YES

Problem:
 f(f(x,a()),a()) -> f(f(f(a(),a()),f(x,a())),a())

Proof:
 Uncurry Processor (mirror):
  a2(a1(x),x6) -> a2(a2(x,a1(a())),x6)
  a1(a1(x)) -> a1(a2(x,a1(a())))
  f(a1(x4),x5) -> a2(x4,x5)
  f(a(),x5) -> a1(x5)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [a1](x0) = 4x0,
    
    [a2](x0, x1) = 4x0 + 4x1,
    
    [f](x0, x1) = 2x0 + 5x1 + 4,
    
    [a] = 0
   orientation:
    a2(a1(x),x6) = 16x + 4x6 >= 16x + 4x6 = a2(a2(x,a1(a())),x6)
    
    a1(a1(x)) = 16x >= 16x = a1(a2(x,a1(a())))
    
    f(a1(x4),x5) = 8x4 + 5x5 + 4 >= 4x4 + 4x5 = a2(x4,x5)
    
    f(a(),x5) = 5x5 + 4 >= 4x5 = a1(x5)
   problem:
    a2(a1(x),x6) -> a2(a2(x,a1(a())),x6)
    a1(a1(x)) -> a1(a2(x,a1(a())))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                [1 1 0]     [0]
     [a1](x0) = [0 0 0]x0 + [1]
                [0 0 0]     [0],
     
                    [1 1 0]     [1 0 0]  
     [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                    [0 0 0]     [0 0 0]  ,
     
           [0]
     [a] = [0]
           [0]
    orientation:
                    [1 1 0]    [1 0 0]     [1]    [1 1 0]    [1 0 0]                         
     a2(a1(x),x6) = [0 0 0]x + [0 0 0]x6 + [0] >= [0 0 0]x + [0 0 0]x6 = a2(a2(x,a1(a())),x6)
                    [0 0 0]    [0 0 0]     [0]    [0 0 0]    [0 0 0]                         
     
                 [1 1 0]    [1]    [1 1 0]    [0]                    
     a1(a1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a1(a2(x,a1(a())))
                 [0 0 0]    [0]    [0 0 0]    [0]                    
    problem:
     
    Qed