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

Q is empty.

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

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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(b(a(x1)))) → a(c(a(b(x1))))
a(c(x1)) → c(c(a(x1)))
c(c(c(x1))) → b(c(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 4 less nodes.

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

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(A1(x1)) = 1 + 2·x1   
POL(C(x1)) = 2·x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   



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

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

A1(b(b(A(x)))) → A1(c(A(x)))
A1(b(b(a(x)))) → A1(c(a(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(b(A(x)))) → A1(c(A(x))) at position [0] we obtained the following new rules:

A1(b(b(A(x0)))) → A1(A(x0))



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

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

A1(b(b(a(x)))) → A1(c(a(x)))
A1(b(b(A(x0)))) → A1(A(x0))

The TRS R consists of the following rules:

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

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

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

A1(b(b(a(x)))) → A1(c(a(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(b(a(x)))) → A1(c(a(x))) at position [0] we obtained the following new rules:

A1(b(b(a(b(b(a(x0))))))) → A1(c(b(a(c(a(x0))))))
A1(b(b(a(b(b(A(x0))))))) → A1(c(b(a(c(A(x0))))))
A1(b(b(a(x0)))) → A1(a(c(c(x0))))



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

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

A1(b(b(a(b(b(a(x0))))))) → A1(c(b(a(c(a(x0))))))
A1(b(b(a(x0)))) → A1(a(c(c(x0))))
A1(b(b(a(b(b(A(x0))))))) → A1(c(b(a(c(A(x0))))))

The TRS R consists of the following rules:

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

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

A1(b(b(a(x0)))) → A1(a(c(c(x0))))

The TRS R consists of the following rules:

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

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

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

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

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

Q is empty.