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:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)
a → b

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a) = 1   
POL(b) = 0   
POL(f(x₁)) = 1 + x₁   
POL(g(x₁)) = x₁   
POL(h(x₁)) = x₁   

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

a → b

(2) Obligation:

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

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

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

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The signature Sigma is {h, g, f}

(4) Obligation:

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

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The set Q consists of the following terms:

h(x0)
g(a)
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:

H(X) → G(X)
G(a) → F(b)
F(X) → H(a)

The TRS R consists of the following rules:

h(X) → g(X)
g(a) → f(b)
f(X) → h(a)

The set Q consists of the following terms:

h(x0)
g(a)
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:

H(X) → G(X)
G(a) → F(b)
F(X) → H(a)

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

h(x0)
g(a)
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].

h(x0)
g(a)
f(x0)

(10) Obligation:

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

H(X) → G(X)
G(a) → F(b)
F(X) → H(a)

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 narrowing to the left:

s = G(a) evaluates to t =G(a)

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

G(a) → F(b)
with rule G(a) → F(b) at position [] and matcher [ ]

F(b) → H(a)
with rule F(X') → H(a) at position [] and matcher [X' / b]

H(a) → G(a)
with rule H(X) → G(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.

(12) NO