YES

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

Proof:
 Uncurry Processor (mirror):
  a2(a1(x),x4) -> a2(a2(a(),a1(x)),x4)
  a1(a1(x)) -> a1(a2(a(),a1(x)))
  f(a1(x2),x3) -> a2(x2,x3)
  f(a(),x3) -> a1(x3)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 1]     [1 0 0]  
    [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
    
               [1 0 1]     [0]
    [a1](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
    
                  [1 0 0]     [1 0 1]     [0]
    [f](x0, x1) = [0 0 0]x0 + [1 0 0]x1 + [0]
                  [0 0 0]     [0 0 0]     [1],
    
          [0]
    [a] = [0]
          [0]
   orientation:
                   [1 0 1]    [1 0 0]     [1]    [1 0 1]    [1 0 0]                         
    a2(a1(x),x4) = [0 0 0]x + [0 0 0]x4 + [0] >= [0 0 0]x + [0 0 0]x4 = a2(a2(a(),a1(x)),x4)
                   [0 0 0]    [0 0 0]     [0]    [0 0 0]    [0 0 0]                         
    
                [1 0 1]    [1]    [1 0 1]    [0]                    
    a1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(a2(a(),a1(x)))
                [0 0 0]    [1]    [0 0 0]    [1]                    
    
                   [1 0 1]     [1 0 1]     [0]    [1 0 1]     [1 0 0]              
    f(a1(x2),x3) = [0 0 0]x2 + [1 0 0]x3 + [0] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3)
                   [0 0 0]     [0 0 0]     [1]    [0 0 0]     [0 0 0]              
    
                [1 0 1]     [0]    [1 0 1]     [0]         
    f(a(),x3) = [1 0 0]x3 + [0] >= [0 0 0]x3 + [0] = a1(x3)
                [0 0 0]     [1]    [0 0 0]     [1]         
   problem:
    f(a1(x2),x3) -> a2(x2,x3)
    f(a(),x3) -> a1(x3)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1 0 0]     [1 0 0]  
     [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                    [0 0 0]     [0 0 0]  ,
     
                [1 0 0]  
     [a1](x0) = [0 0 0]x0
                [0 0 0]  ,
     
                   [1 0 0]     [1 0 0]     [1]
     [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                   [0 0 0]     [0 0 0]     [0],
     
           [0]
     [a] = [0]
           [0]
    orientation:
                    [1 0 0]     [1 0 0]     [1]    [1 0 0]     [1 0 0]              
     f(a1(x2),x3) = [0 0 0]x2 + [0 0 0]x3 + [0] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3)
                    [0 0 0]     [0 0 0]     [0]    [0 0 0]     [0 0 0]              
     
                 [1 0 0]     [1]    [1 0 0]           
     f(a(),x3) = [0 0 0]x3 + [0] >= [0 0 0]x3 = a1(x3)
                 [0 0 0]     [0]    [0 0 0]           
    problem:
     
    Qed