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(x)) → b(b(a(a(x))))

The set Q consists of the following terms:

a(b(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

a(b(x)) → b(b(a(a(x))))

The set Q consists of the following terms:

a(b(x0))


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

a(b(x)) → b(b(a(a(x))))

The set Q is {a(b(x0))}.
We have obtained the following QTRS:

b(a(x)) → a(a(b(b(x))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

b(a(x)) → a(a(b(b(x))))

Q is empty.

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

a(b(x)) → b(b(a(a(x))))

The set Q is {a(b(x0))}.
We have obtained the following QTRS:

b(a(x)) → a(a(b(b(x))))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof

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

b(a(x)) → a(a(b(b(x))))

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(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → b(b(a(a(x))))

The set Q consists of the following terms:

a(b(x0))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ MNOCProof

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

A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → b(b(a(a(x))))

The set Q consists of the following terms:

a(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ MNOCProof
        ↳ QDP
          ↳ NonTerminationProof

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

A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → b(b(a(a(x))))

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(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → b(b(a(a(x))))


s = A(a(b(x'))) evaluates to t =A(a(b(a(a(x')))))

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



Rewriting sequence

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

A(b(b(a(a(x'))))) → A(a(b(a(a(x')))))
with rule A(b(x)) → A(a(x))

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.