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)
The remaining pairs can at least be oriented weakly.

A__LENGTH11(X) -> A__LENGTH1(X)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( A__LENGTH1(x1) ) = x1 + 1


POL( cons2(x1, x2) ) = x2 + 1


POL( A__LENGTH11(x1) ) = x1 + 1



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 oriented strictly and are deleted.


MARK1(from1(X)) -> MARK1(X)
A__FROM1(X) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from1(x1) ) = x1 + 1


POL( mark1(x1) ) = x1


POL( length1(x1) ) = 1


POL( a__length1(x1) ) = 1


POL( MARK1(x1) ) = x1


POL( 0 ) = 1


POL( nil ) = max{0, -1}


POL( cons2(x1, x2) ) = x1


POL( a__length11(x1) ) = 1


POL( a__from1(x1) ) = x1 + 1


POL( A__FROM1(x1) ) = x1 + 1


POL( s1(x1) ) = x1


POL( length11(x1) ) = 1



The following usable rules [14] were oriented:

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



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

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

MARK1(s1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

MARK1(s1(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 oriented strictly and are deleted.


MARK1(s1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK1(x1) ) = x1 + 1


POL( cons2(x1, x2) ) = x1 + x2 + 1


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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