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

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 Strip Symbols Proof (⇐)
    ↳4 QTRS
  ↳5 QTRS Reverse (⇔)
    ↳6 QTRS
    ↳7 Strip Symbols Proof (⇐)
    ↳8 QTRS
  ↳9 DependencyPairsProof (⇔)
    ↳10 QDP
    ↳11 NonTerminationProof (⇔)
    ↳12 FALSE

(0) Obligation:

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

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

Q is empty.

(3) Strip Symbols Proof (SOUND transformation)

We were given the following TRS:

a(b(a(b(x)))) → b(a(b(a(a(b(x))))))

By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]:

a(b(a(x))) → b(a(b(a(a(x)))))

(4) Obligation:

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

a(b(a(x))) → b(a(b(a(a(x)))))

Q is empty.

(5) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(6) Obligation:

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

b(a(b(a(x)))) → b(a(a(b(a(b(x))))))

Q is empty.

(7) Strip Symbols Proof (SOUND transformation)

We were given the following TRS:

b(a(b(a(x)))) → b(a(a(b(a(b(x))))))

By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]:

a(b(a(x))) → a(a(b(a(b(x)))))

(8) Obligation:

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

a(b(a(x))) → a(a(b(a(b(x)))))

Q is empty.

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

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

The TRS R consists of the following rules:

a(b(a(b(x)))) → b(a(b(a(a(b(x))))))

Q is empty.
We have to consider all minimal (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(b(a(a(b(a(b(x'))))))) evaluates to t =A(b(a(a(b(a(b(a(a(b(x'))))))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x' / a(a(b(x')))]
  • Semiunifier: [ ]



Rewriting sequence

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

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