Termination w.r.t. Q of the given QTRS could be proven:

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 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 DependencyGraphProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE

(0) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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:

A__FROM(X) → MARK(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
A__LENGTH1(X) → A__LENGTH(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(length(X)) → A__LENGTH(X)
MARK(length1(X)) → A__LENGTH1(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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 2 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A__LENGTH1(X) → A__LENGTH(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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.


A__LENGTH1(X) → A__LENGTH(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__LENGTH1(x1)  =  A__LENGTH1(x1)
A__LENGTH(x1)  =  x1
cons(x1, x2)  =  cons(x2)
a__from(x1)  =  a__from(x1)
mark(x1)  =  mark(x1)
from(x1)  =  from(x1)
s(x1)  =  s
a__length(x1)  =  a__length
nil  =  nil
0  =  0
a__length1(x1)  =  a__length1
length(x1)  =  length
length1(x1)  =  length1

Recursive path order with status [RPO].
Quasi-Precedence:
mark1 > afrom1 > cons1 > [ALENGTH11, from1, s]
mark1 > alength1 > [alength, length] > 0 > [ALENGTH11, from1, s]
mark1 > alength1 > length1 > [ALENGTH11, from1, s]
nil > 0 > [ALENGTH11, from1, s]

Status:
ALENGTH11: multiset
cons1: multiset
afrom1: multiset
mark1: multiset
from1: multiset
s: multiset
alength: multiset
nil: multiset
0: multiset
alength1: []
length: multiset
length1: multiset


The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

(7) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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:

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(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.


MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  x1
s(x1)  =  x1
a__from(x1)  =  a__from(x1)
a__length(x1)  =  a__length
nil  =  nil
0  =  0
a__length1(x1)  =  a__length1
length(x1)  =  length
length1(x1)  =  length1

Recursive path order with status [RPO].
Quasi-Precedence:
[MARK1, AFROM1] > [from1, mark1, afrom1, nil]
[alength, alength1, length, length1] > 0

Status:
MARK1: multiset
from1: [1]
AFROM1: multiset
mark1: [1]
afrom1: [1]
alength: []
nil: multiset
0: multiset
alength1: []
length: []
length1: []


The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

(12) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(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.


A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1)
s(x1)  =  x1
a__from(x1)  =  a__from(x1)
a__length(x1)  =  a__length
nil  =  nil
0  =  0
a__length1(x1)  =  a__length1
length(x1)  =  length
length1(x1)  =  length1

Recursive path order with status [RPO].
Quasi-Precedence:
[from1, AFROM1, afrom1] > cons1
[alength, alength1, length, length1] > 0
nil > 0

Status:
from1: multiset
AFROM1: multiset
cons1: [1]
afrom1: multiset
alength: multiset
nil: multiset
0: multiset
alength1: multiset
length: multiset
length1: multiset


The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

(14) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(16) Obligation:

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
s(x1)  =  s(x1)
a__from(x1)  =  a__from
cons(x1, x2)  =  cons(x2)
mark(x1)  =  mark(x1)
from(x1)  =  from
a__length(x1)  =  a__length(x1)
nil  =  nil
0  =  0
a__length1(x1)  =  a__length1(x1)
length(x1)  =  x1
length1(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[afrom, mark1, alength1, alength11] > cons1 > [MARK1, s1] > from
[afrom, mark1, alength1, alength11] > nil > from
[afrom, mark1, alength1, alength11] > 0 > from

Status:
MARK1: [1]
s1: [1]
afrom: []
cons1: multiset
mark1: [1]
from: multiset
alength1: [1]
nil: multiset
0: multiset
alength11: [1]


The following usable rules [FROCOS05] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

(18) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__length(nil) → 0
a__length(cons(X, Y)) → s(a__length1(Y))
a__length1(X) → a__length(X)
mark(from(X)) → a__from(mark(X))
mark(length(X)) → a__length(X)
mark(length1(X)) → a__length1(X)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(0) → 0
a__from(X) → from(X)
a__length(X) → length(X)
a__length1(X) → length1(X)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE