YES

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

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