YES

Problem:
 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)

Proof:
 DP Processor:
  DPs:
   a__c#() -> a__f#(g(c()))
   mark#(c()) -> a__c#()
   mark#(f(X)) -> a__f#(X)
  TRS:
   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)
  TDG Processor:
   DPs:
    a__c#() -> a__f#(g(c()))
    mark#(c()) -> a__c#()
    mark#(f(X)) -> a__f#(X)
   TRS:
    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)
   graph:
    mark#(c()) -> a__c#() -> a__c#() -> a__f#(g(c()))
   SCC Processor:
    #sccs: 0
    #rules: 0
    #arcs: 1/9