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:

c → f(g(c))
f(g(X)) → g(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:

c'(x) → c'(g(f(x)))
g(f(x)) → g(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:

c'(x) → c'(g(f(x)))
g(f(x)) → g(x)

Q is empty.

(5) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(6) Obligation:

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

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

c
f(g(x0))

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

C → F(g(c))
C → C

The TRS R consists of the following rules:

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

c
f(g(x0))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

C → C

The TRS R consists of the following rules:

c → f(g(c))
f(g(X)) → g(X)

The set Q consists of the following terms:

c
f(g(x0))

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

(11) 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.

(12) Obligation:

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

C → C

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

c
f(g(x0))

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

(13) 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
f(g(x0))

(14) Obligation:

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

C → C

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

(15) 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.

(16) Obligation:

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

C → C

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

c
f(g(x0))

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

(17) 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
f(g(x0))

(18) Obligation:

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

C → C

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

(19) 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 = C evaluates to t =C

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

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

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

(20) FALSE