Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 Overlay + Local Confluence (⇔)
    ↳6 QTRS
    ↳7 DependencyPairsProof (⇔)
    ↳8 QDP
    ↳9 MNOCProof (⇔)
    ↳10 QDP
    ↳11 NonTerminationProof (⇔)
    ↳12 FALSE

(0) Obligation:

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

a(b(x)) → b(b(a(a(x))))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

b(a(x)) → a(a(b(b(x))))

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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

b(a(x)) → a(a(b(b(x))))

Q is empty.

(5) Overlay + Local Confluence (EQUIVALENT transformation)

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

(6) Obligation:

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))

(7) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(8) Obligation:

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.

(9) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(10) Obligation:

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.

(11) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:
  • Matcher: [x' / a(a(x'))]
  • Semiunifier: [ ]



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.



(12) FALSE