Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
a(a(a(x))) → a(b(a(x)))
a(b(a(x))) → b(b(b(x)))
a(a(a(a(x)))) → a(a(b(a(a(x)))))
a(b(a(a(x)))) → b(a(b(b(a(x)))))
a(a(b(a(x)))) → a(b(b(a(b(x)))))
a(b(b(a(x)))) → b(b(b(b(b(x)))))
a(a(a(a(a(x))))) → a(a(a(b(a(a(a(x)))))))
a(b(a(a(a(x))))) → b(a(a(b(b(a(a(x)))))))
a(a(b(a(a(x))))) → a(b(a(b(a(b(a(x)))))))
a(b(b(a(a(x))))) → b(b(a(b(b(b(a(x)))))))
a(a(a(b(a(x))))) → a(a(b(b(a(a(b(x)))))))
a(b(a(b(a(x))))) → b(a(b(b(b(a(b(x)))))))
a(a(b(b(a(x))))) → a(b(b(b(a(b(b(x)))))))
a(b(b(b(a(x))))) → b(b(b(b(b(b(b(x)))))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

a(a(x)) → b(x)
a(a(a(x))) → a(b(a(x)))
a(b(a(x))) → b(b(b(x)))
a(a(a(a(x)))) → a(a(b(a(a(x)))))
a(b(a(a(x)))) → b(a(b(b(a(x)))))
a(a(b(a(x)))) → a(b(b(a(b(x)))))
a(b(b(a(x)))) → b(b(b(b(b(x)))))
a(a(a(a(a(x))))) → a(a(a(b(a(a(a(x)))))))
a(b(a(a(a(x))))) → b(a(a(b(b(a(a(x)))))))
a(a(b(a(a(x))))) → a(b(a(b(a(b(a(x)))))))
a(b(b(a(a(x))))) → b(b(a(b(b(b(a(x)))))))
a(a(a(b(a(x))))) → a(a(b(b(a(a(b(x)))))))
a(b(a(b(a(x))))) → b(a(b(b(b(a(b(x)))))))
a(a(b(b(a(x))))) → a(b(b(b(a(b(b(x)))))))
a(b(b(b(a(x))))) → b(b(b(b(b(b(b(x)))))))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
a(a(a(x))) → a(b(a(x)))
a(b(a(x))) → b(b(b(x)))
a(a(a(a(x)))) → a(a(b(a(a(x)))))
a(b(a(a(x)))) → b(a(b(b(a(x)))))
a(a(b(a(x)))) → a(b(b(a(b(x)))))
a(b(b(a(x)))) → b(b(b(b(b(x)))))
a(a(a(a(a(x))))) → a(a(a(b(a(a(a(x)))))))
a(b(a(a(a(x))))) → b(a(a(b(b(a(a(x)))))))
a(a(b(a(a(x))))) → a(b(a(b(a(b(a(x)))))))
a(b(b(a(a(x))))) → b(b(a(b(b(b(a(x)))))))
a(a(a(b(a(x))))) → a(a(b(b(a(a(b(x)))))))
a(b(a(b(a(x))))) → b(a(b(b(b(a(b(x)))))))
a(a(b(b(a(x))))) → a(b(b(b(a(b(b(x)))))))
a(b(b(b(a(x))))) → b(b(b(b(b(b(b(x)))))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof

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

a(a(x)) → b(x)
a(a(a(x))) → a(b(a(x)))
a(b(a(x))) → b(b(b(x)))
a(a(a(a(x)))) → a(a(b(a(a(x)))))
a(b(a(a(x)))) → b(a(b(b(a(x)))))
a(a(b(a(x)))) → a(b(b(a(b(x)))))
a(b(b(a(x)))) → b(b(b(b(b(x)))))
a(a(a(a(a(x))))) → a(a(a(b(a(a(a(x)))))))
a(b(a(a(a(x))))) → b(a(a(b(b(a(a(x)))))))
a(a(b(a(a(x))))) → a(b(a(b(a(b(a(x)))))))
a(b(b(a(a(x))))) → b(b(a(b(b(b(a(x)))))))
a(a(a(b(a(x))))) → a(a(b(b(a(a(b(x)))))))
a(b(a(b(a(x))))) → b(a(b(b(b(a(b(x)))))))
a(a(b(b(a(x))))) → a(b(b(b(a(b(b(x)))))))
a(b(b(b(a(x))))) → b(b(b(b(b(b(b(x)))))))

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:

A(b(b(a(a(x1))))) → A(b(b(b(a(x1)))))
A(a(b(b(a(x1))))) → A(b(b(b(a(b(b(x1)))))))
A(a(a(a(x1)))) → A(a(b(a(a(x1)))))
A(a(b(a(a(x1))))) → A(b(a(x1)))
A(b(a(a(a(x1))))) → A(a(b(b(a(a(x1))))))
A(a(b(a(x1)))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(b(a(x1)))))
A(a(a(a(a(x1))))) → A(b(a(a(a(x1)))))
A(a(a(b(a(x1))))) → A(b(x1))
A(b(a(b(a(x1))))) → A(b(b(b(a(b(x1))))))
A(b(a(a(a(x1))))) → A(b(b(a(a(x1)))))
A(a(b(b(a(x1))))) → A(b(b(x1)))
A(a(a(a(a(x1))))) → A(a(b(a(a(a(x1))))))
A(a(a(b(a(x1))))) → A(a(b(x1)))
A(b(a(b(a(x1))))) → A(b(x1))
A(a(a(x1))) → A(b(a(x1)))
A(a(a(b(a(x1))))) → A(b(b(a(a(b(x1))))))
A(a(a(a(x1)))) → A(b(a(a(x1))))
A(a(a(a(a(x1))))) → A(a(a(b(a(a(a(x1)))))))
A(a(b(a(a(x1))))) → A(b(a(b(a(b(a(x1)))))))
A(b(a(a(x1)))) → A(b(b(a(x1))))
A(a(b(a(x1)))) → A(b(b(a(b(x1)))))
A(a(a(b(a(x1))))) → A(a(b(b(a(a(b(x1)))))))

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

A(b(b(a(a(x1))))) → A(b(b(b(a(x1)))))
A(a(b(b(a(x1))))) → A(b(b(b(a(b(b(x1)))))))
A(a(a(a(x1)))) → A(a(b(a(a(x1)))))
A(a(b(a(a(x1))))) → A(b(a(x1)))
A(b(a(a(a(x1))))) → A(a(b(b(a(a(x1))))))
A(a(b(a(x1)))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(b(a(x1)))))
A(a(a(a(a(x1))))) → A(b(a(a(a(x1)))))
A(a(a(b(a(x1))))) → A(b(x1))
A(b(a(b(a(x1))))) → A(b(b(b(a(b(x1))))))
A(b(a(a(a(x1))))) → A(b(b(a(a(x1)))))
A(a(b(b(a(x1))))) → A(b(b(x1)))
A(a(a(a(a(x1))))) → A(a(b(a(a(a(x1))))))
A(a(a(b(a(x1))))) → A(a(b(x1)))
A(b(a(b(a(x1))))) → A(b(x1))
A(a(a(x1))) → A(b(a(x1)))
A(a(a(b(a(x1))))) → A(b(b(a(a(b(x1))))))
A(a(a(a(x1)))) → A(b(a(a(x1))))
A(a(a(a(a(x1))))) → A(a(a(b(a(a(a(x1)))))))
A(a(b(a(a(x1))))) → A(b(a(b(a(b(a(x1)))))))
A(b(a(a(x1)))) → A(b(b(a(x1))))
A(a(b(a(x1)))) → A(b(b(a(b(x1)))))
A(a(a(b(a(x1))))) → A(a(b(b(a(a(b(x1)))))))

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

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 8 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

A(a(a(a(x1)))) → A(a(b(a(a(x1)))))
A(a(b(a(a(x1))))) → A(b(a(x1)))
A(b(a(a(a(x1))))) → A(a(b(b(a(a(x1))))))
A(a(b(a(x1)))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(b(a(x1)))))
A(a(a(a(a(x1))))) → A(b(a(a(a(x1)))))
A(a(a(b(a(x1))))) → A(b(x1))
A(a(a(a(a(x1))))) → A(a(b(a(a(a(x1))))))
A(a(a(b(a(x1))))) → A(a(b(x1)))
A(b(a(b(a(x1))))) → A(b(x1))
A(a(a(x1))) → A(b(a(x1)))
A(a(a(a(x1)))) → A(b(a(a(x1))))
A(a(a(a(a(x1))))) → A(a(a(b(a(a(a(x1)))))))
A(a(b(a(a(x1))))) → A(b(a(b(a(b(a(x1)))))))
A(a(a(b(a(x1))))) → A(a(b(b(a(a(b(x1)))))))

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

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(a(a(a(x1)))) → A(a(b(a(a(x1)))))
A(a(a(a(a(x1))))) → A(b(a(a(a(x1)))))
A(a(a(b(a(x1))))) → A(b(x1))
A(a(a(a(a(x1))))) → A(a(b(a(a(a(x1))))))
A(a(a(b(a(x1))))) → A(a(b(x1)))
A(a(a(x1))) → A(b(a(x1)))
A(a(a(a(x1)))) → A(b(a(a(x1))))
A(a(a(a(a(x1))))) → A(a(a(b(a(a(a(x1)))))))
A(a(a(b(a(x1))))) → A(a(b(b(a(a(b(x1)))))))
The remaining pairs can at least be oriented weakly.

A(a(b(a(a(x1))))) → A(b(a(x1)))
A(b(a(a(a(x1))))) → A(a(b(b(a(a(x1))))))
A(a(b(a(x1)))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(b(a(x1)))))
A(b(a(b(a(x1))))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(b(a(b(a(x1)))))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = max{0, x1 - 1}


POL( a(x1) ) = x1 + 1


POL( b(x1) ) = max{0, -1}



The following usable rules [17] were oriented:

a(b(a(x1))) → b(b(b(x1)))
a(a(a(x1))) → a(b(a(x1)))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(a(x1)) → b(x1)
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

A(b(a(b(a(x1))))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(x1)))
A(a(b(a(x1)))) → A(b(x1))
A(b(a(a(a(x1))))) → A(a(b(b(a(a(x1))))))
A(a(b(a(a(x1))))) → A(b(a(b(a(x1)))))
A(a(b(a(a(x1))))) → A(b(a(b(a(b(a(x1)))))))

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

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(b(a(x1))))) → A(b(x1))
A(a(b(a(a(x1))))) → A(b(a(x1)))
A(a(b(a(x1)))) → A(b(x1))
A(b(a(a(a(x1))))) → A(a(b(b(a(a(x1))))))
A(a(b(a(a(x1))))) → A(b(a(b(a(x1)))))
A(a(b(a(a(x1))))) → A(b(a(b(a(b(a(x1)))))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(a(x1)) = 1 + (3)x_1   
POL(A(x1)) = x_1   
POL(b(x1)) = (1/4)x_1   
The value of delta used in the strict ordering is 7/16.
The following usable rules [17] were oriented:

a(b(a(x1))) → b(b(b(x1)))
a(a(a(x1))) → a(b(a(x1)))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(a(x1)) → b(x1)
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

a(a(x1)) → b(x1)
a(a(a(x1))) → a(b(a(x1)))
a(b(a(x1))) → b(b(b(x1)))
a(a(a(a(x1)))) → a(a(b(a(a(x1)))))
a(a(b(a(x1)))) → a(b(b(a(b(x1)))))
a(b(a(a(x1)))) → b(a(b(b(a(x1)))))
a(b(b(a(x1)))) → b(b(b(b(b(x1)))))
a(a(a(a(a(x1))))) → a(a(a(b(a(a(a(x1)))))))
a(a(a(b(a(x1))))) → a(a(b(b(a(a(b(x1)))))))
a(a(b(a(a(x1))))) → a(b(a(b(a(b(a(x1)))))))
a(a(b(b(a(x1))))) → a(b(b(b(a(b(b(x1)))))))
a(b(a(a(a(x1))))) → b(a(a(b(b(a(a(x1)))))))
a(b(a(b(a(x1))))) → b(a(b(b(b(a(b(x1)))))))
a(b(b(a(a(x1))))) → b(b(a(b(b(b(a(x1)))))))
a(b(b(b(a(x1))))) → b(b(b(b(b(b(b(x1)))))))

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.