Termination w.r.t. Q proof of /tmp/tmpDac8L2/LengthOfFiniteLists_nosorts

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 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 PisEmptyProof (⇔)

    ↳50 TRUE

  ↳51 QDP

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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(length(cons(N, L))) → S(length(L))
ACTIVE(length(cons(N, L))) → LENGTH(L)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → LENGTH(active(X))
ACTIVE(length(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(length(X)) → LENGTH(proper(X))
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
AND(ok(X1), ok(X2)) → AND(X1, X2)
LENGTH(ok(X)) → LENGTH(X)
S(ok(X)) → S(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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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 7 SCCs with 13 less nodes.

(4) Complex Obligation (AND)

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

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)
mark(x1) = x1
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = x2
0 = 0
and(x1, x2) = x2
tt = tt
length(x1) = length(x1)
nil = nil
s(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

proper₁ > zeros > ok₁
proper₁ > tt > ok₁
proper₁ > length₁ > 0 > ok₁
proper₁ > nil > 0 > ok₁
top > active₁ > zeros > ok₁
top > active₁ > length₁ > 0 > ok₁

The following usable rules [FROCOS05] were oriented:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

S₁ > mark₁
active₁ > zeros > mark₁
active₁ > cons₂ > s₁ > mark₁
active₁ > 0 > mark₁
active₁ > and₂ > mark₁
tt > mark₁
proper₁ > zeros > mark₁
proper₁ > cons₂ > s₁ > mark₁
proper₁ > and₂ > mark₁
proper₁ > nil > 0 > mark₁
top > mark₁

The following usable rules [FROCOS05] were oriented:

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

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

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)
mark(x1) = x1
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = x2
0 = 0
and(x1, x2) = x2
tt = tt
length(x1) = length(x1)
nil = nil
s(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Recursive Path Order [RPO].
Precedence:

proper₁ > zeros > ok₁
proper₁ > tt > ok₁
proper₁ > length₁ > 0 > ok₁
proper₁ > nil > 0 > ok₁
top > active₁ > zeros > ok₁
top > active₁ > length₁ > 0 > ok₁

The following usable rules [FROCOS05] were oriented:

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

LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

LENGTH₁ > mark₁
active₁ > zeros > mark₁
active₁ > cons₂ > s₁ > mark₁
active₁ > 0 > mark₁
active₁ > and₂ > mark₁
tt > mark₁
proper₁ > zeros > mark₁
proper₁ > cons₂ > s₁ > mark₁
proper₁ > and₂ > mark₁
proper₁ > nil > 0 > mark₁
top > mark₁

The following usable rules [FROCOS05] were oriented:

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

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

The TRS R consists of the following rules:

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

AND(mark(X1), X2) → AND(X1, X2)

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

Recursive Path Order [RPO].
Precedence:

active₁ > zeros > mark₁
active₁ > cons₂ > mark₁
active₁ > 0
active₁ > and₂ > mark₁
tt > mark₁
nil > mark₁
nil > 0

The following usable rules [FROCOS05] were oriented:

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

AND(ok(X1), ok(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

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

AND(ok(X1), ok(X2)) → AND(X1, X2)

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

Recursive Path Order [RPO].
Precedence:

proper₁ > zeros > ok₁
proper₁ > zeros > 0
proper₁ > tt
proper₁ > nil > 0

The following usable rules [FROCOS05] were oriented:

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

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

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) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
zeros = zeros
cons(x1, x2) = cons(x1, x2)
0 = 0
and(x1, x2) = and(x1, x2)
tt = tt
length(x1) = x1
nil = nil
s(x1) = x1
proper(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > zeros > mark₁
active₁ > cons₂ > mark₁
active₁ > 0
active₁ > and₂ > mark₁
tt > mark₁
nil > mark₁
nil > 0

The following usable rules [FROCOS05] were oriented:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

proper₁ > zeros > ok₁
proper₁ > zeros > 0
proper₁ > tt
proper₁ > nil > 0

The following usable rules [FROCOS05] were oriented:

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

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

and₂ > mark > cons₂ > ok
and₂ > mark > top
length₁ > PROPER₁
length₁ > mark > cons₂ > ok
length₁ > mark > top
length₁ > 0 > ok
zeros > mark > cons₂ > ok
zeros > mark > top
zeros > 0 > ok
tt > mark > cons₂ > ok
tt > mark > top
nil > 0 > ok

The following usable rules [FROCOS05] were oriented:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

s₁ > mark > and₂
s₁ > mark > top
s₁ > ok > and₂
s₁ > ok > top
zeros > mark > and₂
zeros > mark > top
zeros > ok > and₂
zeros > ok > top
cons > mark > and₂
cons > mark > top
cons > ok > and₂
cons > ok > top
tt > mark > and₂
tt > mark > top
tt > ok > and₂
tt > ok > top
nil > mark > and₂
nil > mark > top
nil > 0 > ok > and₂
nil > 0 > ok > top

The following usable rules [FROCOS05] were oriented:

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

ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

ACTIVE(cons(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)
and(x1, x2) = x1
cons(x1, x2) = cons(x1)
length(x1) = x1
s(x1) = x1
active(x1) = active(x1)
zeros = zeros
mark(x1) = mark
0 = 0
tt = tt
nil = nil
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > zeros > cons₁ > ACTIVE₁
active₁ > mark > cons₁ > ACTIVE₁
active₁ > 0 > ACTIVE₁
nil > ACTIVE₁
proper₁ > cons₁ > ACTIVE₁
proper₁ > tt > ACTIVE₁
top > ACTIVE₁

The following usable rules [FROCOS05] were oriented:

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

ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(length(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

proper₁ > length₁ > mark > top
proper₁ > length₁ > 0
proper₁ > zeros > mark > top
proper₁ > tt > mark > top
proper₁ > nil

The following usable rules [FROCOS05] were oriented:

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

ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

ACTIVE(and(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)
and(x1, x2) = and(x1, x2)
s(x1) = x1
active(x1) = active(x1)
zeros = zeros
mark(x1) = x1
cons(x1, x2) = cons(x1, x2)
0 = 0
tt = tt
length(x1) = length
nil = nil
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > and₂
active₁ > zeros
active₁ > cons₂ > length
active₁ > 0

The following usable rules [FROCOS05] were oriented:

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

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

Recursive Path Order [RPO].
Precedence:

s₁ > mark > and₂
s₁ > mark > top
s₁ > ok > and₂
s₁ > ok > top
zeros > mark > and₂
zeros > mark > top
zeros > ok > and₂
zeros > ok > top
cons > mark > and₂
cons > mark > top
cons > ok > and₂
cons > ok > top
tt > mark > and₂
tt > mark > top
tt > ok > and₂
tt > ok > top
nil > mark > and₂
nil > mark > top
nil > 0 > ok > and₂
nil > 0 > ok > top

The following usable rules [FROCOS05] were oriented:

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

(48) Obligation:

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

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) 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(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(cons(X1, X2)) → cons(active(X1), X2)
active(and(X1, X2)) → and(active(X1), X2)
active(length(X)) → length(active(X))
active(s(X)) → s(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
and(mark(X1), X2) → mark(and(X1, X2))
length(mark(X)) → mark(length(X))
s(mark(X)) → mark(s(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
length(ok(X)) → ok(length(X))
s(ok(X)) → ok(s(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.