YES

Problem:
 a__f(a(),X,X) -> a__f(X,a__b(),b())
 a__b() -> a()
 mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
 mark(b()) -> a__b()
 mark(a()) -> a()
 a__f(X1,X2,X3) -> f(X1,X2,X3)
 a__b() -> b()

Proof:
 DP Processor:
  DPs:
   a__f#(a(),X,X) -> a__b#()
   a__f#(a(),X,X) -> a__f#(X,a__b(),b())
   mark#(f(X1,X2,X3)) -> mark#(X2)
   mark#(f(X1,X2,X3)) -> a__f#(X1,mark(X2),X3)
   mark#(b()) -> a__b#()
  TRS:
   a__f(a(),X,X) -> a__f(X,a__b(),b())
   a__b() -> a()
   mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
   mark(b()) -> a__b()
   mark(a()) -> a()
   a__f(X1,X2,X3) -> f(X1,X2,X3)
   a__b() -> b()
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    
   interpretation:
    [mark#](x0) = x0,
    
    [a__b#] = 0,
    
    [a__f#](x0, x1, x2) = x0 + 2x2,
    
    [mark](x0) = x0 + 0,
    
    [f](x0, x1, x2) = 4x0 + 2x1 + 4x2 + 0,
    
    [b] = 1,
    
    [a__b] = 0,
    
    [a__f](x0, x1, x2) = 2x0 + 3x2,
    
    [a] = 4
   orientation:
    a__f#(a(),X,X) = 2X + 4 >= 0 = a__b#()
    
    a__f#(a(),X,X) = 2X + 4 >= X + 3 = a__f#(X,a__b(),b())
    
    mark#(f(X1,X2,X3)) = 4X1 + 2X2 + 4X3 + 0 >= X2 = mark#(X2)
    
    mark#(f(X1,X2,X3)) = 4X1 + 2X2 + 4X3 + 0 >= X1 + 2X3 = a__f#(X1,mark(X2),X3)
    
    mark#(b()) = 1 >= 0 = a__b#()
    
    a__f(a(),X,X) = 3X + 6 >= 2X + 4 = a__f(X,a__b(),b())
    
    a__b() = 0 >= 4 = a()
    
    mark(f(X1,X2,X3)) = 4X1 + 2X2 + 4X3 + 0 >= 2X1 + 3X3 = a__f(X1,mark(X2),X3)
    
    mark(b()) = 1 >= 0 = a__b()
    
    mark(a()) = 4 >= 4 = a()
    
    a__f(X1,X2,X3) = 2X1 + 3X3 >= 4X1 + 2X2 + 4X3 + 0 = f(X1,X2,X3)
    
    a__b() = 0 >= 1 = b()
   problem:
    DPs:
     
    TRS:
     a__f(a(),X,X) -> a__f(X,a__b(),b())
     a__b() -> a()
     mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
     mark(b()) -> a__b()
     mark(a()) -> a()
     a__f(X1,X2,X3) -> f(X1,X2,X3)
     a__b() -> b()
   Qed