YES

We show the termination of the TRS R:

  a__c() -> a__f(g(c()))
  a__f(g(X)) -> g(X)
  mark(c()) -> a__c()
  mark(f(X)) -> a__f(X)
  mark(g(X)) -> g(X)
  a__c() -> c()
  a__f(X) -> f(X)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__c#() -> a__f#(g(c()))
p2: mark#(c()) -> a__c#()
p3: mark#(f(X)) -> a__f#(X)

and R consists of:

r1: a__c() -> a__f(g(c()))
r2: a__f(g(X)) -> g(X)
r3: mark(c()) -> a__c()
r4: mark(f(X)) -> a__f(X)
r5: mark(g(X)) -> g(X)
r6: a__c() -> c()
r7: a__f(X) -> f(X)

The estimated dependency graph contains the following SCCs:

  (no SCCs)