Termination w.r.t. Q proof of TRS/Endrullisquadruple1.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(x3, a₁(x2)), p₂(b₁(a₁(x1)), b₁(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(x3, a₁(x2)), p₂(b₁(a₁(x1)), b₁(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(p₂(x3, a₁(x2)), p₂(b₁(a₁(x1)), b₁(x0)))
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(x3, a₁(x2))
P₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> P₂(b₁(a₁(x1)), b₁(x0))

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

p₂(p₂(b₁(a₁(x0)), x1), p₂(x2, x3)) -> p₂(p₂(x3, a₁(x2)), p₂(b₁(a₁(x1)), b₁(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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