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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

a(a(b(b(x0))))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(b(x0))))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(a(b(b(x0))))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ MNOCProof
            ↳ QDP
              ↳ NonTerminationProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all (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(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(a(a(a(a(x1)))))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
A(a(b(b(x1)))) → A(a(a(a(x1))))

The TRS R consists of the following rules:

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


s = A(a(a(a(b(b(x1')))))) evaluates to t =A(a(a(a(b(b(b(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(b(b(x1')))))) → A(a(b(b(b(b(b(a(a(a(a(a(x1'))))))))))))
with rule a(a(b(b(x1'')))) → b(b(b(b(b(a(a(a(a(a(x1'')))))))))) at position [0,0] and matcher [x1'' / x1']

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