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(X, g(X)) → f(1, g(X))
g(1) → g(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(X, g(X)) → F(1, g(X))
G(1) → G(0)

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(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 1 SCC with 1 less node.

(4) Obligation:

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

F(X, g(X)) → F(1, g(X))

The TRS R consists of the following rules:

f(X, g(X)) → f(1, g(X))
g(1) → g(0)

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

(5) 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.

(6) Obligation:

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

F(X, g(X)) → F(1, g(X))

The TRS R consists of the following rules:

g(1) → g(0)

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

(7) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(X, g(X)) → F(1, g(X)) we obtained the following new rules [LPAR04]:

F(1, g(1)) → F(1, g(1))

(8) Obligation:

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

F(1, g(1)) → F(1, g(1))

The TRS R consists of the following rules:

g(1) → g(0)

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

(9) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

g(1) → g(0)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(1) = 1   
POL(F(x₁, x₂)) = x₁ + x₂   
POL(g(x₁)) = x₁   

(10) Obligation:

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

F(1, g(1)) → F(1, g(1))

R is empty.
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 semiunifying a rule from P directly.

s = F(1, g(1)) evaluates to t =F(1, g(1))

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

• Semiunifier: [ ]
• Matcher: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(1, g(1)) to F(1, g(1)).

(12) FALSE