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:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [mark](x0) = 2x0,
   
   [f](x0, x1, x2) = 2x0 + 2x1 + x2 + 1,
   
   [b] = 0,
   
   [a__b] = 0,
   
   [a__f](x0, x1, x2) = 4x0 + 2x1 + 2x2 + 2,
   
   [a] = 0
  orientation:
   a__f(a(),X,X) = 4X + 2 >= 4X + 2 = a__f(X,a__b(),b())
   
   a__b() = 0 >= 0 = a()
   
   mark(f(X1,X2,X3)) = 4X1 + 4X2 + 2X3 + 2 >= 4X1 + 4X2 + 2X3 + 2 = a__f(X1,mark(X2),X3)
   
   mark(b()) = 0 >= 0 = a__b()
   
   mark(a()) = 0 >= 0 = a()
   
   a__f(X1,X2,X3) = 4X1 + 2X2 + 2X3 + 2 >= 2X1 + 2X2 + X3 + 1 = f(X1,X2,X3)
   
   a__b() = 0 >= 0 = b()
  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__b() -> b()
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [mark](x0) = x0 + 3,
    
    [f](x0, x1, x2) = 3x0 + 4x1 + 4x2 + 6,
    
    [b] = 0,
    
    [a__b] = 0,
    
    [a__f](x0, x1, x2) = x0 + x1 + x2,
    
    [a] = 0
   orientation:
    a__f(a(),X,X) = 2X >= X = a__f(X,a__b(),b())
    
    a__b() = 0 >= 0 = a()
    
    mark(f(X1,X2,X3)) = 3X1 + 4X2 + 4X3 + 9 >= X1 + X2 + X3 + 3 = a__f(X1,mark(X2),X3)
    
    mark(b()) = 3 >= 0 = a__b()
    
    mark(a()) = 3 >= 0 = a()
    
    a__b() = 0 >= 0 = b()
   problem:
    a__f(a(),X,X) -> a__f(X,a__b(),b())
    a__b() -> a()
    a__b() -> b()
   Matrix Interpretation Processor: dim=3
    
    interpretation:
           [0]
     [b] = [0]
           [0],
     
              [1]
     [a__b] = [0]
              [1],
     
                          [1 0 1]     [1 0 0]     [1 0 1]  
     [a__f](x0, x1, x2) = [1 0 0]x0 + [0 0 0]x1 + [1 0 0]x2
                          [0 0 0]     [0 0 0]     [0 0 0]  ,
     
           [0]
     [a] = [0]
           [1]
    orientation:
                     [2 0 1]    [1]    [1 0 1]    [1]                     
     a__f(a(),X,X) = [1 0 0]X + [0] >= [1 0 0]X + [0] = a__f(X,a__b(),b())
                     [0 0 0]    [0]    [0 0 0]    [0]                     
     
              [1]    [0]      
     a__b() = [0] >= [0] = a()
              [1]    [1]      
     
              [1]    [0]      
     a__b() = [0] >= [0] = b()
              [1]    [0]      
    problem:
     a__f(a(),X,X) -> a__f(X,a__b(),b())
    Matrix Interpretation Processor: dim=3
     
     interpretation:
            [0]
      [b] = [0]
            [0],
      
               [0]
      [a__b] = [0]
               [0],
      
                           [1 0 0]     [1 0 0]     [1 0 0]  
      [a__f](x0, x1, x2) = [0 0 0]x0 + [0 0 0]x1 + [0 0 0]x2
                           [0 0 0]     [0 0 0]     [0 0 0]  ,
      
            [1]
      [a] = [0]
            [0]
     orientation:
                      [2 0 0]    [1]    [1 0 0]                      
      a__f(a(),X,X) = [0 0 0]X + [0] >= [0 0 0]X = a__f(X,a__b(),b())
                      [0 0 0]    [0]    [0 0 0]                      
     problem:
      
     Qed