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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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) → b(x)
b(a(x)) → x
a(b(x)) → b(a(c(a(x))))
c(c(x)) → x
a(B(x)) → b(A(x))
a(B(x)) → b(a(c(A(x))))
A(x) → B(x)
a(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:

A1(B(x)) → A2(x)
A1(B(x)) → A1(c(A(x)))
A1(x) → B1(x)
A1(b(x)) → C(a(x))
A1(b(x)) → B1(a(c(a(x))))
A1(b(x)) → A1(c(a(x)))
A1(B(x)) → C(A(x))
A1(B(x)) → B1(a(c(A(x))))
A1(B(x)) → B1(A(x))
A1(b(x)) → A1(x)

The TRS R consists of the following rules:

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

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

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

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

A1(B(x)) → A2(x)
A1(B(x)) → A1(c(A(x)))
A1(x) → B1(x)
A1(b(x)) → C(a(x))
A1(b(x)) → B1(a(c(a(x))))
A1(b(x)) → A1(c(a(x)))
A1(B(x)) → C(A(x))
A1(B(x)) → B1(a(c(A(x))))
A1(B(x)) → B1(A(x))
A1(b(x)) → A1(x)

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

A1(B(x0)) → A1(c(B(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(B(x0))) → A1(c(b(A(x0)))) at position [0] we obtained the following new rules:

A1(b(B(x0))) → A1(c(b(B(x0))))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(x0)) → A1(c(b(x0))) at position [0] we obtained the following new rules:

A1(b(a(x0))) → A1(c(x0))



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(x0))) → A1(c(x0)) at position [0] we obtained the following new rules:

A1(b(a(c(x0)))) → A1(x0)



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

Q is empty.