1) = active(x1)
mark(x1) = mark(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = x2
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [active₁, mark₁] > 0 > [cons, nil]
zeros > [active₁, mark₁] > tt > [cons, nil]
zeros > [active₁, mark₁] > take > [cons, nil]
[length, s] > [active₁, mark₁] > 0 > [cons, nil]
[length, s] > [active₁, mark₁] > tt > [cons, nil]
[length, s] > [active₁, mark₁] > take > [cons, nil]

Status:

active₁: [1]
cons: []
tt: []
zeros: []
mark₁: [1]
s: []
length: []
0: []
nil: []
take: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = AND(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = and(x2)
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [mark₁, 0, take₂] > AND₁ > [nil, s]
zeros > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
zeros > [mark₁, 0, take₂] > tt > active₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > AND₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > tt > active₁ > [nil, s]
length > [mark₁, 0, take₂] > AND₁ > [nil, s]
length > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
length > [mark₁, 0, take₂] > tt > active₁ > [nil, s]

Status:

AND₁: [1]
mark₁: [1]
s: []
length: []
take₂: [2,1]
0: []
active₁: [1]
cons: []
tt: []
zeros: []
and₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(24) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = and(x1, x2)
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [mark₁, 0, and₂] > tt
zeros > [mark₁, 0, and₂] > nil > active₁ > [cons, s]
length > [mark₁, 0, and₂] > tt
length > [mark₁, 0, and₂] > nil > active₁ > [cons, s]
take > [mark₁, 0, and₂] > tt
take > [mark₁, 0, and₂] > nil > active₁ > [cons, s]

Status:

active₁: [1]
cons: []
tt: []
zeros: []
mark₁: [1]
and₂: [2,1]
s: []
length: []
0: []
nil: []
take: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(26) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = and(x2)
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [mark₁, 0, take₂] > CONS₁ > [nil, s]
zeros > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
zeros > [mark₁, 0, take₂] > tt > active₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > CONS₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
and₁ > [mark₁, 0, take₂] > tt > active₁ > [nil, s]
length > [mark₁, 0, take₂] > CONS₁ > [nil, s]
length > [mark₁, 0, take₂] > cons > active₁ > [nil, s]
length > [mark₁, 0, take₂] > tt > active₁ > [nil, s]

Status:

CONS₁: [1]
mark₁: [1]
s: []
length: []
take₂: [2,1]
0: []
active₁: [1]
cons: []
tt: []
zeros: []
and₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(31) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
and(x1, x2) = and(x1, x2)
tt = tt
length(x1) = length
nil = nil
s(x1) = s
take(x1, x2) = take

Lexicographic path order with status [LPO].
Quasi-Precedence:

zeros > [mark₁, 0, and₂] > tt
zeros > [mark₁, 0, and₂] > nil > active₁ > [cons, s]
length > [mark₁, 0, and₂] > tt
length > [mark₁, 0, and₂] > nil > active₁ > [cons, s]
take > [mark₁, 0, and₂] > tt
take > [mark₁, 0, and₂] > nil > active₁ > [cons, s]

Status:

active₁: [1]
cons: []
tt: []
zeros: []
mark₁: [1]
and₂: [2,1]
s: []
length: []
0: []
nil: []
take: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(33) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(and(tt, X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK
cons(x1, x2) = cons
ACTIVE(x1) = x1
mark(x1) = mark
and(x1, x2) = and
tt = tt
zeros = zeros
0 = 0
length(x1) = length
s(x1) = s
take(x1, x2) = take
active(x1) = x1
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

[MARK, mark, and, zeros, length, s, take, nil] > cons
[MARK, mark, and, zeros, length, s, take, nil] > tt
[MARK, mark, and, zeros, length, s, take, nil] > 0

Status:

MARK: []
cons: []
tt: []
zeros: []
mark: []
and: []
s: []
length: []
0: []
take: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(38) Obligation:

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

ACTIVE(and(tt, X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
and(x1, x2) = and
tt = tt
MARK(x1) = MARK
zeros = zeros
cons(x1, x2) = cons
0 = 0
mark(x1) = mark
length(x1) = length
s(x1) = s
take(x1, x2) = take
active(x1) = active
nil = nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

cons > [tt, mark, active] > [and, MARK, zeros, length, take, nil] > 0
cons > [tt, mark, active] > [and, MARK, zeros, length, take, nil] > s

Status:

active: []
MARK: []
s: []
length: []
0: []
take: []
tt: []
cons: []
zeros: []
mark: []
and: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

(40) Obligation:

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

ACTIVE(and(tt, X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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