Termination w.r.t. Q proof of TRS/SK902.43.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:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) → IF(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) → IF(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

MERGE(.(x, y), .(u, v)) → IF(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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) = x1
.(x1, x2) = .(x2)

Recursive Path Order [2].
Precedence:

trivial

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:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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.

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)

The remaining pairs can at least be oriented weakly.

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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.

MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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