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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(x1) → b(x1)
a(b(x1)) → c(a(c(a(x1))))
c(b(c(x1))) → b(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A(b(x1)) → A(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = x1   



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

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

A(b(x1)) → A(c(a(x1)))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

A(b(c(x0))) → A(b(x0))



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

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

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

The TRS R consists of the following rules:

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ Narrowing
                    ↳ 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:

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

B(b(A(x))) → A1(c(c(A(x))))


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2 + x1   
POL(A1(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(a(x1)) = 2·x1   
POL(b(x1)) = 2·x1   
POL(c(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = x1   
POL(A1(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(a(x1)) = 1 + 2·x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = x1   



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

Q is empty.