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:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(x0)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(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:

P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2))
P(a(x0), p(b(a(x1)), x2)) → P(a(b(a(x1))), x2)
A(b(a(x0))) → A(b(x0))
P(a(x0), p(b(a(x1)), x2)) → A(b(a(x1)))

The TRS R consists of the following rules:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(x0)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2))
P(a(x0), p(b(a(x1)), x2)) → P(a(b(a(x1))), x2)
A(b(a(x0))) → A(b(x0))
P(a(x0), p(b(a(x1)), x2)) → A(b(a(x1)))

The TRS R consists of the following rules:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(x0)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2))
P(a(x0), p(b(a(x1)), x2)) → P(a(b(a(x1))), x2)
A(b(a(x0))) → A(b(x0))
P(a(x0), p(b(a(x1)), x2)) → A(b(a(x1)))

The TRS R consists of the following rules:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

A(b(a(x0))) → A(b(x0))

The TRS R consists of the following rules:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(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.

A(b(a(x0))) → A(b(x0))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1) = A(x1)
b(x1) = b(x1)
a(x1) = a(x1)

Lexicographic path order with status [19].
Precedence:

b₁ > A₁
a₁ > A₁

Status:

a₁: multiset
b₁: [1]
A₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2))
P(a(x0), p(b(a(x1)), x2)) → P(a(b(a(x1))), x2)

The TRS R consists of the following rules:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(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.

P(a(x0), p(b(a(x1)), x2)) → P(a(b(a(x1))), x2)

The remaining pairs can at least be oriented weakly.

P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2))

Used ordering: Combined order from the following AFS and order.
P(x1, x2) = P(x2)
a(x1) = a
p(x1, x2) = p(x2)
b(x1) = b

Lexicographic path order with status [19].
Precedence:

P₁ > p₁
a > p₁
b > p₁

Status:

b: []
a: []
P₁: [1]
p₁: [1]

The following usable rules [14] were oriented:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

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

P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2))

The TRS R consists of the following rules:

p(a(x0), p(b(a(x1)), x2)) → p(x1, p(a(b(a(x1))), x2))
a(b(a(x0))) → b(a(b(x0)))

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