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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(c(x1))) → A(a(b(x1))) at position [0] we obtained the following new rules:

A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x0))) → A(x0)



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
QDP
          ↳ QDPToSRSProof
  ↳ QTRS Reverse

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

A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x1))) → A(b(x1))
A(b(c(x0))) → A(x0)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
QTRS
              ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x1))) → A(b(x1))
A(b(c(x0))) → A(x0)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x1))) → A(b(x1))
A(b(c(x0))) → A(x0)

The set Q is empty.
We have obtained the following QTRS:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

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(c(b(A(x)))) → C(b(c(b(A(x)))))
C(c(b(A(x)))) → B(a(a(c(b(c(b(A(x))))))))
C(c(b(A(x)))) → B(c(b(A(x))))
C(b(a(x))) → B(x)
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → B(c(b(x)))
C(b(a(x))) → C(b(x))
C(b(a(x))) → B(a(a(c(b(c(b(x)))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
QDP
                      ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(c(b(A(x)))) → C(b(c(b(A(x)))))
C(c(b(A(x)))) → B(a(a(c(b(c(b(A(x))))))))
C(c(b(A(x)))) → B(c(b(A(x))))
C(b(a(x))) → B(x)
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → B(c(b(x)))
C(b(a(x))) → C(b(x))
C(b(a(x))) → B(a(a(c(b(c(b(x)))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

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 5 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
QDP
                          ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(c(b(A(x)))) → C(b(c(b(A(x)))))
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → C(b(x))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(b(A(x)))) → C(b(c(b(A(x))))) at position [0] we obtained the following new rules:

C(c(b(A(x0)))) → C(b(b(A(x0))))
C(c(b(A(x0)))) → C(b(A(x0)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
QDP
                              ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(c(b(A(x0)))) → C(b(A(x0)))
C(b(a(x))) → C(b(c(b(x))))
C(c(b(A(x0)))) → C(b(b(A(x0))))
C(b(a(x))) → C(b(x))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

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 2 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
QDP
                                  ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → C(b(x))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(a(x))) → C(b(c(b(x)))) at position [0] we obtained the following new rules:

C(b(a(A(x0)))) → C(b(A(x0)))
C(b(a(a(x0)))) → C(b(b(a(a(c(b(c(b(x0)))))))))
C(b(a(A(x0)))) → C(b(b(A(x0))))
C(b(a(a(x0)))) → C(b(c(x0)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
QDP
                                      ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(a(A(x0)))) → C(b(A(x0)))
C(b(a(a(x0)))) → C(b(b(a(a(c(b(c(b(x0)))))))))
C(b(a(A(x0)))) → C(b(b(A(x0))))
C(b(a(a(x0)))) → C(b(c(x0)))
C(b(a(x))) → C(b(x))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

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 2 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(a(a(x0)))) → C(b(b(a(a(c(b(c(b(x0)))))))))
C(b(a(a(x0)))) → C(b(c(x0)))
C(b(a(x))) → C(b(x))

The TRS R consists of the following rules:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

The set Q is empty.
We have obtained the following QTRS:

a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
QTRS
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)

The set Q is empty.
We have obtained the following QTRS:

a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))

Q is empty.