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 → f
g(b) → c
b → c

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(b) = 1   
POL(c) = 0   
POL(f) = 0   
POL(g(x₁)) = x₁   

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

g(b) → c
b → c

(2) Obligation:

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

f → f

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

f → f

The signature Sigma is {f}

(4) Obligation:

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

f → f

The set Q consists of the following terms:

f

(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:

F → F

The TRS R consists of the following rules:

f → f

The set Q consists of the following terms:

f

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:

F → F

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

f

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].

f

(10) Obligation:

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

F → F

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 evaluates to t =F

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F to F.

(12) NO