Termination w.r.t. Q proof of TRS/TRCSREx14_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__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.