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:

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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


Used ordering: POLO with Polynomial interpretation [25]:

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



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

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

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

The TRS R consists of the following rules:

b(a(x1)) → b(b(x1))
b(a(b(x1))) → b(a(a(x1)))
a(a(a(x1))) → a(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 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
  ↳ 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:

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


A(a(a(x1))) → A(b(b(x1)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( A(x1) ) = max{0, x1 - 1}


POL( a(x1) ) = x1 + 1


POL( b(x1) ) = max{0, -1}



The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
  ↳ QTRS Reverse

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

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

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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


B(a(x1)) → B(b(x1))
The remaining pairs can at least be oriented weakly.

B(a(b(x1))) → B(a(a(x1)))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( B(x1) ) = x1 + 1


POL( a(x1) ) = 1


POL( b(x1) ) = 0



The following usable rules [17] were oriented:

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



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

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

B(a(b(x1))) → B(a(a(x1)))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following 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.1(a.0(b.1(a.1(a.1(x0))))) → B.1(a.1(a.0(b.0(b.1(x0)))))
B.1(a.0(b.1(a.1(a.0(x0))))) → B.1(a.1(a.0(b.0(b.0(x0)))))
B.1(a.0(b.1(a.0(x0)))) → B.1(a.0(b.0(b.0(x0))))
B.1(a.0(b.1(a.1(x0)))) → B.1(a.0(b.0(b.1(x0))))

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 4 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.