YES

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

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(f(a(),a()),x)),x1) -> f(f(f(a(),a()),f(a(),f(a(),x))),x1)
  problem:
   a1(a2(a0(),x)) -> a2(a0(),a1(a1(x)))
   a2(a2(a0(),x),x1) -> f(a2(a0(),a1(a1(x))),x1)
   f(a1(x2),x3) -> a2(x2,x3)
   f(a0(),x2) -> a1(x2)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 0]     [1 0 0]  
    [a2](x0, x1) = [0 0 0]x0 + [0 0 1]x1
                   [1 0 0]     [0 0 1]  ,
    
                  [1 0 0]     [1 0 0]  
    [f](x0, x1) = [0 0 0]x0 + [0 0 1]x1
                  [1 0 0]     [0 0 1]  ,
    
               [1 0 0]  
    [a1](x0) = [0 0 1]x0
               [0 0 1]  ,
    
           [1]
    [a0] = [0]
           [0]
   orientation:
                     [1 0 0]    [1]    [1 0 0]    [1]                     
    a1(a2(a0(),x)) = [0 0 1]x + [1] >= [0 0 1]x + [0] = a2(a0(),a1(a1(x)))
                     [0 0 1]    [1]    [0 0 1]    [1]                     
    
                        [1 0 0]    [1 0 0]     [1]    [1 0 0]    [1 0 0]     [1]                           
    a2(a2(a0(),x),x1) = [0 0 0]x + [0 0 1]x1 + [0] >= [0 0 0]x + [0 0 1]x1 + [0] = f(a2(a0(),a1(a1(x))),x1)
                        [1 0 0]    [0 0 1]     [1]    [1 0 0]    [0 0 1]     [1]                           
    
                   [1 0 0]     [1 0 0]      [1 0 0]     [1 0 0]              
    f(a1(x2),x3) = [0 0 0]x2 + [0 0 1]x3 >= [0 0 0]x2 + [0 0 1]x3 = a2(x2,x3)
                   [1 0 0]     [0 0 1]      [1 0 0]     [0 0 1]              
    
                 [1 0 0]     [1]    [1 0 0]           
    f(a0(),x2) = [0 0 1]x2 + [0] >= [0 0 1]x2 = a1(x2)
                 [0 0 1]     [1]    [0 0 1]           
   problem:
    a1(a2(a0(),x)) -> a2(a0(),a1(a1(x)))
    a2(a2(a0(),x),x1) -> f(a2(a0(),a1(a1(x))),x1)
    f(a1(x2),x3) -> a2(x2,x3)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1 1 0]     [1 0 0]     [0]
     [a2](x0, x1) = [0 0 0]x0 + [0 1 1]x1 + [1]
                    [0 0 0]     [0 1 1]     [1],
     
                   [1 0 0]     [1 0 0]     [1]
     [f](x0, x1) = [0 0 0]x0 + [0 1 1]x1 + [1]
                   [0 0 0]     [0 1 1]     [1],
     
                [1 1 0]  
     [a1](x0) = [0 0 1]x0
                [0 1 0]  ,
     
            [0]
     [a0] = [0]
            [0]
    orientation:
                      [1 1 1]    [1]    [1 1 1]    [0]                     
     a1(a2(a0(),x)) = [0 1 1]x + [1] >= [0 1 1]x + [1] = a2(a0(),a1(a1(x)))
                      [0 1 1]    [1]    [0 1 1]    [1]                     
     
                         [1 1 1]    [1 0 0]     [1]    [1 1 1]    [1 0 0]     [1]                           
     a2(a2(a0(),x),x1) = [0 0 0]x + [0 1 1]x1 + [1] >= [0 0 0]x + [0 1 1]x1 + [1] = f(a2(a0(),a1(a1(x))),x1)
                         [0 0 0]    [0 1 1]     [1]    [0 0 0]    [0 1 1]     [1]                           
     
                    [1 1 0]     [1 0 0]     [1]    [1 1 0]     [1 0 0]     [0]            
     f(a1(x2),x3) = [0 0 0]x2 + [0 1 1]x3 + [1] >= [0 0 0]x2 + [0 1 1]x3 + [1] = a2(x2,x3)
                    [0 0 0]     [0 1 1]     [1]    [0 0 0]     [0 1 1]     [1]            
    problem:
     a2(a2(a0(),x),x1) -> f(a2(a0(),a1(a1(x))),x1)
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                     [1 1 0]     [1 0 0]     [0]
      [a2](x0, x1) = [1 0 0]x0 + [0 0 0]x1 + [1]
                     [0 0 0]     [0 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 1]x0 + [0]
                 [0 0 0]     [1],
      
             [0]
      [a0] = [0]
             [0]
     orientation:
                          [1 0 0]    [1 0 0]     [1]    [1 0 0]    [1 0 0]                             
      a2(a2(a0(),x),x1) = [1 0 0]x + [0 0 0]x1 + [1] >= [0 0 0]x + [0 0 0]x1 = f(a2(a0(),a1(a1(x))),x1)
                          [0 0 0]    [0 0 0]     [0]    [0 0 0]    [0 0 0]                             
     problem:
      
     Qed