Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

a(b(a(x1))) → b(a(x1))
b(b(b(x1))) → b(a(b(x1)))
a(a(x1)) → b(b(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 4 less nodes.

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

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

B(b(b(x1))) → B(a(b(x1)))

The TRS R consists of the following rules:

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

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

B(b(b(a(x0)))) → B(b(a(x0)))
B(b(b(b(b(x0))))) → B(a(b(a(b(x0)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ SemLabProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B(b(b(b(b(x0))))) → B(a(b(a(b(x0)))))
B(b(b(a(x0)))) → B(b(a(x0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.B: 0
a: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

B.0(b.0(b.1(a.1(x0)))) → B.0(b.1(a.1(x0)))
B.0(b.0(b.1(a.0(x0)))) → B.0(b.1(a.0(x0)))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.1(a.0(b.1(a.0(b.0(x0)))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.1(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(a.0(b.1(a.0(b.0(x0)))))

The TRS R consists of the following rules:

b.0(b.0(b.0(x1))) → b.1(a.0(b.0(x1)))
b.0(b.0(b.1(x1))) → b.1(a.0(b.1(x1)))
a.0(b.1(a.1(x1))) → b.1(a.1(x1))
a.1(a.1(x1)) → b.0(b.0(b.1(x1)))
a.0(b.1(a.0(x1))) → b.1(a.0(x1))
b.1(x0) → b.0(x0)
a.1(a.0(x1)) → b.0(b.0(b.0(x1)))
a.1(x0) → a.0(x0)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ SemLabProof
                ↳ QDP
                  ↳ DependencyGraphProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B.0(b.0(b.1(a.1(x0)))) → B.0(b.1(a.1(x0)))
B.0(b.0(b.1(a.0(x0)))) → B.0(b.1(a.0(x0)))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.1(a.0(b.1(a.0(b.0(x0)))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.1(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(a.0(b.1(a.0(b.0(x0)))))

The TRS R consists of the following rules:

b.0(b.0(b.0(x1))) → b.1(a.0(b.0(x1)))
b.0(b.0(b.1(x1))) → b.1(a.0(b.1(x1)))
a.0(b.1(a.1(x1))) → b.1(a.1(x1))
a.1(a.1(x1)) → b.0(b.0(b.1(x1)))
a.0(b.1(a.0(x1))) → b.1(a.0(x1))
b.1(x0) → b.0(x0)
a.1(a.0(x1)) → b.0(b.0(b.0(x1)))
a.1(x0) → a.0(x0)

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
          ↳ Narrowing
            ↳ QDP
              ↳ SemLabProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ RuleRemovalProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

B.0(b.0(b.1(a.1(x0)))) → B.0(b.1(a.1(x0)))
B.0(b.0(b.1(a.0(x0)))) → B.0(b.1(a.0(x0)))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(a.0(b.1(a.0(b.0(x0)))))

The TRS R consists of the following rules:

b.0(b.0(b.0(x1))) → b.1(a.0(b.0(x1)))
b.0(b.0(b.1(x1))) → b.1(a.0(b.1(x1)))
a.0(b.1(a.1(x1))) → b.1(a.1(x1))
a.1(a.1(x1)) → b.0(b.0(b.1(x1)))
a.0(b.1(a.0(x1))) → b.1(a.0(x1))
b.1(x0) → b.0(x0)
a.1(a.0(x1)) → b.0(b.0(b.0(x1)))
a.1(x0) → a.0(x0)

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 rules of the TRS R:

a.1(a.1(x1)) → b.0(b.0(b.1(x1)))
a.1(a.0(x1)) → b.0(b.0(b.0(x1)))
a.1(x0) → a.0(x0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(B.0(x1)) = x1   
POL(a.0(x1)) = x1   
POL(a.1(x1)) = 1 + x1   
POL(b.0(x1)) = x1   
POL(b.1(x1)) = x1   



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

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

B.0(b.0(b.1(a.0(x0)))) → B.0(b.1(a.0(x0)))
B.0(b.0(b.1(a.1(x0)))) → B.0(b.1(a.1(x0)))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(a.0(b.1(a.0(b.0(x0)))))

The TRS R consists of the following rules:

b.0(b.0(b.0(x1))) → b.1(a.0(b.0(x1)))
b.0(b.0(b.1(x1))) → b.1(a.0(b.1(x1)))
a.0(b.1(a.1(x1))) → b.1(a.1(x1))
a.0(b.1(a.0(x1))) → b.1(a.0(x1))
b.1(x0) → b.0(x0)

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.0(b.0(b.1(a.0(x0)))) → B.0(b.1(a.0(x0)))
B.0(b.0(b.1(a.1(x0)))) → B.0(b.1(a.1(x0)))
B.0(b.0(b.0(b.0(b.1(x0))))) → B.0(a.0(b.1(a.0(b.1(x0)))))
B.0(b.0(b.0(b.0(b.0(x0))))) → B.0(a.0(b.1(a.0(b.0(x0)))))

Strictly oriented rules of the TRS R:

b.0(b.0(b.0(x1))) → b.1(a.0(b.0(x1)))
b.0(b.0(b.1(x1))) → b.1(a.0(b.1(x1)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(B.0(x1)) = x1   
POL(a.0(x1)) = x1   
POL(a.1(x1)) = x1   
POL(b.0(x1)) = 1 + x1   
POL(b.1(x1)) = 1 + x1   



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a.0(b.1(a.1(x1))) → b.1(a.1(x1))
a.0(b.1(a.0(x1))) → b.1(a.0(x1))
b.1(x0) → b.0(x0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We have reversed the following QTRS:
The set of rules R is

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

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

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

Q is empty.

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

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

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

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

Q is empty.