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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


C(c(x1)) → A(x1)
C(c(x1)) → C(a(x1))
The remaining pairs can at least be oriented weakly.

C(c(x1)) → A(b(c(a(x1))))
A(b(b(x1))) → C(x1)
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = 4 + x1   
POL(C(x1)) = 8 + x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 2 + x1   
POL(c(x1)) = 6 + x1   

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

a(x1) → b(x1)
a(b(b(x1))) → c(x1)
c(c(x1)) → a(b(c(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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ 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(b(x1))) → c(x1)
c(c(x1)) → a(b(c(a(x1))))
A(b(b(x1))) → C(x1)
C(c(b(b(x0)))) → A(b(c(c(x0))))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ 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) → b(x)
b(b(a(x))) → c(x)
c(c(x)) → a(c(b(a(x))))
b(b(A(x))) → C(x)
b(b(c(C(x)))) → c(c(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:

C1(c(x)) → A1(c(b(a(x))))
B(b(c(C(x)))) → B(A(x))
C1(c(x)) → C1(b(a(x)))
A1(x) → B(x)
B(b(c(C(x)))) → C1(b(A(x)))
C1(c(x)) → A1(x)
C1(c(x)) → B(a(x))
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ 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:

C1(c(x)) → A1(c(b(a(x))))
B(b(c(C(x)))) → B(A(x))
C1(c(x)) → C1(b(a(x)))
A1(x) → B(x)
B(b(c(C(x)))) → C1(b(A(x)))
C1(c(x)) → A1(x)
C1(c(x)) → B(a(x))
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))

The TRS R consists of the following rules:

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

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

C1(c(x)) → A1(c(b(a(x))))
C1(c(x)) → C1(b(a(x)))
A1(x) → B(x)
C1(c(x)) → A1(x)
C1(c(x)) → B(a(x))
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


C1(c(x)) → C1(b(a(x)))
C1(c(x)) → A1(x)
C1(c(x)) → B(a(x))
The remaining pairs can at least be oriented weakly.

C1(c(x)) → A1(c(b(a(x))))
A1(x) → B(x)
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = 2   
POL(c(x1)) = 3/2 + x_1   
POL(A1(x1)) = 7/2 + (1/2)x_1   
POL(B(x1)) = 7/2 + (1/2)x_1   
POL(a(x1)) = 1/2 + x_1   
POL(A(x1)) = 1   
POL(b(x1)) = 1/2 + x_1   
POL(C1(x1)) = 4 + (1/2)x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



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

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

C1(c(x)) → A1(c(b(a(x))))
A1(x) → B(x)
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


C1(c(x)) → C1(b(a(x)))
C1(c(x)) → A1(x)
C1(c(x)) → B(a(x))
The remaining pairs can at least be oriented weakly.

C1(c(x)) → A1(c(b(a(x))))
A1(x) → B(x)
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = 0   
POL(A1(x1)) = 8 + x1   
POL(B(x1)) = 8 + x1   
POL(C(x1)) = 3   
POL(C1(x1)) = 11 + 4·x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = 7 + 8·x1   

The following usable rules [17] were oriented:

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



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

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

C1(c(x)) → A1(c(b(a(x))))
A1(x) → B(x)
B(b(a(x))) → C1(x)
B(b(c(C(x)))) → C1(c(b(A(x))))

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.