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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

The TRS R consists of the following rules:

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

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

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

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

A1(c(c(x))) → A1(a(x))
A1(c(C(x))) → A2(x)
A1(c(C(x))) → A1(A(x))
A1(c(c(x))) → A1(x)

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

A1(c(C(x0))) → A1(C(x0))



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

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

A1(c(c(x))) → A1(a(x))
A1(c(c(x))) → A1(x)
A1(c(C(x0))) → A1(C(x0))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(x) → x
a(x) → c(b(b(x)))
a(c(c(x))) → c(c(a(a(x))))
a(c(C(x))) → c(c(A(x)))
A(x) → C(x)
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(c(C(x))) → c(c(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 3 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.