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)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [mark#](x0) = 1x0 + 4,
    
    [a__f#](x0) = 0,
    
    [a__c#] = 1,
    
    [f](x0) = x0,
    
    [mark](x0) = 4x0 + 12,
    
    [a__f](x0) = 1x0 + 4,
    
    [g](x0) = -1x0 + 0,
    
    [c] = 3,
    
    [a__c] = 8
   orientation:
    a__c#() = 1 >= 0 = a__f#(g(c()))
    
    mark#(c()) = 4 >= 1 = a__c#()
    
    mark#(f(X)) = 1X + 4 >= 0 = a__f#(X)
    
    a__c() = 8 >= 4 = a__f(g(c()))
    
    a__f(g(X)) = X + 4 >= -1X + 0 = g(X)
    
    mark(c()) = 12 >= 8 = a__c()
    
    mark(f(X)) = 4X + 12 >= 1X + 4 = a__f(X)
    
    mark(g(X)) = 3X + 12 >= -1X + 0 = g(X)
    
    a__c() = 8 >= 3 = c()
    
    a__f(X) = 1X + 4 >= X = f(X)
   problem:
    DPs:
     
    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)
   Qed