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 [1,15] 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))
A(b(a(x0))) → A(b(x0))
P(a(x0), p(b(a(x1)), x2)) → A(b(a(x1)))
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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))
A(b(a(x0))) → A(b(x0))
P(a(x0), p(b(a(x1)), x2)) → A(b(a(x1)))
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.
The approximation of the Dependency Graph [15,17,22] 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:

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 [15].


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: Polynomial interpretation [25,35]:

POL(a(x1)) = 1/4 + (11/4)x_1   
POL(A(x1)) = (2)x_1   
POL(b(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] 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:

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
      ↳ 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 [15].


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: Polynomial interpretation [25,35]:

POL(P(x1, x2)) = (1/4)x_2   
POL(a(x1)) = 1/2 + (5/4)x_1   
POL(p(x1, x2)) = 1/4 + (11/4)x_2   
POL(b(x1)) = 4   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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.