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) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.

Found the self-embedding DerivationStructure:

c' → c' g

c' → c' g
by OverlapClosure OC 3

c' → n__c' activate g
by OverlapClosure OC 2

c' → n__c' n__g f
by original rule (OC 1)

n__g f → activate g
by original rule (OC 1)

n__c' activate → c'
by original rule (OC 1)

(4) NO