Termination w.r.t. Q proof of /tmp/tmpBStlpg/multum4.srs

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

b(b(a(b(b(a(b(b(b(b(b(b(a(b(x1)))))))))))))) → b(b(b(b(b(a(b(b(a(b(b(a(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 11 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(B(x₁)) = (1/4)x_1
POL(a(x₁)) = 1/4 + x_1
POL(b(x₁)) = x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

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

The TRS R consists of the following rules:

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

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