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:

g(X) → h(X)
c → d
h(d) → g(c)

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:

g(x) → h(x)
c'(x) → d'(x)
d'(h(x)) → c'(g(x))

Q is empty.

(3) AAECC Innermost (EQUIVALENT transformation)

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

g(x) → h(x)

The TRS R 2 is

c'(x) → d'(x)
d'(h(x)) → c'(g(x))

The signature Sigma is {c', d'}

(4) Obligation:

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

g(x) → h(x)
c'(x) → d'(x)
d'(h(x)) → c'(g(x))

The set Q consists of the following terms:

g(x0)
c'(x0)
d'(h(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:

C'(x) → D'(x)
D'(h(x)) → C'(g(x))
D'(h(x)) → G(x)

The TRS R consists of the following rules:

g(x) → h(x)
c'(x) → d'(x)
d'(h(x)) → c'(g(x))

The set Q consists of the following terms:

g(x0)
c'(x0)
d'(h(x0))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

D'(h(x)) → C'(g(x))
C'(x) → D'(x)

The TRS R consists of the following rules:

g(x) → h(x)
c'(x) → d'(x)
d'(h(x)) → c'(g(x))

The set Q consists of the following terms:

g(x0)
c'(x0)
d'(h(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:

D'(h(x)) → C'(g(x))
C'(x) → D'(x)

The TRS R consists of the following rules:

g(x) → h(x)

The set Q consists of the following terms:

g(x0)
c'(x0)
d'(h(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].

c'(x0)
d'(h(x0))

(12) Obligation:

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

D'(h(x)) → C'(g(x))
C'(x) → D'(x)

The TRS R consists of the following rules:

g(x) → h(x)

The set Q consists of the following terms:

g(x0)

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

(13) MNOCProof (EQUIVALENT transformation)

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

(14) Obligation:

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

D'(h(x)) → C'(g(x))
C'(x) → D'(x)

The TRS R consists of the following rules:

g(x) → h(x)

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

(15) 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 = C'(g(x')) evaluates to t =C'(g(x'))

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

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

C'(h(x')) → D'(h(x'))
with rule C'(x'') → D'(x'') at position [] and matcher [x'' / h(x')]

D'(h(x')) → C'(g(x'))
with rule D'(h(x)) → C'(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.

(16) NO