Termination w.r.t. Q proof of /tmp/tmpJDdgAa/4.24.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:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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:

REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV(++(x, y)) → REV1(x, y)
REV1(x, ++(y, z)) → REV1(y, z)
REV2(x, ++(y, z)) → REV2(y, z)
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV(++(x, y)) → REV2(x, y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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:

REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV(++(x, y)) → REV1(x, y)
REV1(x, ++(y, z)) → REV1(y, z)
REV2(x, ++(y, z)) → REV2(y, z)
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV(++(x, y)) → REV2(x, y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

REV1(x, ++(y, z)) → REV1(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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.

REV1(x, ++(y, z)) → REV1(y, z)

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

POL(++(x₁, x₂)) = 1 + (4)x_1 + (4)x_2
POL(REV1(x₁, x₂)) = (3)x_1 + (4)x_2

The value of delta used in the strict ordering is 4.
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:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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:

REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) → REV2(y, z)
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV(++(x, y)) → REV2(x, y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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.

REV2(x, ++(y, z)) → REV2(y, z)
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV(++(x, y)) → REV2(x, y)

The remaining pairs can at least be oriented weakly.

REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))

Used ordering: Polynomial interpretation [25,35]:

POL(REV2(x₁, x₂)) = (2)x_2
POL(++(x₁, x₂)) = 1 + (2)x_2
POL(rev2(x₁, x₂)) = x_2
POL(REV(x₁)) = (2)x_1
POL(rev(x₁)) = x_1
POL(nil) = 2
POL(rev1(x₁, x₂)) = 4

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

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.