Termination w.r.t. Q proof of TRS/Rubiorevlist.trs

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:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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:

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV₁(cons₂(X, L)) -> REV1₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))
REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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:

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV₁(cons₂(X, L)) -> REV1₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))
REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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.

REV1₂(X, cons₂(Y, L)) -> REV1₂(Y, L)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(REV1₂(x₁, x₂)) = x₁ + 3·x₂
POL(cons₂(x₁, x₂)) = 1 + x₁ + 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:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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:

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))

The TRS R consists of the following rules:

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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.

REV2₂(X, cons₂(Y, L)) -> REV₁(cons₂(X, rev₁(rev2₂(Y, L))))
REV2₂(X, cons₂(Y, L)) -> REV2₂(Y, L)
REV₁(cons₂(X, L)) -> REV2₂(X, L)
REV2₂(X, cons₂(Y, L)) -> REV₁(rev2₂(Y, L))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(REV₁(x₁)) = 3 + 2·x₁
POL(REV2₂(x₁, x₂)) = 1 + 3·x₂
POL(cons₂(x₁, x₂)) = 3 + 2·x₂
POL(nil) = 2
POL(rev₁(x₁)) = x₁
POL(rev1₂(x₁, x₂)) = 3 + 3·x₂
POL(rev2₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 2

The following usable rules [14] were oriented:

rev₁(nil) -> nil
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

rev1₂(0, nil) -> 0
rev1₂(s₁(X), nil) -> s₁(X)
rev1₂(X, cons₂(Y, L)) -> rev1₂(Y, L)
rev₁(nil) -> nil
rev₁(cons₂(X, L)) -> cons₂(rev1₂(X, L), rev2₂(X, L))
rev2₂(X, nil) -> nil
rev2₂(X, cons₂(Y, L)) -> rev₁(cons₂(X, rev₁(rev2₂(Y, L))))

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.