Termination w.r.t. Q proof of TRS/AG01#3.38.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:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV₁(cons₂(x, l)) -> REV1₂(x, l)
REV1₂(x, cons₂(y, l)) -> REV1₂(y, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV₁(cons₂(x, l)) -> REV1₂(x, l)
REV1₂(x, cons₂(y, l)) -> REV1₂(y, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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)) = 2·x₂
POL(cons₂(x₁, x₂)) = 1 + 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:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)

The remaining pairs can at least be oriented weakly.

REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented:

rev2₂(x, nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, 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 0 SCCs with 1 less node.