Termination w.r.t. Q proof of TRS/SK904.26.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(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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:

REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
++¹(x, ++(y, z)) → ++¹(++(x, y), z)
REV(++(x, y)) → ++¹(rev(y), rev(x))
++¹(x, ++(y, z)) → ++¹(x, y)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
++¹(x, ++(y, z)) → ++¹(++(x, y), z)
REV(++(x, y)) → ++¹(rev(y), rev(x))
++¹(x, ++(y, z)) → ++¹(x, y)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
++¹(x, ++(y, z)) → ++¹(++(x, y), z)
REV(++(x, y)) → ++¹(rev(y), rev(x))
++¹(x, ++(y, z)) → ++¹(x, y)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

++¹(x, ++(y, z)) → ++¹(++(x, y), z)
++¹(x, ++(y, z)) → ++¹(x, y)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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.

++¹(x, ++(y, z)) → ++¹(++(x, y), z)
++¹(x, ++(y, z)) → ++¹(x, y)

The remaining pairs can at least be oriented weakly.

++¹(.(x, y), z) → ++¹(y, z)

Used ordering: Combined order from the following AFS and order.
++¹(x1, x2) = ++¹(x2)
++(x1, x2) = ++(x1, x2)
.(x1, x2) = .(x1)
nil = nil

Recursive Path Order [2].
Precedence:

.₁ > ++₂ > ++^1₁
nil > ++^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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.

++¹(.(x, y), z) → ++¹(y, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2) = ++¹(x1)
.(x1, x2) = .(x1, x2)

Recursive Path Order [2].
Precedence:

.₂ > ++^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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.

REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)

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

Recursive Path Order [2].
Precedence:

++₂ > REV₁

The following usable rules [14] were oriented: none

↳ 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(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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.