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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → x1
a(c(x1)) → c(c(a(a(b(x1)))))
b(b(x1)) → 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
          ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(c(b(x0))) → A(x0)



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

A(c(b(x0))) → A(x0)



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

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

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

The TRS R consists of the following rules:

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → x
c(a(x)) → b(a(a(c(c(x)))))
b(b(x)) → x
c(A(x)) → A(x)
b(c(A(x))) → A(x)
b(c(A(x))) → a(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 4 less nodes.

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

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

C(a(x)) → C(c(x))
C(a(x)) → C(x)

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.