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:

ap(ap(g, x), y) → y
ap(f, x) → ap(f, app(g, x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(ap(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(app(x₁, x₂)) = x₁ + x₂   
POL(f) = 0   
POL(g) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

ap(ap(g, x), y) → y

(2) Obligation:

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

ap(f, x) → ap(f, app(g, x))

Q is empty.

(3) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

ap(f, x) → ap(f, app(g, x))

The signature Sigma is {ap}

(4) Obligation:

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

ap(f, x) → ap(f, app(g, x))

The set Q consists of the following terms:

ap(f, x0)

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

AP(f, x) → AP(f, app(g, x))

The TRS R consists of the following rules:

ap(f, x) → ap(f, app(g, x))

The set Q consists of the following terms:

ap(f, x0)

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

(7) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(8) Obligation:

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

AP(f, x) → AP(f, app(g, x))

R is empty.
The set Q consists of the following terms:

ap(f, x0)

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

(9) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

ap(f, x0)

(10) Obligation:

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

AP(f, x) → AP(f, app(g, x))

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

(11) ATransformationProof (EQUIVALENT transformation)

We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.

(12) Obligation:

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

f(x) → f(g(x))

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

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule f(x) → f(g(x)) we obtained the following new rules [LPAR04]:

f(g(z0)) → f(g(g(z0)))

(14) Obligation:

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

f(g(z0)) → f(g(g(z0)))

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

(15) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule f(g(z0)) → f(g(g(z0))) we obtained the following new rules [LPAR04]:

f(g(g(z0))) → f(g(g(g(z0))))

(16) Obligation:

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

f(g(g(z0))) → f(g(g(g(z0))))

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

(17) 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(g(g(z0))) evaluates to t =f(g(g(g(z0))))

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

• Matcher: [z0 / g(z0)]
• Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from f(g(g(z0))) to f(g(g(g(z0)))).

(18) NO