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(h(x)) → f(i(x))
f(i(x)) → a
i(x) → h(x)

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

h(f(x)) → i(f(x))
i(f(x)) → a'(x)
i(x) → h(x)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a'(x₁)) = x₁   
POL(f(x₁)) = 1 + x₁   
POL(h(x₁)) = x₁   
POL(i(x₁)) = x₁   

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

i(f(x)) → a'(x)

(4) Obligation:

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

h(f(x)) → i(f(x))
i(x) → h(x)

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

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

The TRS R 2 is

h(f(x)) → i(f(x))
i(x) → h(x)

The signature Sigma is {h, i}

(6) Obligation:

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

h(f(x)) → i(f(x))
i(x) → h(x)

The set Q consists of the following terms:

h(f(x0))
i(x0)

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

H(f(x)) → I(f(x))
I(x) → H(x)

The TRS R consists of the following rules:

h(f(x)) → i(f(x))
i(x) → h(x)

The set Q consists of the following terms:

h(f(x0))
i(x0)

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

(9) 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.

(10) Obligation:

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

H(f(x)) → I(f(x))
I(x) → H(x)

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

h(f(x0))
i(x0)

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

(11) 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].

h(f(x0))
i(x0)

(12) Obligation:

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

H(f(x)) → I(f(x))
I(x) → H(x)

R is empty.
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 = I(f(x)) evaluates to t =I(f(x))

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

I(f(x)) → H(f(x))
with rule I(x') → H(x') at position [] and matcher [x' / f(x)]

H(f(x)) → I(f(x))
with rule H(f(x)) → I(f(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