Termination w.r.t. Q proof of /tmp/tmpy8H2vl/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 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 QDPOrderProof (⇔)

    ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 PisEmptyProof (⇔)

    ↳51 TRUE

  ↳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(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) = x1
mark(x1) = mark(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

Recursive path order with status [RPO].
Precedence:

zeros > mark₁ > s
zeros > mark₁ > nil > 0
tt > U12 > mark₁ > s
tt > U12 > mark₁ > nil > 0
length > U11 > U12 > mark₁ > s
length > U11 > U12 > mark₁ > nil > 0

Status:

LENGTH₁: multiset
mark₁: [1]
zeros: multiset
0: multiset
U11: multiset
tt: multiset
U12: multiset
s: multiset
length: []
nil: multiset

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:
The TRS P consists of the following rules:

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.

(8) 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)

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

Recursive path order with status [RPO].
Precedence:

LENGTH₁ > active₁
zeros > cons > mark₁ > active₁
zeros > 0 > active₁
length > 0 > active₁
length > U11 > tt > mark₁ > active₁
length > U11 > U12 > s > mark₁ > active₁
nil > mark₁ > active₁
nil > 0 > active₁

Status:

LENGTH₁: multiset
active₁: multiset
zeros: multiset
mark₁: multiset
cons: []
0: multiset
U11: []
tt: multiset
U12: []
s: []
length: []
nil: multiset

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)

(9) 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.

(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(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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) = x1
mark(x1) = mark(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

Recursive path order with status [RPO].
Precedence:

zeros > mark₁ > s
zeros > mark₁ > nil > 0
tt > U12 > mark₁ > s
tt > U12 > mark₁ > nil > 0
length > U11 > U12 > mark₁ > s
length > U11 > U12 > mark₁ > nil > 0

Status:

S₁: multiset
mark₁: [1]
zeros: multiset
0: multiset
U11: multiset
tt: multiset
U12: multiset
s: multiset
length: []
nil: multiset

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)

(14) Obligation:

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

S(active(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.

(15) 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)

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

Recursive path order with status [RPO].
Precedence:

S₁ > active₁
zeros > cons > mark₁ > active₁
zeros > 0 > active₁
length > 0 > active₁
length > U11 > tt > mark₁ > active₁
length > U11 > U12 > s > mark₁ > active₁
nil > mark₁ > active₁
nil > 0 > active₁

Status:

S₁: multiset
active₁: multiset
zeros: multiset
mark₁: multiset
cons: []
0: multiset
U11: []
tt: multiset
U12: []
s: []
length: []
nil: multiset

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)

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.

(20) 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) = U12¹(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

Recursive path order with status [RPO].
Precedence:

zeros > cons > tt > U12 > mark₁ > length₁
zeros > cons > tt > U12 > s
zeros > 0
nil > mark₁ > length₁
nil > 0

Status:

U12^1₁: [1]
mark₁: [1]
zeros: multiset
cons: []
0: multiset
tt: multiset
U12: multiset
s: multiset
length₁: multiset
nil: multiset

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(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.

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

Recursive path order with status [RPO].
Precedence:

U12^1₂ > length₁
zeros > cons > tt > U12 > mark₁ > length₁
zeros > cons > tt > U12 > s > length₁
zeros > 0 > length₁
nil > mark₁ > length₁
nil > 0 > length₁

Status:

U12^1₂: [2,1]
mark₁: [1]
zeros: multiset
cons: []
0: multiset
tt: multiset
U12: []
s: multiset
length₁: multiset
nil: multiset

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:
The TRS P consists of the following rules:

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.

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

Recursive path order with status [RPO].
Precedence:

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

Status:

active₁: multiset
zeros: multiset
mark₁: [1]
cons₁: [1]
0: multiset
U11₁: multiset
tt: multiset
U12: []
s: multiset
length: []
nil: multiset

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)

(25) Obligation:

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

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.

(26) 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) = U12¹(x1, x2)
active(x1) = active(x1)
zeros = zeros
mark(x1) = mark(x1)
cons(x1, x2) = cons(x1)
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12(x1)
s(x1) = s
length(x1) = length
nil = nil

Recursive path order with status [RPO].
Precedence:

zeros > mark₁ > 0 > active₁ > cons₁ > U12^1₂
zeros > mark₁ > 0 > active₁ > U12₁ > U12^1₂
zeros > mark₁ > s > active₁ > cons₁ > U12^1₂
zeros > mark₁ > s > active₁ > U12₁ > U12^1₂
zeros > mark₁ > nil > U12^1₂
length > U11 > tt > mark₁ > 0 > active₁ > cons₁ > U12^1₂
length > U11 > tt > mark₁ > 0 > active₁ > U12₁ > U12^1₂
length > U11 > tt > mark₁ > s > active₁ > cons₁ > U12^1₂
length > U11 > tt > mark₁ > s > active₁ > U12₁ > U12^1₂
length > U11 > tt > mark₁ > nil > U12^1₂

Status:

U12^1₂: [1,2]
active₁: multiset
zeros: multiset
mark₁: [1]
cons₁: [1]
0: multiset
U11: []
tt: multiset
U12₁: multiset
s: multiset
length: multiset
nil: multiset

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)

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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.

(31) 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) = U11¹(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

Recursive path order with status [RPO].
Precedence:

zeros > cons > tt > U12 > mark₁ > length₁
zeros > cons > tt > U12 > s
zeros > 0
nil > mark₁ > length₁
nil > 0

Status:

U11^1₁: [1]
mark₁: [1]
zeros: multiset
cons: []
0: multiset
tt: multiset
U12: multiset
s: multiset
length₁: multiset
nil: multiset

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(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.

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

Recursive path order with status [RPO].
Precedence:

U11^1₂ > length₁
zeros > cons > tt > U12 > mark₁ > length₁
zeros > cons > tt > U12 > s > length₁
zeros > 0 > length₁
nil > mark₁ > length₁
nil > 0 > length₁

Status:

U11^1₂: [2,1]
mark₁: [1]
zeros: multiset
cons: []
0: multiset
tt: multiset
U12: []
s: multiset
length₁: multiset
nil: multiset

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:
The TRS P consists of the following rules:

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.

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

Recursive path order with status [RPO].
Precedence:

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

Status:

active₁: multiset
zeros: multiset
mark₁: [1]
cons₁: [1]
0: multiset
U11₁: multiset
tt: multiset
U12: []
s: multiset
length: []
nil: multiset

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)

(36) Obligation:

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

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.

(37) 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) = U11¹(x1, x2)
active(x1) = active(x1)
zeros = zeros
mark(x1) = mark(x1)
cons(x1, x2) = cons(x1)
0 = 0
U11(x1, x2) = U11
tt = tt
U12(x1, x2) = U12(x1)
s(x1) = s
length(x1) = length
nil = nil

Recursive path order with status [RPO].
Precedence:

zeros > mark₁ > 0 > active₁ > cons₁ > U11^1₂
zeros > mark₁ > 0 > active₁ > U12₁ > U11^1₂
zeros > mark₁ > s > active₁ > cons₁ > U11^1₂
zeros > mark₁ > s > active₁ > U12₁ > U11^1₂
zeros > mark₁ > nil > U11^1₂
length > U11 > tt > mark₁ > 0 > active₁ > cons₁ > U11^1₂
length > U11 > tt > mark₁ > 0 > active₁ > U12₁ > U11^1₂
length > U11 > tt > mark₁ > s > active₁ > cons₁ > U11^1₂
length > U11 > tt > mark₁ > s > active₁ > U12₁ > U11^1₂
length > U11 > tt > mark₁ > nil > U11^1₂

Status:

U11^1₂: [1,2]
active₁: multiset
zeros: multiset
mark₁: [1]
cons₁: [1]
0: multiset
U11: []
tt: multiset
U12₁: multiset
s: multiset
length: multiset
nil: multiset

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)

(38) 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.

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) 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.

(42) 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) = CONS(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

Recursive path order with status [RPO].
Precedence:

zeros > cons > tt > U12 > mark₁ > length₁
zeros > cons > tt > U12 > s
zeros > 0
nil > mark₁ > length₁
nil > 0

Status:

CONS₁: [1]
mark₁: [1]
zeros: multiset
cons: []
0: multiset
tt: multiset
U12: multiset
s: multiset
length₁: multiset
nil: multiset

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(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.

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

Recursive path order with status [RPO].
Precedence:

CONS₂ > length₁
zeros > cons > tt > U12 > mark₁ > length₁
zeros > cons > tt > U12 > s > length₁
zeros > 0 > length₁
nil > mark₁ > length₁
nil > 0 > length₁

Status:

CONS₂: [2,1]
mark₁: [1]
zeros: multiset
cons: []
0: multiset
tt: multiset
U12: []
s: multiset
length₁: multiset
nil: multiset

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:
The TRS P consists of the following rules:

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.

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

Recursive path order with status [RPO].
Precedence:

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

Status:

active₁: multiset
zeros: multiset
mark₁: [1]
cons₁: [1]
0: multiset
U11₁: multiset
tt: multiset
U12: []
s: multiset
length: []
nil: multiset

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)

(47) Obligation:

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

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.

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

Recursive path order with status [RPO].
Precedence:

zeros > mark₁ > 0 > active₁ > cons₁ > CONS₂
zeros > mark₁ > 0 > active₁ > U12₁ > CONS₂
zeros > mark₁ > s > active₁ > cons₁ > CONS₂
zeros > mark₁ > s > active₁ > U12₁ > CONS₂
zeros > mark₁ > nil > CONS₂
length > U11 > tt > mark₁ > 0 > active₁ > cons₁ > CONS₂
length > U11 > tt > mark₁ > 0 > active₁ > U12₁ > CONS₂
length > U11 > tt > mark₁ > s > active₁ > cons₁ > CONS₂
length > U11 > tt > mark₁ > s > active₁ > U12₁ > CONS₂
length > U11 > tt > mark₁ > nil > CONS₂

Status:

CONS₂: [1,2]
active₁: multiset
zeros: multiset
mark₁: [1]
cons₁: [1]
0: multiset
U11: []
tt: multiset
U12₁: multiset
s: multiset
length: multiset
nil: multiset

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)

(49) 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.

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) TRUE

(52) 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.