YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   a ==> a0/0 a1/1 a2/2
   
  uncurry-rules:
   f(a1(x2),x3) -> a2(x2,x3)
   f(a0(),x2) -> a1(x2)
  eta-rules:
   f(f(a(),f(a(),x)),x1) -> f(f(a(),f(f(a(),x),f(a(),a()))),x1)
  problem:
   a1(a1(x)) -> a1(a2(x,a1(a0())))
   a2(a1(x),x1) -> a2(a2(x,a1(a0())),x1)
   f(a1(x2),x3) -> a2(x2,x3)
   f(a0(),x2) -> a1(x2)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 1 0]     [1 0 0]  
    [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [1 0 0]  ,
    
                  [1 0 0]     [1 1 1]     [1]
    [f](x0, x1) = [0 0 0]x0 + [0 1 0]x1 + [1]
                  [0 0 0]     [1 0 0]     [0],
    
               [1 1 0]     [0]
    [a1](x0) = [0 0 0]x0 + [1]
               [1 0 0]     [0],
    
           [0]
    [a0] = [0]
           [0]
   orientation:
                [1 1 0]    [1]    [1 1 0]    [0]                     
    a1(a1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a1(a2(x,a1(a0())))
                [1 1 0]    [0]    [1 1 0]    [0]                     
    
                   [1 1 0]    [1 0 0]     [1]    [1 1 0]    [1 0 0]                          
    a2(a1(x),x1) = [0 0 0]x + [0 0 0]x1 + [0] >= [0 0 0]x + [0 0 0]x1 = a2(a2(x,a1(a0())),x1)
                   [0 0 0]    [1 0 0]     [0]    [0 0 0]    [1 0 0]                          
    
                   [1 1 0]     [1 1 1]     [1]    [1 1 0]     [1 0 0]              
    f(a1(x2),x3) = [0 0 0]x2 + [0 1 0]x3 + [1] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3)
                   [0 0 0]     [1 0 0]     [0]    [0 0 0]     [1 0 0]              
    
                 [1 1 1]     [1]    [1 1 0]     [0]         
    f(a0(),x2) = [0 1 0]x2 + [1] >= [0 0 0]x2 + [1] = a1(x2)
                 [1 0 0]     [0]    [1 0 0]     [0]         
   problem:
    
   Qed