Termination w.r.t. Q proof of /tmp/tmpbOL2Dy/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

  ↳ 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 [19] 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 [1,15] 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))))

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

          ↳ 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))))

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 [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

REV1(x, ++(y, z)) → REV1(y, z)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

REV1(x, ++(y, z)) → REV1(y, z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

REV1(x, ++(y, z)) → REV1(y, z)
The graph contains the following edges 2 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))))

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 [15].

The following pairs can be oriented strictly and are deleted.

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))

The remaining pairs can at least be oriented weakly.

REV(++(x, y)) → REV2(x, y)

Used ordering: Polynomial interpretation [25]:

POL(++(x₁, x₂)) = 1 + x₂
POL(REV(x₁)) = x₁
POL(REV2(x₁, x₂)) = 1 + x₂
POL(nil) = 1
POL(rev(x₁)) = x₁
POL(rev1(x₁, x₂)) = 0
POL(rev2(x₁, x₂)) = x₂

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

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 [15,17,22] contains 0 SCCs with 1 less node.