YES

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

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(a0(),x) -> a1(f(x,a1(a1(a0()))))
   f(a1(x1),x2) -> a2(x1,x2)
   f(a0(),x1) -> a1(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 1]     [1 1 1]  
    [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
    
                  [1 1 1]     [1 1 1]  
    [f](x0, x1) = [0 0 0]x0 + [0 1 0]x1
                  [0 0 0]     [0 0 0]  ,
    
               [1 0 1]  
    [a1](x0) = [0 1 0]x0
               [0 0 0]  ,
    
           [0]
    [a0] = [0]
           [1]
   orientation:
                 [1 1 1]    [1]    [1 1 1]    [1]                        
    a2(a0(),x) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(f(x,a1(a1(a0()))))
                 [0 0 0]    [0]    [0 0 0]    [0]                        
    
                   [1 1 1]     [1 1 1]      [1 0 1]     [1 1 1]              
    f(a1(x1),x2) = [0 0 0]x1 + [0 1 0]x2 >= [0 0 0]x1 + [0 0 0]x2 = a2(x1,x2)
                   [0 0 0]     [0 0 0]      [0 0 0]     [0 0 0]              
    
                 [1 1 1]     [1]    [1 0 1]           
    f(a0(),x1) = [0 1 0]x1 + [0] >= [0 1 0]x1 = a1(x1)
                 [0 0 0]     [0]    [0 0 0]           
   problem:
    a2(a0(),x) -> a1(f(x,a1(a1(a0()))))
    f(a1(x1),x2) -> a2(x1,x2)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1 1 0]     [1 0 1]     [1]
     [a2](x0, x1) = [0 0 0]x0 + [0 1 0]x1 + [0]
                    [0 1 0]     [0 0 0]     [1],
     
                   [1 0 1]     [1 0 1]     [1]
     [f](x0, x1) = [0 0 0]x0 + [0 1 0]x1 + [1]
                   [0 0 1]     [0 0 0]     [0],
     
                [1 0 0]     [0]
     [a1](x0) = [0 0 0]x0 + [0]
                [0 1 0]     [1],
     
            [0]
     [a0] = [1]
            [0]
    orientation:
                  [1 0 1]    [2]    [1 0 1]    [2]                        
     a2(a0(),x) = [0 1 0]x + [0] >= [0 0 0]x + [0] = a1(f(x,a1(a1(a0()))))
                  [0 0 0]    [2]    [0 0 0]    [2]                        
     
                    [1 1 0]     [1 0 1]     [2]    [1 1 0]     [1 0 1]     [1]            
     f(a1(x1),x2) = [0 0 0]x1 + [0 1 0]x2 + [1] >= [0 0 0]x1 + [0 1 0]x2 + [0] = a2(x1,x2)
                    [0 1 0]     [0 0 0]     [1]    [0 1 0]     [0 0 0]     [1]            
    problem:
     a2(a0(),x) -> a1(f(x,a1(a1(a0()))))
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                     [1 0 1]     [1 0 0]  
      [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                     [0 0 1]     [0 0 0]  ,
      
                    [1 0 0]     [1 0 0]  
      [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                    [0 0 0]     [0 0 0]  ,
      
                 [1 0 0]     [0]
      [a1](x0) = [0 0 0]x0 + [0]
                 [0 0 0]     [1],
      
             [0]
      [a0] = [0]
             [1]
     orientation:
                   [1 0 0]    [1]    [1 0 0]    [0]                        
      a2(a0(),x) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(f(x,a1(a1(a0()))))
                   [0 0 0]    [1]    [0 0 0]    [1]                        
     problem:
      
     Qed