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(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
c → n__c
activate(n__g(X)) → g(X)
activate(n__c) → c
activate(X) → 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) → n__c'(n__g(f(x)))
n__g(f(x)) → activate(g(x))
g(x) → n__g(x)
c'(x) → n__c'(x)
n__g(activate(x)) → g(x)
n__c'(activate(x)) → c'(x)
activate(x) → 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) → n__c'(n__g(f(x)))
n__g(f(x)) → activate(g(x))
g(x) → n__g(x)
c'(x) → n__c'(x)
n__g(activate(x)) → g(x)
n__c'(activate(x)) → c'(x)
activate(x) → x

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x₁)) = 1 + x₁   
POL(c) = 1   
POL(f(x₁)) = 1 + x₁   
POL(g(x₁)) = x₁   
POL(n__c) = 0   
POL(n__g(x₁)) = x₁   

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

c → n__c
activate(n__g(X)) → g(X)
activate(X) → X

(6) Obligation:

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

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

Q is empty.

(7) Overlay + Local Confluence (EQUIVALENT transformation)

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

(8) Obligation:

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

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

c
f(n__g(x0))
g(x0)
activate(n__c)

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

C → F(n__g(n__c))
F(n__g(X)) → G(activate(X))
F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C

The TRS R consists of the following rules:

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

c
f(n__g(x0))
g(x0)
activate(n__c)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Obligation:

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

The TRS R consists of the following rules:

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

c
f(n__g(x0))
g(x0)
activate(n__c)

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

(13) 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.

(14) Obligation:

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

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

c
f(n__g(x0))
g(x0)
activate(n__c)

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

(15) 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(n__g(x0))
g(x0)
activate(n__c)

(16) Obligation:

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

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

(17) 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 = ACTIVATE(n__c) evaluates to t =ACTIVATE(n__c)

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

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

ACTIVATE(n__c) → C
with rule ACTIVATE(n__c) → C at position [] and matcher [ ]

C → F(n__g(n__c))
with rule C → F(n__g(n__c)) at position [] and matcher [ ]

F(n__g(n__c)) → ACTIVATE(n__c)
with rule F(n__g(X)) → ACTIVATE(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.

(18) FALSE

(19) 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.

(20) Obligation:

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

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

c
f(n__g(x0))
g(x0)
activate(n__c)

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

(21) 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(n__g(x0))
g(x0)
activate(n__c)

(22) Obligation:

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

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