Termination w.r.t. Q proof of TRS/Endrullispair2hard.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(x0), p₂(b₁(a₁(x1)), x2)) -> p₂(x1, p₂(a₁(b₁(a₁(x1))), x2))
a₁(b₁(a₁(x0))) -> b₁(a₁(b₁(x0)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> p₂(x1, p₂(a₁(b₁(a₁(x1))), x2))
a₁(b₁(a₁(x0))) -> b₁(a₁(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(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> P₂(x1, p₂(a₁(b₁(a₁(x1))), x2))
P₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> P₂(a₁(b₁(a₁(x1))), x2)
A₁(b₁(a₁(x0))) -> A₁(b₁(x0))
P₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> A₁(b₁(a₁(x1)))

The TRS R consists of the following rules:

p₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> p₂(x1, p₂(a₁(b₁(a₁(x1))), x2))
a₁(b₁(a₁(x0))) -> b₁(a₁(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(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> P₂(x1, p₂(a₁(b₁(a₁(x1))), x2))
P₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> P₂(a₁(b₁(a₁(x1))), x2)
A₁(b₁(a₁(x0))) -> A₁(b₁(x0))
P₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> A₁(b₁(a₁(x1)))

The TRS R consists of the following rules:

p₂(a₁(x0), p₂(b₁(a₁(x1)), x2)) -> p₂(x1, p₂(a₁(b₁(a₁(x1))), x2))
a₁(b₁(a₁(x0))) -> b₁(a₁(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 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

A₁(b₁(a₁(x0))) -> A₁(b₁(x0))

The TRS R consists of the following rules:

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

A₁(b₁(a₁(x0))) -> A₁(b₁(x0))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(A₁(x₁)) = 2·x₁ + 2·x₁²
POL(a₁(x₁)) = 2 + 2·x₁
POL(b₁(x₁)) = 2 + 2·x₁ + 2·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

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

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(P₂(x₁, x₂)) = x₂
POL(a₁(x₁)) = 0
POL(b₁(x₁)) = 0
POL(p₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

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

The TRS R consists of the following rules:

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

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