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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x1) → b(x1)
b(a(b(c(x1)))) → b(c(c(a(a(a(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(x1) → B(x1) we obtained the following new rules:

A(a(y_1)) → B(a(y_1))



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

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

B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(x1)))) → A(x1)
A(a(y_1)) → B(a(y_1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(a(b(c(x1)))) → A(x1) we obtained the following new rules:

B(a(b(c(a(y_0))))) → A(a(y_0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
QDP
                  ↳ QDPToSRSProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(a(y_0))))) → A(a(y_0))
A(a(y_1)) → B(a(y_1))

The TRS R consists of the following rules:

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

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

a(x1) → b(x1)
b(a(b(c(x1)))) → b(c(c(a(a(a(x1))))))
B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(a(y_0))))) → A(a(y_0))
A(a(y_1)) → B(a(y_1))

Q is empty.

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

a(x1) → b(x1)
b(a(b(c(x1)))) → b(c(c(a(a(a(x1))))))
B(a(b(c(x1)))) → A(a(a(x1)))
B(a(b(c(x1)))) → A(a(x1))
B(a(b(c(a(y_0))))) → A(a(y_0))
A(a(y_1)) → B(a(y_1))

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
QTRS
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
QDP
                              ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(x) → b(x)
c(b(a(b(x)))) → a(a(a(c(c(b(x))))))
c(b(a(B(x)))) → a(a(A(x)))
c(b(a(B(x)))) → a(A(x))
a(c(b(a(B(x))))) → a(A(x))
a(A(x)) → a(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 7 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

Q is empty.