YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   a ==> a0/0 a1/1 a2/2
   
  uncurry-rules:
   f(a1(x1),x2) -> a2(x1,x2)
   f(a0(),x1) -> a1(x1)
  eta-rules:
   
  problem:
   a2(x,a0()) -> a2(a1(a0()),x)
   f(a1(x1),x2) -> a2(x1,x2)
   f(a0(),x1) -> a1(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 1 0]     [1 1 0]  
    [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 1 1]     [0 1 1]  ,
    
                  [1 0 1]     [1 1 1]     [1]
    [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                  [0 0 1]     [0 1 1]     [0],
    
               [1 0 0]  
    [a1](x0) = [0 0 0]x0
               [0 1 1]  ,
    
           [0]
    [a0] = [1]
           [0]
   orientation:
                 [1 1 0]    [1]    [1 1 0]    [0]                 
    a2(x,a0()) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a2(a1(a0()),x)
                 [0 1 1]    [1]    [0 1 1]    [1]                 
    
                   [1 1 1]     [1 1 1]     [1]    [1 1 0]     [1 1 0]              
    f(a1(x1),x2) = [0 0 0]x1 + [0 0 0]x2 + [0] >= [0 0 0]x1 + [0 0 0]x2 = a2(x1,x2)
                   [0 1 1]     [0 1 1]     [0]    [0 1 1]     [0 1 1]              
    
                 [1 1 1]     [1]    [1 0 0]           
    f(a0(),x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 = a1(x1)
                 [0 1 1]     [0]    [0 1 1]           
   problem:
    
   Qed