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:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


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

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



Rewriting sequence

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

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

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

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

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

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

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

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

b(b(a(x))) → a(a(b(b(x))))
a(a(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:

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

Q is empty.

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

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

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

b(b(a(x))) → a(a(b(b(x))))
a(a(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:

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

Q is empty.