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) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) 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.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a) = 0   
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:

f(i(x)) → a

(6) Obligation:

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

f(h(x)) → f(i(x))
i(x) → h(x)

Q is empty.

(7) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

i(x) → h(x)

The TRS R 2 is

f(h(x)) → f(i(x))

The signature Sigma is {f}

(8) Obligation:

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

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

f(h(x0))
i(x0)

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

F(h(x)) → F(i(x))
F(h(x)) → I(x)

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

f(h(x0))
i(x0)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(12) Obligation:

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

f(h(x)) → f(i(x))
i(x) → h(x)

The set Q consists of the following terms:

f(h(x0))
i(x0)

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

(13) 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.

(14) Obligation:

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

f(h(x0))
i(x0)

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

(15) 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].

f(h(x0))

(16) Obligation:

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

i(x0)

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

(17) 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.

(18) Obligation:

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

f(h(x0))
i(x0)

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

(19) 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].

f(h(x0))

(20) Obligation:

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

The set Q consists of the following terms:

i(x0)

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

(21) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(22) Obligation:

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

F(h(x)) → F(i(x))

The TRS R consists of the following rules:

i(x) → h(x)

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

(23) 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(i(x')) evaluates to t =F(i(x'))

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

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

F(i(x')) → F(h(x'))
with rule i(x'') → h(x'') at position [0] and matcher [x'' / x']

F(h(x')) → F(i(x'))
with rule F(h(x)) → F(i(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.

(24) FALSE