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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

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(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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:

REV2(x, cons(y, l)) → REV2(y, l)
REV(cons(x, l)) → REV2(x, l)
REV(cons(x, l)) → REV1(x, l)
REV1(x, cons(y, l)) → REV1(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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:

REV2(x, cons(y, l)) → REV2(y, l)
REV(cons(x, l)) → REV2(x, l)
REV(cons(x, l)) → REV1(x, l)
REV1(x, cons(y, l)) → REV1(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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:

REV2(x, cons(y, l)) → REV2(y, l)
REV(cons(x, l)) → REV2(x, l)
REV(cons(x, l)) → REV1(x, l)
REV1(x, cons(y, l)) → REV1(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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, cons(y, l)) → REV1(y, l)

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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, cons(y, l)) → REV1(y, l)

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)
cons(x1, x2) = cons(x2)

Recursive path order with status [2].
Quasi-Precedence:

[REV1₁, cons_1]

Status:

REV1₁: [1]
cons₁: multiset

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(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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, cons(y, l)) → REV2(y, l)
REV(cons(x, l)) → REV2(x, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(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, cons(y, l)) → REV2(y, l)
REV(cons(x, l)) → REV2(x, l)

The remaining pairs can at least be oriented weakly.

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

Used ordering: Combined order from the following AFS and order.
REV2(x1, x2) = x2
cons(x1, x2) = cons(x2)
REV(x1) = x1
rev2(x1, x2) = x2
rev(x1) = x1
rev1(x1, x2) = rev1
0 = 0
nil = nil
s(x1) = x1

Recursive path order with status [2].
Quasi-Precedence:

cons₁ > rev1
0 > rev1
nil > rev1

Status:

rev1: []
0: multiset
nil: multiset
cons₁: multiset

The following usable rules [14] were oriented:

rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
rev2(x, nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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.