YES

Problem:
 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)

Proof:
 DP Processor:
  DPs:
   mark#(f(X)) -> mark#(X)
   mark#(f(X)) -> a__f#(mark(X))
   mark#(g(X)) -> mark#(X)
  TRS:
   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)
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    
   interpretation:
    [mark#](x0) = x0,
    
    [a__f#](x0) = 0,
    
    [mark](x0) = x0 + 1,
    
    [c](x0) = x0 + 3,
    
    [g](x0) = 4x0,
    
    [a__f](x0) = x0 + 1,
    
    [f](x0) = 1x0 + 1,
    
    [a] = 0
   orientation:
    mark#(f(X)) = 1X + 1 >= X = mark#(X)
    
    mark#(f(X)) = 1X + 1 >= 0 = a__f#(mark(X))
    
    mark#(g(X)) = 4X >= X = mark#(X)
    
    a__f(f(a())) = 1 >= 6 = c(f(g(f(a()))))
    
    mark(f(X)) = 1X + 1 >= X + 1 = a__f(mark(X))
    
    mark(a()) = 1 >= 0 = a()
    
    mark(c(X)) = X + 3 >= X + 3 = c(X)
    
    mark(g(X)) = 4X + 1 >= 4X + 5 = g(mark(X))
    
    a__f(X) = X + 1 >= 1X + 1 = f(X)
   problem:
    DPs:
     
    TRS:
     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)
   Qed