Termination w.r.t. Q proof of TRS/SK902.42.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:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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:

FLATTEN(++(x, y)) → FLATTEN(x)
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
++¹(++(x, y), z) → ++¹(x, ++(y, z))
FLATTEN(++(x, y)) → ++¹(flatten(x), flatten(y))
REV(++(x, y)) → ++¹(rev(y), rev(x))
FLATTEN(++(unit(x), y)) → ++¹(flatten(x), flatten(y))
FLATTEN(unit(x)) → FLATTEN(x)
++¹(++(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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:

FLATTEN(++(x, y)) → FLATTEN(x)
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
++¹(++(x, y), z) → ++¹(x, ++(y, z))
FLATTEN(++(x, y)) → ++¹(flatten(x), flatten(y))
REV(++(x, y)) → ++¹(rev(y), rev(x))
FLATTEN(++(unit(x), y)) → ++¹(flatten(x), flatten(y))
FLATTEN(unit(x)) → FLATTEN(x)
++¹(++(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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:

FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
++¹(++(x, y), z) → ++¹(x, ++(y, z))
FLATTEN(++(x, y)) → ++¹(flatten(x), flatten(y))
REV(++(x, y)) → ++¹(rev(y), rev(x))
++¹(++(x, y), z) → ++¹(y, z)
FLATTEN(++(x, y)) → FLATTEN(y)
REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(unit(x), y)) → ++¹(flatten(x), flatten(y))
FLATTEN(unit(x)) → FLATTEN(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ 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(y, z)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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(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) = x1
++(x1, x2) = ++(x1, x2)
nil = nil

Lexicographic path order with status [19].
Precedence:

nil > ++₂

Status:

++₂: [2,1]
nil: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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)

Lexicographic path order with status [19].
Precedence:

++₂ > REV₁

Status:

++₂: multiset
REV₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(unit(x)) → FLATTEN(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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.

FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)

The remaining pairs can at least be oriented weakly.

FLATTEN(unit(x)) → FLATTEN(x)

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

Lexicographic path order with status [19].
Precedence:

++₂ > FLATTEN₁

Status:

++₂: multiset
FLATTEN₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

FLATTEN(unit(x)) → FLATTEN(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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.

FLATTEN(unit(x)) → FLATTEN(x)

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

Lexicographic path order with status [19].
Precedence:

unit₁ > FLATTEN₁

Status:

unit₁: multiset
FLATTEN₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(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.