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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
      ↳ NonTerminationProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
      ↳ NonTerminationProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

B(b(a(a(x0)))) → B(b(b(b(x0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
      ↳ NonTerminationProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


s = A(b(b(a(b(b(x1)))))) evaluates to t =A(b(b(a(b(b(b(a(x1))))))))

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



Rewriting sequence

A(b(b(a(b(b(x1)))))) → A(b(b(a(a(b(a(x1)))))))
with rule b(b(x1')) → a(b(a(x1'))) at position [0,0,0,0] and matcher [x1' / x1]

A(b(b(a(a(b(a(x1))))))) → A(a(b(a(a(a(b(a(x1))))))))
with rule b(b(x1')) → a(b(a(x1'))) at position [0] and matcher [x1' / a(a(b(a(x1))))]

A(a(b(a(a(a(b(a(x1)))))))) → A(a(b(b(b(b(b(a(x1))))))))
with rule a(a(a(x1'))) → b(b(b(x1'))) at position [0,0,0] and matcher [x1' / b(a(x1))]

A(a(b(b(b(b(b(a(x1)))))))) → A(a(a(b(a(b(b(b(a(x1)))))))))
with rule b(b(x1')) → a(b(a(x1'))) at position [0,0] and matcher [x1' / b(b(b(a(x1))))]

A(a(a(b(a(b(b(b(a(x1))))))))) → B(b(b(b(a(b(b(b(a(x1)))))))))
with rule A(a(a(x1'))) → B(b(b(x1'))) at position [] and matcher [x1' / b(a(b(b(b(a(x1))))))]

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

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.