Termination w.r.t. Q proof of /tmp/tmp0yI9e2/Ex14_AEGL02

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 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 PisEmptyProof (⇔)

    ↳31 TRUE

  ↳32 QDP

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length(cons(X, Y))) → S(length1(Y))
ACTIVE(length(cons(X, Y))) → LENGTH1(Y)
ACTIVE(length1(X)) → MARK(length(X))
ACTIVE(length1(X)) → LENGTH(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(nil) → ACTIVE(nil)
MARK(0) → ACTIVE(0)
MARK(length1(X)) → ACTIVE(length1(X))
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
LENGTH1(mark(X)) → LENGTH1(X)
LENGTH1(active(X)) → LENGTH1(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 12 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

LENGTH1(active(X)) → LENGTH1(X)
LENGTH1(mark(X)) → LENGTH1(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(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.

LENGTH1(active(X)) → LENGTH1(X)
LENGTH1(mark(X)) → LENGTH1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH1(x1) = LENGTH1(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
length(x1) = length(x1)
nil = nil
0 = 0
length1(x1) = length1(x1)

Lexicographic path order with status [LPO].
Precedence:

from > cons > mark₁ > length₁ > active₁
from > cons > mark₁ > length₁ > s
nil > mark₁ > length₁ > active₁
nil > mark₁ > length₁ > s
nil > 0 > active₁
length1₁ > mark₁ > length₁ > active₁
length1₁ > mark₁ > length₁ > s

Status:

from: []
active₁: [1]
LENGTH1₁: [1]
length1₁: [1]
cons: []
mark₁: [1]
s: []
length₁: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

(7) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = LENGTH(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
length(x1) = length(x1)
nil = nil
0 = 0
length1(x1) = length1(x1)

Lexicographic path order with status [LPO].
Precedence:

from > cons > mark₁ > length₁ > active₁
from > cons > mark₁ > length₁ > s
nil > mark₁ > length₁ > active₁
nil > mark₁ > length₁ > s
nil > 0 > active₁
length1₁ > mark₁ > length₁ > active₁
length1₁ > mark₁ > length₁ > s

Status:

from: []
LENGTH₁: [1]
active₁: [1]
length1₁: [1]
cons: []
mark₁: [1]
s: []
length₁: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

(12) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(X)) → S(X)
S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
length(x1) = length(x1)
nil = nil
0 = 0
length1(x1) = length1(x1)

Lexicographic path order with status [LPO].
Precedence:

from > cons > mark₁ > length₁ > active₁
from > cons > mark₁ > length₁ > s
nil > mark₁ > length₁ > active₁
nil > mark₁ > length₁ > s
nil > 0 > active₁
length1₁ > mark₁ > length₁ > active₁
length1₁ > mark₁ > length₁ > s

Status:

from: []
active₁: [1]
length1₁: [1]
cons: []
mark₁: [1]
s: []
length₁: [1]
0: []
nil: []
S₁: [1]

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

(17) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(21) 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)
CONS(active(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
length(x1) = x1
nil = nil
0 = 0
length1(x1) = length1(x1)

Lexicographic path order with status [LPO].
Precedence:

CONS₁ > active₁
from > cons > mark₁ > active₁
from > cons > s > active₁
nil > mark₁ > active₁
nil > 0 > active₁
length1₁ > mark₁ > active₁

Status:

from: []
active₁: [1]
length1₁: [1]
cons: []
CONS₁: [1]
mark₁: [1]
s: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

(22) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
length(x1) = length
nil = nil
0 = 0
length1(x1) = length1

Lexicographic path order with status [LPO].
Precedence:

from > mark₁ > active₁ > CONS₂
from > mark₁ > active₁ > cons
from > mark₁ > s
nil > mark₁ > active₁ > CONS₂
nil > mark₁ > active₁ > cons
nil > mark₁ > s
length1 > length > mark₁ > active₁ > CONS₂
length1 > length > mark₁ > active₁ > cons
length1 > length > mark₁ > s
length1 > length > 0 > active₁ > CONS₂
length1 > length > 0 > active₁ > cons

Status:

from: []
active₁: [1]
cons: []
length1: []
CONS₂: [1,2]
mark₁: [1]
s: []
length: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

(24) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
from(x1) = from
cons(x1, x2) = cons
s(x1) = s
length(x1) = length(x1)
nil = nil
0 = 0
length1(x1) = length1(x1)

Lexicographic path order with status [LPO].
Precedence:

from > cons > mark₁ > length₁ > active₁
from > cons > mark₁ > length₁ > s
nil > mark₁ > length₁ > active₁
nil > mark₁ > length₁ > s
nil > 0 > active₁
length1₁ > mark₁ > length₁ > active₁
length1₁ > mark₁ > length₁ > s

Status:

from: []
active₁: [1]
length1₁: [1]
cons: []
mark₁: [1]
s: []
length₁: [1]
0: []
nil: []
FROM₁: [1]

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

(29) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)

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