Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 UsableRulesProof (⇔)
↳8 QDP
↳9 NonTerminationProof (⇔)
↳10 NO

(0) Obligation:

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

f(g(a)) → a
f(f(x)) → b
g(x) → f(g(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a) = 0   
POL(b) = 0   
POL(f(x1)) = x1   
POL(g(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(g(a)) → a


(2) Obligation:

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

f(f(x)) → b
g(x) → f(g(x))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

G(x) → F(g(x))
G(x) → G(x)

The TRS R consists of the following rules:

f(f(x)) → b
g(x) → f(g(x))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

G(x) → G(x)

The TRS R consists of the following rules:

f(f(x)) → b
g(x) → f(g(x))

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

(7) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(8) Obligation:

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

G(x) → G(x)

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

(9) 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 = G(x) evaluates to t =G(x)

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 G(x) to G(x).



(10) NO