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(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

C(R(x1)) → B(a(R(x1)))
L1(a(a(x1))) → C(x1)
C(a(x1)) → C(x1)
L1(a(a(x1))) → L1(a(b(c(x1))))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
L1(a(a(x1))) → B(c(x1))
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

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

C(R(x1)) → B(a(R(x1)))
L1(a(a(x1))) → C(x1)
C(a(x1)) → C(x1)
L1(a(a(x1))) → L1(a(b(c(x1))))
B(a(a(x1))) → B(c(x1))
B(a(a(x1))) → C(x1)
L1(a(a(x1))) → B(c(x1))
C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(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 3 less nodes.

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

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

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

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(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.


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

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

POL(C(x1)) = 4 + (4)x_1   
POL(c(x1)) = 1 + x_1   
POL(B(x1)) = (4)x_1   
POL(a(x1)) = 1 + x_1   
POL(b(x1)) = x_1   
POL(R(x1)) = 0   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

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



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

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

C(b(x1)) → B(a(x1))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

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

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

L1(a(a(x1))) → L1(a(b(c(x1))))

The TRS R consists of the following rules:

b(a(a(x1))) → a(b(c(x1)))
c(a(x1)) → a(c(x1))
c(b(x1)) → b(a(x1))
L(a(a(x1))) → L(a(b(c(x1))))
c(R(x1)) → b(a(R(x1)))

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