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)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [mark#](x0) = 3x0 + 6,
    
    [a__f#](x0) = x0 + 7,
    
    [mark](x0) = 3x0 + 7,
    
    [c](x0) = x0 + 1,
    
    [g](x0) = x0 + 3,
    
    [a__f](x0) = x0 + 7,
    
    [f](x0) = x0 + 3,
    
    [a] = 6
   orientation:
    mark#(f(X)) = 3X + 15 >= 3X + 6 = mark#(X)
    
    mark#(f(X)) = 3X + 15 >= 3X + 14 = a__f#(mark(X))
    
    mark#(g(X)) = 3X + 15 >= 3X + 6 = mark#(X)
    
    a__f(f(a())) = 16 >= 16 = c(f(g(f(a()))))
    
    mark(f(X)) = 3X + 16 >= 3X + 14 = a__f(mark(X))
    
    mark(a()) = 25 >= 6 = a()
    
    mark(c(X)) = 3X + 10 >= X + 1 = c(X)
    
    mark(g(X)) = 3X + 16 >= 3X + 10 = g(mark(X))
    
    a__f(X) = X + 7 >= X + 3 = 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