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

Lexicographic path order with status [LPO].
Precedence:
LENGTH1 > cons
zeros > mark1 > s > active1 > 0 > cons
tt > U12 > mark1 > s > active1 > 0 > cons
length > U11 > U12 > mark1 > s > active1 > 0 > cons
nil > mark1 > s > active1 > 0 > cons

Status:
LENGTH1: [1]
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
active1: [1]
cons: []
tt: []
zeros: []
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)

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

Lexicographic path order with status [LPO].
Precedence:
S1 > cons
zeros > mark1 > s > active1 > 0 > cons
tt > U12 > mark1 > s > active1 > 0 > cons
length > U11 > U12 > mark1 > s > active1 > 0 > cons
nil > mark1 > s > active1 > 0 > cons

Status:
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
active1: [1]
cons: []
tt: []
zeros: []
nil: []
S1: [1]

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(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)
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 with status [LPO].
Precedence:
zeros > mark1 > tt
zeros > mark1 > s
zeros > 0
length > 0
length > U11 > U12 > mark1 > tt
length > U11 > U12 > mark1 > s
nil > mark1 > tt
nil > mark1 > s
nil > 0

Status:
U12^11: [1]
tt: []
zeros: []
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
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)

(17) Obligation:

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

U121(X1, mark(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, 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(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1)
0  =  0
U11(x1, x2)  =  U11
tt  =  tt
U12(x1, x2)  =  U12
s(x1)  =  s
length(x1)  =  length
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
U12^12 > cons1
zeros > mark1 > cons1
zeros > 0 > cons1
tt > length > 0 > cons1
tt > length > U11 > U12 > mark1 > cons1
tt > length > U11 > U12 > s > cons1
nil > mark1 > cons1

Status:
U12^12: [1,2]
cons1: [1]
tt: []
zeros: []
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
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)

(19) 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.

(20) 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)  =  U121(x1)
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
s(x1)  =  s
length(x1)  =  length
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
U12^11 > tt
zeros > mark1 > s > active1 > cons1 > tt
zeros > mark1 > s > active1 > 0 > tt
zeros > mark1 > nil > tt
length > U11 > U12 > mark1 > s > active1 > cons1 > tt
length > U11 > U12 > mark1 > s > active1 > 0 > tt
length > U11 > U12 > mark1 > nil > tt

Status:
cons1: [1]
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
active1: [1]
U12^11: [1]
tt: []
zeros: []
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)

(21) 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.

(22) 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
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(x1)
s(x1)  =  s
length(x1)  =  length(x1)
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
zeros > cons > U11 > tt > mark1 > U121 > active1
zeros > cons > U11 > tt > mark1 > s > active1
zeros > cons > U11 > tt > mark1 > length1 > 0 > active1
nil > mark1 > U121 > active1
nil > mark1 > s > active1
nil > mark1 > length1 > 0 > active1

Status:
active1: [1]
cons: []
tt: []
zeros: []
mark1: [1]
U11: []
U121: [1]
s: []
length1: [1]
0: []
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)

(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(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)
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 with status [LPO].
Precedence:
zeros > mark1 > tt
zeros > mark1 > s
zeros > 0
length > 0
length > U11 > U12 > mark1 > tt
length > U11 > U12 > mark1 > s
nil > mark1 > tt
nil > mark1 > s
nil > 0

Status:
U11^11: [1]
tt: []
zeros: []
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
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)

(28) Obligation:

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

U111(X1, mark(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, 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(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1)
0  =  0
U11(x1, x2)  =  U11
tt  =  tt
U12(x1, x2)  =  U12
s(x1)  =  s
length(x1)  =  length
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
U11^12 > cons1
zeros > mark1 > cons1
zeros > 0 > cons1
tt > length > 0 > cons1
tt > length > U11 > U12 > mark1 > cons1
tt > length > U11 > U12 > s > cons1
nil > mark1 > cons1

Status:
cons1: [1]
tt: []
U11^12: [1,2]
zeros: []
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
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)

(30) 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.

(31) 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)  =  U111(x1)
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
s(x1)  =  s
length(x1)  =  length
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
U11^11 > tt
zeros > mark1 > s > active1 > cons1 > tt
zeros > mark1 > s > active1 > 0 > tt
zeros > mark1 > nil > tt
length > U11 > U12 > mark1 > s > active1 > cons1 > tt
length > U11 > U12 > mark1 > s > active1 > 0 > tt
length > U11 > U12 > mark1 > nil > tt

Status:
cons1: [1]
U11^11: [1]
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
active1: [1]
tt: []
zeros: []
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)

(32) 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.

(33) 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
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(x1)
s(x1)  =  s
length(x1)  =  length(x1)
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
zeros > cons > U11 > tt > mark1 > U121 > active1
zeros > cons > U11 > tt > mark1 > s > active1
zeros > cons > U11 > tt > mark1 > length1 > 0 > active1
nil > mark1 > U121 > active1
nil > mark1 > s > active1
nil > mark1 > length1 > 0 > active1

Status:
active1: [1]
cons: []
tt: []
zeros: []
mark1: [1]
U11: []
U121: [1]
s: []
length1: [1]
0: []
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)

(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(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)
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 with status [LPO].
Precedence:
zeros > mark1 > tt
zeros > mark1 > s
zeros > 0
length > 0
length > U11 > U12 > mark1 > tt
length > U11 > U12 > mark1 > s
nil > mark1 > tt
nil > mark1 > s
nil > 0

Status:
tt: []
CONS1: [1]
zeros: []
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
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)

(39) Obligation:

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

CONS(X1, mark(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, 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(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1)
0  =  0
U11(x1, x2)  =  U11
tt  =  tt
U12(x1, x2)  =  U12
s(x1)  =  s
length(x1)  =  length
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
CONS2 > cons1
zeros > mark1 > cons1
zeros > 0 > cons1
tt > length > 0 > cons1
tt > length > U11 > U12 > mark1 > cons1
tt > length > U11 > U12 > s > cons1
nil > mark1 > cons1

Status:
cons1: [1]
tt: []
zeros: []
CONS2: [1,2]
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
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)

(41) 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.

(42) 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)  =  CONS(x1)
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
s(x1)  =  s
length(x1)  =  length
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
CONS1 > tt
zeros > mark1 > s > active1 > cons1 > tt
zeros > mark1 > s > active1 > 0 > tt
zeros > mark1 > nil > tt
length > U11 > U12 > mark1 > s > active1 > cons1 > tt
length > U11 > U12 > mark1 > s > active1 > 0 > tt
length > U11 > U12 > mark1 > nil > tt

Status:
cons1: [1]
CONS1: [1]
U12: []
mark1: [1]
U11: []
s: []
length: []
0: []
active1: [1]
tt: []
zeros: []
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)

(43) 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.

(44) 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
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(x1)
s(x1)  =  s
length(x1)  =  length(x1)
nil  =  nil

Lexicographic path order with status [LPO].
Precedence:
zeros > cons > U11 > tt > mark1 > U121 > active1
zeros > cons > U11 > tt > mark1 > s > active1
zeros > cons > U11 > tt > mark1 > length1 > 0 > active1
nil > mark1 > U121 > active1
nil > mark1 > s > active1
nil > mark1 > length1 > 0 > active1

Status:
active1: [1]
cons: []
tt: []
zeros: []
mark1: [1]
U11: []
U121: [1]
s: []
length1: [1]
0: []
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)

(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.