Termination w.r.t. Q proof of TRS/SK902.38.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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(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:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(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)
++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, x2)
++(x1, x2) = ++(x1, x2)
.(x1, x2) = .(x2)
nil = nil

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

++^1₂ > ++₂ > .₁
nil > .₁

Status:

++₂: [1,2]
.₁: multiset
nil: multiset
++^1₂: [1,2]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

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