Termination w.r.t. Q proof of /tmp/tmpM4SqVJ/Ex14_AEGL02

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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(length1(X)) → A__LENGTH1(X)
A__LENGTH(cons(X, Y)) → A__LENGTH1(Y)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)
MARK(length(X)) → A__LENGTH(X)
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 [15,17,22] 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__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 [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(A__LENGTH(x₁)) = (2)x_1
POL(cons(x₁, x₂)) = 4 + x_2
POL(A__LENGTH1(x₁)) = 2 + (2)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] 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__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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(a__length1(x₁)) = 0
POL(from(x₁)) = 4 + (2)x_1
POL(length1(x₁)) = 0
POL(mark(x₁)) = x_1
POL(a__from(x₁)) = 4 + (2)x_1
POL(0) = 0
POL(cons(x₁, x₂)) = x_1
POL(MARK(x₁)) = 1 + (4)x_1
POL(a__length(x₁)) = 0
POL(A__FROM(x₁)) = 4 + (4)x_1
POL(s(x₁)) = x_1
POL(length(x₁)) = 0
POL(nil) = 0

The value of delta used in the strict ordering is 3.
The following usable rules [17] 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(cons(X1, X2)) → cons(mark(X1), X2)
mark(length1(X)) → a__length1(X)
mark(nil) → nil
mark(s(X)) → s(mark(X))
a__from(X) → from(X)
mark(0) → 0
a__length1(X) → length1(X)
a__length(X) → length(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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(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 [15].

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: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 4 + x_1
POL(MARK(x₁)) = (4)x_1
POL(s(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] 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__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.