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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(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 3 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

B(b(a(a(b(x1))))) → B(b(x1))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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