Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MARK(s(X)) → MARK(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
MARK(length(X)) → A__LENGTH(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
A__LENGTH1(X) → A__LENGTH(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length1(X)) → A__LENGTH1(X)
A__FROM(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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
A__LENGTH1(X) → A__LENGTH(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
The remaining pairs can at least be oriented weakly.

A__LENGTH1(X) → A__LENGTH(X)
Used ordering: Combined order from the following AFS and order.
A__LENGTH(x1)  =  A__LENGTH(x1)
cons(x1, x2)  =  cons(x2)
A__LENGTH1(x1)  =  A__LENGTH1(x1)

Recursive path order with status [2].
Quasi-Precedence:
[ALENGTH1, cons1, ALENGTH11]

Status:
ALENGTH11: multiset
cons1: multiset
ALENGTH1: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ DependencyGraphProof
              ↳ QDP

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

A__LENGTH1(X) → A__LENGTH(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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
s(x1)  =  x1
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  x1
a__from(x1)  =  a__from(x1)
a__length(x1)  =  a__length
length(x1)  =  length
0  =  0
length1(x1)  =  length1
a__length1(x1)  =  a__length1
nil  =  nil

Recursive path order with status [2].
Quasi-Precedence:
[MARK1, AFROM1] > [from1, mark1, afrom1, 0, nil]
[alength, length, length1, alength1] > [from1, mark1, afrom1, 0, nil]

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


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof

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

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK(from(X)) → A__FROM(mark(X))
The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
s(x1)  =  x1
from(x1)  =  from(x1)
A__FROM(x1)  =  A__FROM(x1)
mark(x1)  =  x1
cons(x1, x2)  =  x1
a__from(x1)  =  a__from(x1)
a__length(x1)  =  a__length
length(x1)  =  length
0  =  0
length1(x1)  =  length1
a__length1(x1)  =  a__length1
nil  =  nil

Recursive path order with status [2].
Quasi-Precedence:
[MARK1, AFROM1]
[from1, afrom1]
[alength, length, 0, length1, alength1]

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


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
s1: multiset
cons2: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ PisEmptyProof

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