YES

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

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