YES

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

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