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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(a(a(a(x1)))) → a(a(b(b(b(b(b(b(x1))))))))
b(b(a(a(x1)))) → b(b(b(b(c(c(x1))))))
a(a(b(b(b(b(c(c(x1)))))))) → a(a(a(a(a(a(b(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:

A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(b(b(x1)))))
B(b(a(a(x1)))) → B(b(b(c(c(x1)))))
B(b(a(a(x1)))) → B(c(c(x1)))
A(a(a(a(x1)))) → B(b(x1))
B(b(a(a(x1)))) → B(b(c(c(x1))))
A(a(b(b(b(b(c(c(x1)))))))) → B(x1)
A(a(a(a(x1)))) → B(b(b(b(x1))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(b(b(x1))))))
A(a(b(b(b(b(c(c(x1)))))))) → B(b(x1))
A(a(b(b(b(b(c(c(x1)))))))) → A(b(b(x1)))
A(a(a(a(x1)))) → B(b(b(b(b(b(x1))))))
A(a(a(a(x1)))) → A(b(b(b(b(b(b(x1)))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(a(a(b(b(x1))))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(b(b(x1))))
A(a(a(a(x1)))) → A(a(b(b(b(b(b(b(x1))))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(a(b(b(x1)))))))
A(a(a(a(x1)))) → B(x1)
A(a(a(a(x1)))) → B(b(b(b(b(x1)))))
B(b(a(a(x1)))) → B(b(b(b(c(c(x1))))))
A(a(a(a(x1)))) → B(b(b(x1)))

The TRS R consists of the following rules:

a(a(a(a(x1)))) → a(a(b(b(b(b(b(b(x1))))))))
b(b(a(a(x1)))) → b(b(b(b(c(c(x1))))))
a(a(b(b(b(b(c(c(x1)))))))) → a(a(a(a(a(a(b(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:

A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(b(b(x1)))))
B(b(a(a(x1)))) → B(b(b(c(c(x1)))))
B(b(a(a(x1)))) → B(c(c(x1)))
A(a(a(a(x1)))) → B(b(x1))
B(b(a(a(x1)))) → B(b(c(c(x1))))
A(a(b(b(b(b(c(c(x1)))))))) → B(x1)
A(a(a(a(x1)))) → B(b(b(b(x1))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(b(b(x1))))))
A(a(b(b(b(b(c(c(x1)))))))) → B(b(x1))
A(a(b(b(b(b(c(c(x1)))))))) → A(b(b(x1)))
A(a(a(a(x1)))) → B(b(b(b(b(b(x1))))))
A(a(a(a(x1)))) → A(b(b(b(b(b(b(x1)))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(a(a(b(b(x1))))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(b(b(x1))))
A(a(a(a(x1)))) → A(a(b(b(b(b(b(b(x1))))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(a(b(b(x1)))))))
A(a(a(a(x1)))) → B(x1)
A(a(a(a(x1)))) → B(b(b(b(b(x1)))))
B(b(a(a(x1)))) → B(b(b(b(c(c(x1))))))
A(a(a(a(x1)))) → B(b(b(x1)))

The TRS R consists of the following rules:

a(a(a(a(x1)))) → a(a(b(b(b(b(b(b(x1))))))))
b(b(a(a(x1)))) → b(b(b(b(c(c(x1))))))
a(a(b(b(b(b(c(c(x1)))))))) → a(a(a(a(a(a(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 12 less nodes.

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

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

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

The TRS R consists of the following rules:

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

A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(b(b(x1)))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(a(a(b(b(x1))))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(b(b(x1))))
A(a(a(a(x1)))) → A(a(b(b(b(b(b(b(x1))))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(a(b(b(x1)))))))
A(a(b(b(b(b(c(c(x1)))))))) → A(a(a(a(b(b(x1))))))
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = 0   
POL(a(x1)) = 2   
POL(A(x1)) = (2)x_1   
POL(b(x1)) = 0   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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