Termination w.r.t. Q proof of /tmp/tmpiNmhj8/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 PisEmptyProof (⇔)

    ↳23 TRUE

  ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 PisEmptyProof (⇔)

    ↳28 TRUE

  ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳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 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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) → CONS(0, zeros)
ACTIVE(U11(tt, L)) → U12¹(tt, L)
ACTIVE(U12(tt, L)) → S(length(L))
ACTIVE(U12(tt, L)) → LENGTH(L)
ACTIVE(length(cons(N, L))) → U11¹(tt, L)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(U11(X1, X2)) → U11¹(active(X1), X2)
ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → U12¹(active(X1), X2)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
U11¹(mark(X1), X2) → U11¹(X1, X2)
U12¹(mark(X1), X2) → U12¹(X1, X2)
S(mark(X)) → S(X)
LENGTH(mark(X)) → LENGTH(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(U11(X1, X2)) → U11¹(proper(X1), proper(X2))
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U12(X1, X2)) → U12¹(proper(X1), proper(X2))
PROPER(U12(X1, X2)) → PROPER(X1)
PROPER(U12(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
U11¹(ok(X1), ok(X2)) → U11¹(X1, X2)
U12¹(ok(X1), ok(X2)) → U12¹(X1, X2)
S(ok(X)) → S(X)
LENGTH(ok(X)) → LENGTH(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 8 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

LENGTH(ok(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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)
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1, x2)
0 = 0
U11(x1, x2) = U11(x1, x2)
tt = tt
U12(x1, x2) = U12(x1, x2)
s(x1) = x1
length(x1) = length(x1)
nil = nil
proper(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

LENGTH₁ > mark₁
active₁ > zeros > mark₁
active₁ > cons₂ > U11₂ > tt > mark₁
active₁ > cons₂ > U11₂ > U12₂ > mark₁
active₁ > 0 > mark₁
active₁ > length₁ > U11₂ > tt > mark₁
active₁ > length₁ > U11₂ > U12₂ > mark₁
nil > mark₁
top > mark₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

LENGTH(ok(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X)) → LENGTH(X)

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

Recursive Path Order [RPO].
Precedence:

top > active₁ > zeros
top > proper > zeros
top > proper > tt > ok₁
top > proper > nil > 0 > ok₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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)
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1, x2)
0 = 0
U11(x1, x2) = U11(x1, x2)
tt = tt
U12(x1, x2) = U12(x1, x2)
s(x1) = x1
length(x1) = length(x1)
nil = nil
proper(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

S₁ > mark₁
active₁ > zeros > mark₁
active₁ > cons₂ > U11₂ > tt > mark₁
active₁ > cons₂ > U11₂ > U12₂ > mark₁
active₁ > 0 > mark₁
active₁ > length₁ > U11₂ > tt > mark₁
active₁ > length₁ > U11₂ > U12₂ > mark₁
nil > mark₁
top > mark₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) Obligation:

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

S(ok(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X)) → S(X)

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

Recursive Path Order [RPO].
Precedence:

top > active₁ > zeros
top > proper > zeros
top > proper > tt > ok₁
top > proper > nil > 0 > ok₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X1), ok(X2)) → U12¹(X1, X2)
U12¹(mark(X1), X2) → U12¹(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(ok(X1), ok(X2)) → U12¹(X1, X2)
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)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1, x2)
0 = 0
U11(x1, x2) = U11(x1, x2)
tt = tt
U12(x1, x2) = U12(x1, x2)
s(x1) = s(x1)
length(x1) = x1
nil = nil
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > zeros > ok₁
active₁ > zeros > mark₁
active₁ > cons₂ > U11₂ > ok₁
active₁ > cons₂ > U11₂ > mark₁
active₁ > 0 > ok₁
active₁ > tt > U12₂ > ok₁
active₁ > tt > U12₂ > mark₁
active₁ > tt > s₁ > ok₁
active₁ > tt > s₁ > mark₁
proper₁ > zeros > ok₁
proper₁ > zeros > mark₁
proper₁ > cons₂ > U11₂ > ok₁
proper₁ > cons₂ > U11₂ > mark₁
proper₁ > 0 > ok₁
proper₁ > tt > U12₂ > ok₁
proper₁ > tt > U12₂ > mark₁
proper₁ > tt > s₁ > ok₁
proper₁ > tt > s₁ > mark₁
proper₁ > nil > ok₁
proper₁ > nil > mark₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) 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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(ok(X1), ok(X2)) → U11¹(X1, X2)
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)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1, x2)
0 = 0
U11(x1, x2) = U11(x1, x2)
tt = tt
U12(x1, x2) = U12(x1, x2)
s(x1) = s(x1)
length(x1) = x1
nil = nil
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > zeros > ok₁
active₁ > zeros > mark₁
active₁ > cons₂ > U11₂ > ok₁
active₁ > cons₂ > U11₂ > mark₁
active₁ > 0 > ok₁
active₁ > tt > U12₂ > ok₁
active₁ > tt > U12₂ > mark₁
active₁ > tt > s₁ > ok₁
active₁ > tt > s₁ > mark₁
proper₁ > zeros > ok₁
proper₁ > zeros > mark₁
proper₁ > cons₂ > U11₂ > ok₁
proper₁ > cons₂ > U11₂ > mark₁
proper₁ > 0 > ok₁
proper₁ > tt > U12₂ > ok₁
proper₁ > tt > U12₂ > mark₁
proper₁ > tt > s₁ > ok₁
proper₁ > tt > s₁ > mark₁
proper₁ > nil > ok₁
proper₁ > nil > mark₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(26) 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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
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)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1, x2)
0 = 0
U11(x1, x2) = U11(x1, x2)
tt = tt
U12(x1, x2) = U12(x1, x2)
s(x1) = s(x1)
length(x1) = x1
nil = nil
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > zeros > ok₁
active₁ > zeros > mark₁
active₁ > cons₂ > U11₂ > ok₁
active₁ > cons₂ > U11₂ > mark₁
active₁ > 0 > ok₁
active₁ > tt > U12₂ > ok₁
active₁ > tt > U12₂ > mark₁
active₁ > tt > s₁ > ok₁
active₁ > tt > s₁ > mark₁
proper₁ > zeros > ok₁
proper₁ > zeros > mark₁
proper₁ > cons₂ > U11₂ > ok₁
proper₁ > cons₂ > U11₂ > mark₁
proper₁ > 0 > ok₁
proper₁ > tt > U12₂ > ok₁
proper₁ > tt > U12₂ > mark₁
proper₁ > tt > s₁ > ok₁
proper₁ > tt > s₁ > mark₁
proper₁ > nil > ok₁
proper₁ > nil > mark₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(31) 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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U12(X1, X2)) → PROPER(X1)
PROPER(U12(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U12(X1, X2)) → PROPER(X1)
PROPER(U12(X1, X2)) → PROPER(X2)

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

Recursive Path Order [RPO].
Precedence:

tt > mark > U11₂ > ok > U12₂ > PROPER₁
nil > mark > U11₂ > ok > U12₂ > PROPER₁
nil > 0 > PROPER₁
top > proper₁ > cons₂ > mark > U11₂ > ok > U12₂ > PROPER₁
top > proper₁ > zeros > mark > U11₂ > ok > U12₂ > PROPER₁
top > proper₁ > zeros > 0 > PROPER₁

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(36) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(s(X)) → PROPER(X)
PROPER(length(X)) → PROPER(X)

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

Recursive Path Order [RPO].
Precedence:

PROPER₁ > mark
zeros > mark
tt > length₁ > 0 > mark
tt > length₁ > U11₂ > U12 > s₁ > mark
nil > 0 > mark
top > mark

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → ACTIVE(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(U12(X1, X2)) → ACTIVE(X1)

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

Recursive Path Order [RPO].
Precedence:

U11₂ > U12₂ > ACTIVE₁ > mark
zeros > cons₂ > mark
0 > mark
tt > U12₂ > ACTIVE₁ > mark
nil > mark
top > mark

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(43) Obligation:

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

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → ACTIVE(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(length(X)) → ACTIVE(X)

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

Recursive Path Order [RPO].
Precedence:

ACTIVE₁ > mark
zeros > mark
tt > length₁ > 0 > mark
tt > length₁ > U11₂ > U12 > s₁ > mark
nil > 0 > mark
top > mark

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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(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))
active(cons(X1, X2)) → cons(active(X1), X2)
active(U11(X1, X2)) → U11(active(X1), X2)
active(U12(X1, X2)) → U12(active(X1), X2)
active(s(X)) → s(active(X))
active(length(X)) → length(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
U11(mark(X1), X2) → mark(U11(X1, X2))
U12(mark(X1), X2) → mark(U12(X1, X2))
s(mark(X)) → mark(s(X))
length(mark(X)) → mark(length(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U12(X1, X2)) → U12(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U12(ok(X1), ok(X2)) → ok(U12(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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