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)  =  x1
b(x1)  =  x1
a(x1)  =  a(x1)

Recursive Path Order [2].
Precedence:
trivial

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)  =  x2
p(x1, x2)  =  p(x2)

Recursive Path Order [2].
Precedence:
trivial

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.