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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(b(a(x1))) → b(b(b(x1)))
b(a(b(x1))) → a(b(a(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)
A(b(a(x1))) → B(b(x1))
B(a(b(x1))) → B(a(x1))
A(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
      ↳ NonTerminationProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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


s = A(b(a(a(b(b(x1')))))) evaluates to t =A(b(a(a(b(b(x1'))))))

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



Rewriting sequence

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

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

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

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

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.