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:

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → 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:

C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(x1))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(B(B(c(c(b(b(x1)))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(x1)))
B1(b(b(b(x1)))) → B2(B(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
B2(B(B(B(x1)))) → B1(x1)
C1(C(x1)) → C2(x1)
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(c(c(B(B(c(c(b(b(x1))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(B(B(c(c(b(b(x1))))))))
B2(B(B(B(x1)))) → B1(b(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(x1))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(x1)
B1(b(b(b(x1)))) → B2(x1)
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(c(c(B(B(c(c(b(b(x1)))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
C1(C(x1)) → C2(c(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(c(c(B(B(c(c(b(b(x1)))))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(x1)))))

The TRS R consists of the following rules:

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → 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:

C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(x1))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(B(B(c(c(b(b(x1)))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(x1)))
B1(b(b(b(x1)))) → B2(B(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
B2(B(B(B(x1)))) → B1(x1)
C1(C(x1)) → C2(x1)
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(c(c(B(B(c(c(b(b(x1))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(B(B(c(c(b(b(x1))))))))
B2(B(B(B(x1)))) → B1(b(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(x1))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(x1)
B1(b(b(b(x1)))) → B2(x1)
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(b(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B1(c(c(B(B(c(c(b(b(x1)))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
C1(C(x1)) → C2(c(x1))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(c(c(B(B(c(c(b(b(x1)))))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → B2(c(c(b(b(x1)))))

The TRS R consists of the following rules:

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 10 less nodes.

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

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

B2(B(B(B(x1)))) → B1(b(x1))
B1(b(b(b(x1)))) → B2(B(x1))
B1(b(b(b(x1)))) → B2(x1)
B2(B(B(B(x1)))) → B1(x1)

The TRS R consists of the following rules:

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → 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.


B2(B(B(B(x1)))) → B1(b(x1))
B1(b(b(b(x1)))) → B2(B(x1))
B1(b(b(b(x1)))) → B2(x1)
B2(B(B(B(x1)))) → B1(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(B2(x1)) = 1/2 + (4)x_1   
POL(B(x1)) = 4 + (4)x_1   
POL(B1(x1)) = (4)x_1   
POL(b(x1)) = 1/4 + (4)x_1   
The value of delta used in the strict ordering is 9/2.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → 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.

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

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

C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(B(B(c(c(b(b(x1)))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(x1)))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(x1))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(B(B(c(c(b(b(x1))))))))

The TRS R consists of the following rules:

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → 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.


C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(B(B(c(c(b(b(x1)))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(x1)))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(x1))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(b(b(c(c(B(B(c(c(b(b(x1)))))))))))
C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(B(B(c(c(b(b(x1))))))))
The remaining pairs can at least be oriented weakly.

C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (4)x_1   
POL(c(x1)) = 1/4 + x_1   
POL(B(x1)) = x_1   
POL(C2(x1)) = (2)x_1   
POL(b(x1)) = x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

c(c(c(c(x1)))) → x1
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
B(B(B(B(x1)))) → b(b(x1))
b(b(b(b(x1)))) → B(B(x1))
B(B(b(b(x1)))) → x1
b(b(B(B(x1)))) → x1
c(c(C(C(x1)))) → x1



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

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

C2(c(B(B(c(c(b(b(c(c(x1)))))))))) → C2(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))

The TRS R consists of the following rules:

C(C(x1)) → c(c(x1))
c(c(c(c(x1)))) → x1
b(b(b(b(x1)))) → B(B(x1))
B(B(B(B(x1)))) → b(b(x1))
c(c(B(B(c(c(b(b(c(c(x1)))))))))) → B(B(c(c(b(b(c(c(B(B(c(c(b(b(x1))))))))))))))
b(b(B(B(x1)))) → x1
B(B(b(b(x1)))) → x1
c(c(C(C(x1)))) → x1
C(C(c(c(x1)))) → x1

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