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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

a(a(a(x1))) → a(b(b(x1)))
b(a(b(x1))) → a(b(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
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Narrowing
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A(a(a(x1))) → A(b(b(x1)))

The TRS R consists of the following rules:

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

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

A(a(a(a(b(x0))))) → A(b(a(b(a(x0)))))



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

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

A(a(a(a(b(x0))))) → A(b(a(b(a(x0)))))

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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


s = B(b(a(b(a(A(x)))))) evaluates to t =B(b(a(b(a(A(x))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

B(b(a(b(a(A(x)))))) → B(a(b(a(a(A(x))))))
with rule b(a(b(x'))) → a(b(a(x'))) at position [0] and matcher [x' / a(A(x))]

B(a(b(a(a(A(x)))))) → A1(b(a(a(a(A(x))))))
with rule B(a(b(x'))) → A1(b(a(x'))) at position [] and matcher [x' / a(a(A(x)))]

A1(b(a(a(a(A(x)))))) → A1(a(b(a(b(A(x))))))
with rule b(a(a(a(A(x'))))) → a(b(a(b(A(x'))))) at position [0] and matcher [x' / x]

A1(a(b(a(b(A(x)))))) → A1(a(a(b(a(A(x))))))
with rule b(a(b(x'))) → a(b(a(x'))) at position [0,0] and matcher [x' / A(x)]

A1(a(a(b(a(A(x)))))) → B(b(a(b(a(A(x))))))
with rule A1(a(a(x))) → B(b(a(x)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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

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

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

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

The set Q is empty.

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.