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__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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:

MARK1(s1(X)) -> MARK1(X)
A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
MARK1(length1(X)) -> A__LENGTH1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__LENGTH11(X) -> A__LENGTH1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(length11(X)) -> A__LENGTH11(X)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(X)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

MARK1(s1(X)) -> MARK1(X)
A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
MARK1(length1(X)) -> A__LENGTH1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__LENGTH11(X) -> A__LENGTH1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
MARK1(length11(X)) -> A__LENGTH11(X)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
A__LENGTH11(X) -> A__LENGTH1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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 strictly oriented and are deleted.


A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
The remaining pairs can at least by weakly be oriented.

A__LENGTH11(X) -> A__LENGTH1(X)
Used ordering: Combined order from the following AFS and order.
A__LENGTH1(x1)  =  A__LENGTH1(x1)
cons2(x1, x2)  =  cons1(x2)
A__LENGTH11(x1)  =  A__LENGTH11(x1)

Lexicographic Path Order [19].
Precedence:
[ALENGTH1, ALENGTH11]


The following usable rules [14] were oriented: none



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

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

A__LENGTH11(X) -> A__LENGTH1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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 strictly oriented and are deleted.


MARK1(from1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
s1(x1)  =  x1
from1(x1)  =  from1(x1)
A__FROM1(x1)  =  A__FROM1(x1)
mark1(x1)  =  mark1(x1)
cons2(x1, x2)  =  x1
a__from1(x1)  =  a__from1(x1)
length1(x1)  =  length
a__length1(x1)  =  a__length
length11(x1)  =  length1
a__length11(x1)  =  a__length1
nil  =  nil
0  =  0

Lexicographic Path Order [19].
Precedence:
[from1, mark1, afrom1] > [MARK1, AFROM1]
[from1, mark1, afrom1] > [alength, length1, alength1] > length
[from1, mark1, afrom1] > [alength, length1, alength1] > 0
[from1, mark1, afrom1] > nil


The following usable rules [14] were oriented:

mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__length11(X) -> a__length1(X)
a__length11(X) -> length11(X)
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length1(X) -> length1(X)
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__from1(X) -> from1(X)



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

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

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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 strictly oriented and are deleted.


MARK1(from1(X)) -> A__FROM1(mark1(X))
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least by weakly be oriented.

MARK1(s1(X)) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  from1(x1)
A__FROM1(x1)  =  A__FROM1(x1)
mark1(x1)  =  x1
cons2(x1, x2)  =  cons1(x1)
a__from1(x1)  =  a__from1(x1)
length1(x1)  =  length
a__length1(x1)  =  a__length
length11(x1)  =  length1
a__length11(x1)  =  a__length1
nil  =  nil
0  =  0

Lexicographic Path Order [19].
Precedence:
[from1, cons1, afrom1] > AFROM1
[length, alength, length1, alength1, nil, 0]


The following usable rules [14] were oriented:

mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__length11(X) -> a__length1(X)
a__length11(X) -> length11(X)
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length1(X) -> length1(X)
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__from1(X) -> from1(X)



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

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

MARK1(s1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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 strictly oriented and are deleted.


MARK1(s1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > MARK1


The following usable rules [14] were oriented: none



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

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

a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__length1(nil) -> 0
a__length1(cons2(X, Y)) -> s1(a__length11(Y))
a__length11(X) -> a__length1(X)
mark1(from1(X)) -> a__from1(mark1(X))
mark1(length1(X)) -> a__length1(X)
mark1(length11(X)) -> a__length11(X)
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(s1(X)) -> s1(mark1(X))
mark1(nil) -> nil
mark1(0) -> 0
a__from1(X) -> from1(X)
a__length1(X) -> length1(X)
a__length11(X) -> length11(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.