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

(0) Obligation:

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

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

mark₁ > a_f1 > f₁
mark₁ > a_f1 > c₁
mark₁ > a_f1 > g₁

Status:

trivial

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

a__f(f(a)) → c(f(g(f(a))))
mark(f(X)) → a__f(mark(X))
mark(a) → a
mark(c(X)) → c(X)
mark(g(X)) → g(mark(X))
a__f(X) → f(X)

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE