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(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, 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(p(b(a(x0)), x1), p(x2, x3)) → P(b(x2), a(a(b(x1))))
P(p(b(a(x0)), x1), p(x2, x3)) → P(x3, x0)
P(p(b(a(x0)), x1), p(x2, x3)) → P(p(b(x2), a(a(b(x1)))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, 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(p(b(a(x0)), x1), p(x2, x3)) → P(b(x2), a(a(b(x1))))
P(p(b(a(x0)), x1), p(x2, x3)) → P(x3, x0)
P(p(b(a(x0)), x1), p(x2, x3)) → P(p(b(x2), a(a(b(x1)))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof

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

P(p(b(a(x0)), x1), p(x2, x3)) → P(x3, x0)
P(p(b(a(x0)), x1), p(x2, x3)) → P(p(b(x2), a(a(b(x1)))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

P(p(b(a(x0)), x1), p(x2, x3)) → P(x3, x0)


Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1, x2)) = 2·x1 + 2·x2   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(p(x1, x2)) = 1 + 2·x1 + 2·x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Instantiation

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

P(p(b(a(x0)), x1), p(x2, x3)) → P(p(b(x2), a(a(b(x1)))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule P(p(b(a(x0)), x1), p(x2, x3)) → P(p(b(x2), a(a(b(x1)))), p(x3, x0)) we obtained the following new rules:

P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) → P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation

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

P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) → P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule P(p(b(a(x0)), a(a(b(y_1)))), p(x2, x3)) → P(p(b(x2), a(a(b(a(a(b(y_1))))))), p(x3, x0)) we obtained the following new rules:

P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) → P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ ForwardInstantiation

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

P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) → P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule P(p(b(a(x0)), a(a(b(a(a(b(y_1))))))), p(x2, x3)) → P(p(b(x2), a(a(b(a(a(b(a(a(b(y_1)))))))))), p(x3, x0)) we obtained the following new rules:

P(p(b(a(x0)), a(a(b(a(a(b(x1))))))), p(a(y_0), x3)) → P(p(b(a(y_0)), a(a(b(a(a(b(a(a(b(x1)))))))))), p(x3, x0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP

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

P(p(b(a(x0)), a(a(b(a(a(b(x1))))))), p(a(y_0), x3)) → P(p(b(a(y_0)), a(a(b(a(a(b(a(a(b(x1)))))))))), p(x3, x0))

The TRS R consists of the following rules:

p(p(b(a(x0)), x1), p(x2, x3)) → p(p(b(x2), a(a(b(x1)))), p(x3, x0))

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