YES

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

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