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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(x, f(a, y)) → F(a, a)
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)

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(x, f(a, y)) → F(a, a)
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)

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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing

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)

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.
By narrowing [15] 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:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ NonTerminationProof

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

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))
F(x, f(a, y)) → 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.
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:

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))
F(x, f(a, y)) → 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))


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:



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, y0)) → f(a, f(f(f(a, a), y0), x'''')) at position [1,1] and matcher [y0 / y, x'''' / x'']

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'0)) → f(a, f(f(f(a, a), y'0), x')) at position [1] and matcher [y'0 / f(f(f(a, a), y), x''), x' / f(a, y'')]

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.