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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 Narrowing (⇔)
↳6 QDP
↳7 NonTerminationProof (⇔)
↳8 NO

(0) Obligation:

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

f(x, f(a, y)) → f(a, f(f(f(a, a), y), x))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

F(x, f(a, y)) → F(a, f(f(f(a, a), y), x))
F(x, f(a, y)) → F(f(f(a, a), y), x)
F(x, f(a, y)) → F(f(a, a), y)
F(x, f(a, y)) → F(a, a)

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, a), y), x))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

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

F(x, f(a, y)) → F(f(f(a, a), y), x)
F(x, f(a, y)) → F(a, f(f(f(a, a), y), x))
F(x, f(a, y)) → F(f(a, a), y)

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, a), y), x))

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

(5) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(x, f(a, y)) → F(a, f(f(f(a, a), y), x)) at position [1] we obtained the following new rules [LPAR04]:

F(f(a, x1), f(a, y1)) → F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1))))
F(y0, f(a, f(a, x1))) → F(a, f(f(a, f(f(f(a, a), x1), f(a, a))), y0))

(6) Obligation:

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

F(x, f(a, y)) → F(f(f(a, a), y), x)
F(x, f(a, y)) → F(f(a, a), y)
F(f(a, x1), f(a, y1)) → F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1))))
F(y0, f(a, f(a, x1))) → F(a, f(f(a, f(f(f(a, a), x1), f(a, a))), y0))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, a), y), x))

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

(7) 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 = F(x, f(x', f(x'', f(a, y)))) evaluates to t =F(f(f(a, a), f(f(f(a, a), f(f(f(a, a), y), x'')), x')), x)

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



Rewriting sequence

F(f(f(a, y'''), f(x''', f(a, y'))), f(f(a, y''), f(x'', f(a, y)))) → F(f(f(a, y'''), f(x''', f(a, y'))), f(f(a, y''), f(a, f(f(f(a, a), y), x''))))
with rule f(x'''', f(a, y1)) → f(a, f(f(f(a, a), y1), x'''')) at position [1,1] and matcher [x'''' / x'', y1 / y]

F(f(f(a, y'''), f(x''', f(a, y'))), f(f(a, y''), f(a, f(f(f(a, a), y), x'')))) → F(f(f(a, y'''), f(x''', f(a, y'))), f(a, f(f(f(a, a), f(f(f(a, a), y), x'')), f(a, y''))))
with rule f(x', f(a, y'1)) → f(a, f(f(f(a, a), y'1), x')) at position [1] and matcher [x' / f(a, y''), y'1 / f(f(f(a, a), y), x'')]

F(f(f(a, y'''), f(x''', f(a, y'))), f(a, f(f(f(a, a), f(f(f(a, a), y), x'')), f(a, y'')))) → F(f(f(a, a), f(f(f(a, a), f(f(f(a, a), y), x'')), f(a, y''))), f(f(a, y'''), f(x''', f(a, y'))))
with rule F(x, f(a, y)) → F(f(f(a, a), y), 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.



(8) NO