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

(0) Obligation:

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

f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0

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(g(x), s(0), y) → F(g(s(0)), y, g(x))
F(g(x), s(0), y) → G(s(0))
G(s(x)) → G(x)

The TRS R consists of the following rules:

f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0

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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

G(s(x)) → G(x)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• G(s(x)) → G(x)
  The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

F(g(x), s(0), y) → F(g(s(0)), y, g(x))

The TRS R consists of the following rules:

f(g(x), s(0), y) → f(g(s(0)), y, g(x))
g(s(x)) → s(g(x))
g(0) → 0

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

F(g(x), s(0), y) → F(g(s(0)), y, g(x))

The TRS R consists of the following rules:

g(s(x)) → s(g(x))
g(0) → 0

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

(13) 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(g(x), g(s(0)), y) evaluates to t =F(g(s(0)), y, g(x))

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

• Matcher: [ ]
• Semiunifier: [x / s(0), y / g(s(0))]

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Rewriting sequence

F(g(s(0)), g(s(0)), g(s(0))) → F(g(s(0)), s(g(0)), g(s(0)))
with rule g(s(x')) → s(g(x')) at position [1] and matcher [x' / 0]

F(g(s(0)), s(g(0)), g(s(0))) → F(g(s(0)), s(0), g(s(0)))
with rule g(0) → 0 at position [1,0] and matcher [ ]

F(g(s(0)), s(0), g(s(0))) → F(g(s(0)), g(s(0)), g(s(0)))
with rule F(g(x), s(0), y) → F(g(s(0)), y, g(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.

(14) NO