Termination w.r.t. Q proof of TRS/D3309.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:

.(.(x, y), z) → .(x, .(y, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

.(.(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(y, z)
.¹(.(x, y), z) → .¹(x, .(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)

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

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

.(.(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.