YES

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

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