Termination w.r.t. Q proof of /tmp/tmpRggTdS/LengthOfFiniteLists_nosorts_noand

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

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(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:

ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
ACTIVE(U11(tt, L)) → U12¹(tt, L)
ACTIVE(U12(tt, L)) → MARK(s(length(L)))
ACTIVE(U12(tt, L)) → S(length(L))
ACTIVE(U12(tt, L)) → LENGTH(L)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
ACTIVE(length(cons(N, L))) → U11¹(tt, L)
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(U11(X1, X2)) → ACTIVE(U11(mark(X1), X2))
MARK(U11(X1, X2)) → U11¹(mark(X1), X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(U12(X1, X2)) → ACTIVE(U12(mark(X1), X2))
MARK(U12(X1, X2)) → U12¹(mark(X1), X2)
MARK(U12(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
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)
U11¹(mark(X1), X2) → U11¹(X1, X2)
U11¹(X1, mark(X2)) → U11¹(X1, X2)
U11¹(active(X1), X2) → U11¹(X1, X2)
U11¹(X1, active(X2)) → U11¹(X1, X2)
U12¹(mark(X1), X2) → U12¹(X1, X2)
U12¹(X1, mark(X2)) → U12¹(X1, X2)
U12¹(active(X1), X2) → U12¹(X1, X2)
U12¹(X1, active(X2)) → U12¹(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(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 6 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) 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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(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.

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) = LENGTH(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
U11(x1, x2) = U11(x1)
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

LENGTH₁ > s
zeros > [mark₁, cons] > [active₁, 0] > U11₁ > s
zeros > [mark₁, cons] > nil > s
length > tt > U12 > [mark₁, cons] > [active₁, 0] > U11₁ > s
length > tt > U12 > [mark₁, cons] > nil > s

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(7) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(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:

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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(11) 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) = S(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
zeros = zeros
cons(x1, x2) = cons
0 = 0
U11(x1, x2) = U11(x1)
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

S₁ > s
zeros > [mark₁, cons] > [active₁, 0] > U11₁ > s
zeros > [mark₁, cons] > nil > s
length > tt > U12 > [mark₁, cons] > [active₁, 0] > U11₁ > s
length > tt > U12 > [mark₁, cons] > nil > s

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(12) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

U12¹(X1, mark(X2)) → U12¹(X1, X2)
U12¹(mark(X1), X2) → U12¹(X1, X2)
U12¹(active(X1), X2) → U12¹(X1, X2)
U12¹(X1, active(X2)) → U12¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U12¹(X1, mark(X2)) → U12¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U12¹(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = x1
zeros = zeros
cons(x1, x2) = cons
0 = 0
U11(x1, x2) = x1
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length(x1)
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > [cons, tt] > U12 > s > [mark₁, length₁, nil] > 0

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(17) Obligation:

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

U12¹(mark(X1), X2) → U12¹(X1, X2)
U12¹(active(X1), X2) → U12¹(X1, X2)
U12¹(X1, active(X2)) → U12¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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.

U12¹(X1, active(X2)) → U12¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U12¹(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1)
0 = 0
U11(x1, x2) = U11(x1)
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > mark₁ > [cons₁, U11₁] > [active₁, 0]
zeros > mark₁ > s > [active₁, 0]
length > tt > U12 > mark₁ > [cons₁, U11₁] > [active₁, 0]
length > tt > U12 > mark₁ > s > [active₁, 0]
nil > mark₁ > [cons₁, U11₁] > [active₁, 0]
nil > mark₁ > s > [active₁, 0]

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(19) Obligation:

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

U12¹(mark(X1), X2) → U12¹(X1, X2)
U12¹(active(X1), X2) → U12¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U12¹(mark(X1), X2) → U12¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U12¹(x1, x2) = U12¹(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
zeros = zeros
cons(x1, x2) = x1
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > [mark₁, tt, s] > 0 > U12^1₂
zeros > [mark₁, tt, s] > nil > U12^1₂
length > U11 > U12 > [mark₁, tt, s] > 0 > U12^1₂
length > U11 > U12 > [mark₁, tt, s] > nil > U12^1₂

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(21) Obligation:

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

U12¹(active(X1), X2) → U12¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U12¹(active(X1), X2) → U12¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U12¹(x1, x2) = x1
active(x1) = active(x1)
zeros = zeros
mark(x1) = mark(x1)
cons(x1, x2) = x2
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

[active₁, mark₁, 0, U11, length] > zeros > tt
[active₁, mark₁, 0, U11, length] > U12 > s > tt
[active₁, mark₁, 0, U11, length] > nil > tt

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(23) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

U11¹(X1, mark(X2)) → U11¹(X1, X2)
U11¹(mark(X1), X2) → U11¹(X1, X2)
U11¹(active(X1), X2) → U11¹(X1, X2)
U11¹(X1, active(X2)) → U11¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(X1, mark(X2)) → U11¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = x1
zeros = zeros
cons(x1, x2) = cons
0 = 0
U11(x1, x2) = x1
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length(x1)
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > [cons, tt] > U12 > s > [mark₁, length₁, nil] > 0

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(28) Obligation:

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

U11¹(mark(X1), X2) → U11¹(X1, X2)
U11¹(active(X1), X2) → U11¹(X1, X2)
U11¹(X1, active(X2)) → U11¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(X1, active(X2)) → U11¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2) = x2
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1)
0 = 0
U11(x1, x2) = U11(x1)
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > mark₁ > [cons₁, U11₁] > [active₁, 0]
zeros > mark₁ > s > [active₁, 0]
length > tt > U12 > mark₁ > [cons₁, U11₁] > [active₁, 0]
length > tt > U12 > mark₁ > s > [active₁, 0]
nil > mark₁ > [cons₁, U11₁] > [active₁, 0]
nil > mark₁ > s > [active₁, 0]

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(30) Obligation:

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

U11¹(mark(X1), X2) → U11¹(X1, X2)
U11¹(active(X1), X2) → U11¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(mark(X1), X2) → U11¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2) = U11¹(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
zeros = zeros
cons(x1, x2) = x1
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > [mark₁, tt, s] > 0 > U11^1₂
zeros > [mark₁, tt, s] > nil > U11^1₂
length > U11 > U12 > [mark₁, tt, s] > 0 > U11^1₂
length > U11 > U12 > [mark₁, tt, s] > nil > U11^1₂

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(32) Obligation:

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

U11¹(active(X1), X2) → U11¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(active(X1), X2) → U11¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2) = x1
active(x1) = active(x1)
zeros = zeros
mark(x1) = mark(x1)
cons(x1, x2) = x2
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

[active₁, mark₁, 0, U11, length] > zeros > tt
[active₁, mark₁, 0, U11, length] > U12 > s > tt
[active₁, mark₁, 0, U11, length] > nil > tt

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(34) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) 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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(38) 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)

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) = x1
zeros = zeros
cons(x1, x2) = cons
0 = 0
U11(x1, x2) = x1
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length(x1)
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > [cons, tt] > U12 > s > [mark₁, length₁, nil] > 0

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(39) Obligation:

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

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(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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(x1)
0 = 0
U11(x1, x2) = U11(x1)
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > mark₁ > [cons₁, U11₁] > [active₁, 0]
zeros > mark₁ > s > [active₁, 0]
length > tt > U12 > mark₁ > [cons₁, U11₁] > [active₁, 0]
length > tt > U12 > mark₁ > s > [active₁, 0]
nil > mark₁ > [cons₁, U11₁] > [active₁, 0]
nil > mark₁ > s > [active₁, 0]

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(41) Obligation:

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

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(42) 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)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
zeros = zeros
cons(x1, x2) = x1
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

zeros > [mark₁, tt, s] > 0 > CONS₂
zeros > [mark₁, tt, s] > nil > CONS₂
length > U11 > U12 > [mark₁, tt, s] > 0 > CONS₂
length > U11 > U12 > [mark₁, tt, s] > nil > CONS₂

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(43) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x1
active(x1) = active(x1)
zeros = zeros
mark(x1) = mark(x1)
cons(x1, x2) = x2
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
s(x1) = s
length(x1) = length
nil = nil

Lexicographic Path Order [LPO].
Precedence:

[active₁, mark₁, 0, U11, length] > zeros > tt
[active₁, mark₁, 0, U11, length] > U12 > s > tt
[active₁, mark₁, 0, U11, length] > nil > tt

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(45) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(U11(X1, X2)) → ACTIVE(U11(mark(X1), X2))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))
MARK(U11(X1, X2)) → MARK(X1)
MARK(U12(X1, X2)) → ACTIVE(U12(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(U12(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(49) 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.
MARK(x1) = MARK
cons(x1, x2) = cons
ACTIVE(x1) = x1
mark(x1) = mark
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12
zeros = zeros
0 = 0
s(x1) = s
length(x1) = length
active(x1) = x1
nil = nil

Lexicographic Path Order [LPO].
Precedence:

[MARK, cons, mark, U11, U12, zeros, 0, length, nil] > tt
[MARK, cons, mark, U11, U12, zeros, 0, length, nil] > s

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(50) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(U11(X1, X2)) → ACTIVE(U11(mark(X1), X2))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))
MARK(U11(X1, X2)) → MARK(X1)
MARK(U12(X1, X2)) → ACTIVE(U12(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(U12(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(51) 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))

[MARK, mark, U11, U12, zeros, 0, length] > cons
[MARK, mark, U11, U12, zeros, 0, length] > tt > s
[MARK, mark, U11, U12, zeros, 0, length] > nil

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

(52) Obligation:

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

ACTIVE(U11(tt, L)) → MARK(U12(tt, L))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(U11(X1, X2)) → ACTIVE(U11(mark(X1), X2))
ACTIVE(U12(tt, L)) → MARK(s(length(L)))
MARK(U11(X1, X2)) → MARK(X1)
MARK(U12(X1, X2)) → ACTIVE(U12(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(U11(tt, L))
MARK(U12(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(U11(X1, X2)) → active(U11(mark(X1), X2))
mark(tt) → active(tt)
mark(U12(X1, X2)) → active(U12(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
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)
U11(mark(X1), X2) → U11(X1, X2)
U11(X1, mark(X2)) → U11(X1, X2)
U11(active(X1), X2) → U11(X1, X2)
U11(X1, active(X2)) → U11(X1, X2)
U12(mark(X1), X2) → U12(X1, X2)
U12(X1, mark(X2)) → U12(X1, X2)
U12(active(X1), X2) → U12(X1, X2)
U12(X1, active(X2)) → U12(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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