Termination w.r.t. Q proof of TRS/SK904.24.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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

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

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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.

REV1(x, ++(y, z)) → REV1(y, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
REV1(x1, x2) = REV1(x2)
++(x1, x2) = ++(x2)

Recursive Path Order [2].
Precedence:

++₁ > REV1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

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

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
REV2(x1, x2) = REV2(x2)
++(x1, x2) = ++(x2)
REV(x1) = x1
rev2(x1, x2) = x2
rev(x1) = x1
rev1(x1, x2) = rev1(x1, x2)
nil = nil

Recursive Path Order [2].
Precedence:

++₁ > REV2₁
++₁ > rev1₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.