Termination w.r.t. Q proof of /tmp/tmpGUIAn4/LengthOfFiniteLists_nosorts_noand

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 PisEmptyProof (⇔)

    ↳39 TRUE

  ↳40 QDP

    ↳41 QDPOrderProof (⇔)

    ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 PisEmptyProof (⇔)

    ↳48 TRUE

  ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 PisEmptyProof (⇔)

    ↳55 TRUE

  ↳56 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) = 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) = s(x1)
length(x1) = length(x1)
nil = nil
proper(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

active₁ > [zeros, cons₂] > [mark₁, s₁, length₁]
active₁ > 0 > [mark₁, s₁, length₁]
active₁ > U11₂ > [mark₁, s₁, length₁]
active₁ > tt > [mark₁, s₁, length₁]
active₁ > U12₂ > [mark₁, s₁, length₁]
nil > 0 > [mark₁, s₁, length₁]
top > [mark₁, s₁, length₁]

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

Lexicographic Path Order [LPO].
Precedence:

LENGTH₁ > tt
proper₁ > [ok₁, top] > [active, 0] > zeros > mark > tt
proper₁ > nil > [active, 0] > zeros > mark > tt

The following usable rules [FROCOS05] were oriented:

active(zeros) → mark(cons(0, zeros))
active(U11(tt, L)) → mark(U12(tt, L))
active(U12(tt, L)) → mark(s(length(L)))
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(U11(tt, L))
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) = 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) = s(x1)
length(x1) = length(x1)
nil = nil
proper(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

active₁ > [zeros, cons₂] > [mark₁, s₁, length₁]
active₁ > 0 > [mark₁, s₁, length₁]
active₁ > U11₂ > [mark₁, s₁, length₁]
active₁ > tt > [mark₁, s₁, length₁]
active₁ > U12₂ > [mark₁, s₁, length₁]
nil > 0 > [mark₁, s₁, length₁]
top > [mark₁, s₁, length₁]

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

Lexicographic Path Order [LPO].
Precedence:

S₁ > tt
proper₁ > [ok₁, top] > [active, 0] > zeros > mark > tt
proper₁ > nil > [active, 0] > zeros > mark > tt

The following usable rules [FROCOS05] were oriented:

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

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

Lexicographic Path Order [LPO].
Precedence:

[U11₂, proper₁] > 0 > [ok₁, nil] > [mark, zeros]
[U11₂, proper₁] > tt > U12₁ > [ok₁, nil] > [mark, zeros]
top > [mark, zeros]

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:
The TRS P consists of the following rules:

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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U12¹(mark(X1), X2) → U12¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U12¹(x1, x2) = U12¹(x1, x2)
mark(x1) = mark(x1)
active(x1) = 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) = length(x1)
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

U12^1₂ > [mark₁, tt, length₁, nil]
proper₁ > zeros > [active₁, cons₂, 0] > U11₂ > [mark₁, tt, length₁, nil]
proper₁ > zeros > [active₁, cons₂, 0] > [U12₂, s₁] > [mark₁, tt, length₁, nil]
top > [active₁, cons₂, 0] > U11₂ > [mark₁, tt, length₁, nil]
top > [active₁, cons₂, 0] > [U12₂, s₁] > [mark₁, tt, length₁, 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))
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))

(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))
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.

(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(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.

(27) 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)

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

Lexicographic Path Order [LPO].
Precedence:

[U11₂, proper₁] > 0 > [ok₁, nil] > [mark, zeros]
[U11₂, proper₁] > tt > U12₁ > [ok₁, nil] > [mark, zeros]
top > [mark, zeros]

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))

(28) Obligation:

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

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.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(mark(X1), X2) → U11¹(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2) = U11¹(x1, x2)
mark(x1) = mark(x1)
active(x1) = 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) = length(x1)
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

U11^1₂ > [mark₁, tt, length₁, nil]
proper₁ > zeros > [active₁, cons₂, 0] > U11₂ > [mark₁, tt, length₁, nil]
proper₁ > zeros > [active₁, cons₂, 0] > [U12₂, s₁] > [mark₁, tt, length₁, nil]
top > [active₁, cons₂, 0] > U11₂ > [mark₁, tt, length₁, nil]
top > [active₁, cons₂, 0] > [U12₂, s₁] > [mark₁, tt, length₁, 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))
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))

(30) 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.

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) 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.

(34) 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)

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

Lexicographic Path Order [LPO].
Precedence:

[U11₂, proper₁] > 0 > [ok₁, nil] > [mark, zeros]
[U11₂, proper₁] > tt > U12₁ > [ok₁, nil] > [mark, zeros]
top > [mark, zeros]

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))

(35) Obligation:

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

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.

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
mark(x1) = mark(x1)
active(x1) = 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) = length(x1)
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

CONS₂ > [mark₁, tt, length₁, nil]
proper₁ > zeros > [active₁, cons₂, 0] > U11₂ > [mark₁, tt, length₁, nil]
proper₁ > zeros > [active₁, cons₂, 0] > [U12₂, s₁] > [mark₁, tt, length₁, nil]
top > [active₁, cons₂, 0] > U11₂ > [mark₁, tt, length₁, nil]
top > [active₁, cons₂, 0] > [U12₂, s₁] > [mark₁, tt, length₁, 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))
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))

(37) 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.

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) 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.

(41) 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) = active(x1)
zeros = zeros
mark(x1) = mark
0 = 0
tt = tt
nil = nil
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

PROPER₁ > cons₂
proper₁ > [U12₂, active₁, zeros, ok, top] > U11₂ > mark > cons₂
proper₁ > [U12₂, active₁, zeros, ok, top] > 0 > cons₂
proper₁ > [U12₂, active₁, zeros, ok, top] > tt > mark > cons₂
proper₁ > nil > cons₂

The following usable rules [FROCOS05] were oriented:

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

(42) 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.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

Lexicographic Path Order [LPO].
Precedence:

PROPER₁ > [mark, 0, U11, U12₂, ok]
[length₁, tt] > [mark, 0, U11, U12₂, ok]
zeros > [mark, 0, U11, U12₂, ok]
nil > [mark, 0, U11, U12₂, ok]
top > [mark, 0, U11, U12₂, ok]

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))

(44) Obligation:

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

PROPER(s(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.

(45) 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)

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

Lexicographic Path Order [LPO].
Precedence:

zeros > [mark, cons₁] > tt
zeros > [mark, cons₁] > top
zeros > 0
[U11₂, U12₂, length₁] > [PROPER₁, s₁] > [mark, cons₁] > tt
[U11₂, U12₂, length₁] > [PROPER₁, s₁] > [mark, cons₁] > top
[U11₂, U12₂, length₁] > 0
nil > 0

The following usable rules [FROCOS05] were oriented:

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

(46) 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.

(47) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE

(49) 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.

(50) 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)
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)
U11(x1, x2) = U11(x1)
cons(x1, x2) = cons(x1)
U12(x1, x2) = U12(x1)
s(x1) = x1
length(x1) = length(x1)
active(x1) = x1
zeros = zeros
mark(x1) = mark
0 = 0
tt = tt
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

zeros > [cons₁, mark, proper₁] > [ACTIVE₁, U12₁] > [U11₁, length₁, 0, top]
tt > [cons₁, mark, proper₁] > [ACTIVE₁, U12₁] > [U11₁, length₁, 0, top]
nil > [cons₁, mark, proper₁] > [ACTIVE₁, U12₁] > [U11₁, length₁, 0, top]

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))

(51) Obligation:

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

ACTIVE(s(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.

(52) 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)

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

Lexicographic Path Order [LPO].
Precedence:

zeros > [mark, cons₁] > tt
zeros > [mark, cons₁] > top
zeros > 0
[U11₂, U12₂, length₁] > [ACTIVE₁, s₁] > [mark, cons₁] > tt
[U11₂, U12₂, length₁] > [ACTIVE₁, s₁] > [mark, cons₁] > top
[U11₂, U12₂, length₁] > 0
nil > 0

The following usable rules [FROCOS05] were oriented:

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

(53) 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.

(54) PisEmptyProof (EQUIVALENT transformation)

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

(55) TRUE

(56) 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.