YES

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

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