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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(B(x1)) = 1 + 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

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

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

A(a(b(a(b(x0))))) → B(a(c(b(a(a(x0))))))
A(a(b(b(x0)))) → B(c(b(a(a(x0)))))
A(a(b(y0))) → B(a(y0))



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

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

A(a(b(a(b(x0))))) → B(a(c(b(a(a(x0))))))
A(a(b(b(x0)))) → B(c(b(a(a(x0)))))
A(a(b(y0))) → B(a(y0))
B(c(x1)) → B(x1)
B(c(x1)) → A(b(x1))

The TRS R consists of the following rules:

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

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

B(c(c(x0))) → A(a(b(x0)))



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

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

B(c(c(x0))) → A(a(b(x0)))
A(a(b(a(b(x0))))) → B(a(c(b(a(a(x0))))))
A(a(b(b(x0)))) → B(c(b(a(a(x0)))))
A(a(b(y0))) → B(a(y0))
B(c(x1)) → B(x1)

The TRS R consists of the following rules:

a(x1) → x1
a(a(b(x1))) → c(b(a(a(x1))))
b(c(x1)) → 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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
QTRS
                      ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(x1) → x1
a(a(b(x1))) → c(b(a(a(x1))))
b(c(x1)) → a(b(x1))
B(c(c(x0))) → A(a(b(x0)))
A(a(b(a(b(x0))))) → B(a(c(b(a(a(x0))))))
A(a(b(b(x0)))) → B(c(b(a(a(x0)))))
A(a(b(y0))) → B(a(y0))
B(c(x1)) → B(x1)

Q is empty.

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

a(x1) → x1
a(a(b(x1))) → c(b(a(a(x1))))
b(c(x1)) → a(b(x1))
B(c(c(x0))) → A(a(b(x0)))
A(a(b(a(b(x0))))) → B(a(c(b(a(a(x0))))))
A(a(b(b(x0)))) → B(c(b(a(a(x0)))))
A(a(b(y0))) → B(a(y0))
B(c(x1)) → B(x1)

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

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

The set Q is empty.

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

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

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

B1(a(a(x))) → A1(b(c(x)))
B1(b(a(A(x)))) → C(B(x))
C(b(x)) → A1(x)
B1(a(b(a(A(x))))) → A1(a(b(c(a(B(x))))))
C(c(B(x))) → A1(A(x))
B1(a(a(x))) → A1(a(b(c(x))))
B1(a(b(a(A(x))))) → A1(B(x))
C(b(x)) → B1(a(x))
B1(a(b(a(A(x))))) → B1(c(a(B(x))))
C(c(B(x))) → B1(a(A(x)))
B1(a(b(a(A(x))))) → C(a(B(x)))
B1(a(b(a(A(x))))) → A1(b(c(a(B(x)))))
B1(b(a(A(x)))) → A1(b(c(B(x))))
B1(a(a(x))) → C(x)
B1(a(a(x))) → B1(c(x))
B1(b(a(A(x)))) → B1(c(B(x)))
B1(b(a(A(x)))) → A1(a(b(c(B(x)))))
B1(a(A(x))) → A1(B(x))

The TRS R consists of the following rules:

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

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

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

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

B1(a(a(x))) → A1(b(c(x)))
B1(b(a(A(x)))) → C(B(x))
C(b(x)) → A1(x)
B1(a(b(a(A(x))))) → A1(a(b(c(a(B(x))))))
C(c(B(x))) → A1(A(x))
B1(a(a(x))) → A1(a(b(c(x))))
B1(a(b(a(A(x))))) → A1(B(x))
C(b(x)) → B1(a(x))
B1(a(b(a(A(x))))) → B1(c(a(B(x))))
C(c(B(x))) → B1(a(A(x)))
B1(a(b(a(A(x))))) → C(a(B(x)))
B1(a(b(a(A(x))))) → A1(b(c(a(B(x)))))
B1(b(a(A(x)))) → A1(b(c(B(x))))
B1(a(a(x))) → C(x)
B1(a(a(x))) → B1(c(x))
B1(b(a(A(x)))) → B1(c(B(x)))
B1(b(a(A(x)))) → A1(a(b(c(B(x)))))
B1(a(A(x))) → A1(B(x))

The TRS R consists of the following rules:

a(x) → x
b(a(a(x))) → a(a(b(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
b(a(b(a(A(x))))) → a(a(b(c(a(B(x))))))
b(b(a(A(x)))) → a(a(b(c(B(x)))))
b(a(A(x))) → a(B(x))
c(B(x)) → B(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 11 less nodes.

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

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

B1(a(b(a(A(x))))) → B1(c(a(B(x))))
C(b(x)) → B1(a(x))
B1(a(a(x))) → C(x)
B1(a(a(x))) → B1(c(x))
C(c(B(x))) → B1(a(A(x)))
B1(b(a(A(x)))) → B1(c(B(x)))
B1(a(b(a(A(x))))) → C(a(B(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

C(b(x)) → B1(a(x))
C(c(B(x))) → B1(a(A(x)))


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + x1   
POL(B(x1)) = 2 + x1   
POL(B1(x1)) = 1 + x1   
POL(C(x1)) = 1 + x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = x1   



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

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

B1(a(a(x))) → C(x)
B1(a(b(a(A(x))))) → B1(c(a(B(x))))
B1(a(a(x))) → B1(c(x))
B1(b(a(A(x)))) → B1(c(B(x)))
B1(a(b(a(A(x))))) → C(a(B(x)))

The TRS R consists of the following rules:

a(x) → x
b(a(a(x))) → a(a(b(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
b(a(b(a(A(x))))) → a(a(b(c(a(B(x))))))
b(b(a(A(x)))) → a(a(b(c(B(x)))))
b(a(A(x))) → a(B(x))
c(B(x)) → B(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                                          ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B1(a(b(a(A(x))))) → B1(c(a(B(x))))
B1(a(a(x))) → B1(c(x))
B1(b(a(A(x)))) → B1(c(B(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(x))))) → B1(c(a(B(x)))) at position [0] we obtained the following new rules:

B1(a(b(a(A(y0))))) → B1(c(B(y0)))



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

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

B1(a(a(x))) → B1(c(x))
B1(b(a(A(x)))) → B1(c(B(x)))
B1(a(b(a(A(y0))))) → B1(c(B(y0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(A(x)))) → B1(c(B(x))) at position [0] we obtained the following new rules:

B1(b(a(A(x0)))) → B1(B(x0))



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

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

B1(b(a(A(x0)))) → B1(B(x0))
B1(a(a(x))) → B1(c(x))
B1(a(b(a(A(y0))))) → B1(c(B(y0)))

The TRS R consists of the following rules:

a(x) → x
b(a(a(x))) → a(a(b(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
b(a(b(a(A(x))))) → a(a(b(c(a(B(x))))))
b(b(a(A(x)))) → a(a(b(c(B(x)))))
b(a(A(x))) → a(B(x))
c(B(x)) → B(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 1 less node.

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

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

B1(a(a(x))) → B1(c(x))
B1(a(b(a(A(y0))))) → B1(c(B(y0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(y0))))) → B1(c(B(y0))) at position [0] we obtained the following new rules:

B1(a(b(a(A(x0))))) → B1(B(x0))



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

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

B1(a(b(a(A(x0))))) → B1(B(x0))
B1(a(a(x))) → B1(c(x))

The TRS R consists of the following rules:

a(x) → x
b(a(a(x))) → a(a(b(c(x))))
c(b(x)) → b(a(x))
c(c(B(x))) → b(a(A(x)))
b(a(b(a(A(x))))) → a(a(b(c(a(B(x))))))
b(b(a(A(x)))) → a(a(b(c(B(x)))))
b(a(A(x))) → a(B(x))
c(B(x)) → B(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 1 less node.

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

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

B1(a(a(x))) → B1(c(x))

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.