YES

Problem:
 a__f(f(a())) -> a__f(g(f(a())))
 mark(f(X)) -> a__f(X)
 mark(a()) -> a()
 mark(g(X)) -> g(mark(X))
 a__f(X) -> f(X)

Proof:
 DP Processor:
  DPs:
   a__f#(f(a())) -> a__f#(g(f(a())))
   mark#(f(X)) -> a__f#(X)
   mark#(g(X)) -> mark#(X)
  TRS:
   a__f(f(a())) -> a__f(g(f(a())))
   mark(f(X)) -> a__f(X)
   mark(a()) -> a()
   mark(g(X)) -> g(mark(X))
   a__f(X) -> f(X)
  Usable Rule Processor:
   DPs:
    a__f#(f(a())) -> a__f#(g(f(a())))
    mark#(f(X)) -> a__f#(X)
    mark#(g(X)) -> mark#(X)
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [mark#](x0) = [1 0 1 1]x0 + [1],
     
     [a__f#](x0) = [0 0 0 1]x0,
     
               [1 1 1 0]     [1]
               [1 0 0 1]     [0]
     [g](x0) = [0 0 1 1]x0 + [0]
               [1 0 0 0]     [0],
     
               [1 0 0 0]     [0]
               [0 0 0 0]     [0]
     [f](x0) = [0 0 0 1]x0 + [0]
               [0 0 1 0]     [1],
     
           [1]
           [0]
     [a] = [1]
           [0]
    orientation:
     a__f#(f(a())) = 2 >= 1 = a__f#(g(f(a())))
     
     mark#(f(X)) = [1 0 1 1]X + [2] >= [0 0 0 1]X = a__f#(X)
     
     mark#(g(X)) = [2 1 2 1]X + [2] >= [1 0 1 1]X + [1] = mark#(X)
    problem:
     DPs:
      
     TRS:
      
    Qed