(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) 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

(7) YES

(8) 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.

(9) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [ ] on the rule
F(g(s(0)), s(0), s(0))[ ]n[x0 / s(0)] → F(g(s(0)), s(0), s(0))[ ]n[x0 / s(0)]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
F(g(s(0)), s(0), x1)[ ]n[ ] → F(g(s(0)), x1, s(0))[ ]n[ ]
    by Narrowing at position: [2,0]
        intermediate steps: Instantiation - Instantiation
        F(g(s(y0)), s(0), x0)[ ]n[ ] → F(g(s(0)), x0, s(g(y0)))[ ]n[ ]
            by Narrowing at position: [2]
                intermediate steps: Instantiation - Instantiation
                F(g(x), s(0), y)[ ]n[ ] → F(g(s(0)), y, g(x))[ ]n[ ]
                    by OriginalRule from TRS P

                intermediate steps: Instantiation
                g(s(x))[ ]n[ ] → s(g(x))[ ]n[ ]
                    by OriginalRule from TRS R

        g(0)[ ]n[ ] → 0[ ]n[ ]
            by OriginalRule from TRS R

(10) NO