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
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
          ↳ QDP

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

A(b(a(x0))) → A(b(x0))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

A(b(a(x0))) → A(b(x0))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
R is empty.
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]:

POL(P(x1, x2)) = x2   
POL(a(x1)) = 0   
POL(b(x1)) = 0   
POL(p(x1, x2)) = 1 + x2   

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
                ↳ ForwardInstantiation

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.
By forward instantiating [14] the rule P(a(x0), p(b(a(x1)), x2)) → P(x1, p(a(b(a(x1))), x2)) we obtained the following new rules:

P(a(x0), p(b(a(a(y_0))), x2)) → P(a(y_0), p(a(b(a(a(y_0)))), x2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ ForwardInstantiation
                  ↳ QDP

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

P(a(x0), p(b(a(a(y_0))), x2)) → P(a(y_0), p(a(b(a(a(y_0)))), 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.