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 oriented strictly and are deleted.


A__LENGTH1(cons2(X, Y)) -> A__LENGTH11(Y)
A__LENGTH11(X) -> A__LENGTH1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(A__LENGTH1(x1)) = 1 + 2·x1   
POL(A__LENGTH11(x1)) = 2 + 2·x1   
POL(cons2(x1, x2)) = 2 + 2·x2   

The following usable rules [14] were oriented: none



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

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.

↳ 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 oriented strictly and are deleted.


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

MARK1(s1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__FROM1(x1)) = 2 + x1   
POL(MARK1(x1)) = 1 + x1   
POL(a__from1(x1)) = 2 + 2·x1   
POL(a__length1(x1)) = 0   
POL(a__length11(x1)) = 0   
POL(cons2(x1, x2)) = 1 + x1   
POL(from1(x1)) = 2 + 2·x1   
POL(length1(x1)) = 0   
POL(length11(x1)) = 0   
POL(mark1(x1)) = 2·x1   
POL(nil) = 0   
POL(s1(x1)) = 2·x1   

The following usable rules [14] were oriented:

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



↳ 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)

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 oriented strictly and are deleted.


MARK1(s1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ 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.