Termination w.r.t. Q proof of TRS/Endrullisquadruple2.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

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:

P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(x3, x0)
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(b₁(x2), a₁(a₁(b₁(x1))))
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

The TRS R consists of the following rules:

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(x3, x0)
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(b₁(x2), a₁(a₁(b₁(x1))))
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

The TRS R consists of the following rules:

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(x3, x0)
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

The TRS R consists of the following rules:

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

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.

P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(x3, x0)

The remaining pairs can at least be oriented weakly.

P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

Used ordering: Polynomial interpretation [21]:

POL(P₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(a₁(x₁)) = x₁
POL(b₁(x₁)) = x₁
POL(p₂(x₁, x₂)) = 2 + x₁ + x₂

The following usable rules [14] were oriented:

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

The TRS R consists of the following rules:

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(b₁(x2), a₁(a₁(b₁(x1)))), p₂(x3, x0))

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