Termination w.r.t. Q proof of TRS/Endrullispair3swap.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(a₁(a₁(x0)), p₂(x1, p₂(a₁(x2), x3))) -> p₂(x2, p₂(a₁(a₁(b₁(x1))), p₂(a₁(a₁(x0)), x3)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(P₂(x₁, x₂)) = 2·x₂
POL(a₁(x₁)) = 0
POL(b₁(x₁)) = 0
POL(p₂(x₁, x₂)) = 2 + 2·x₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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