YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   a ==> a0/0 a1/1 a2/2 a3/3
   
  uncurry-rules:
   f(a2(x2,x3),x4) -> a3(x2,x3,x4)
   f(a1(x2),x3) -> a2(x2,x3)
   f(a0(),x2) -> a1(x2)
  eta-rules:
   f(f(f(a(),x),a()),x1) -> f(f(f(f(a(),a()),f(x,a())),a()),x1)
  problem:
   a2(x,a0()) -> a3(a0(),f(x,a0()),a0())
   a3(x,a0(),x1) -> f(a3(a0(),f(x,a0()),a0()),x1)
   f(a2(x2,x3),x4) -> a3(x2,x3,x4)
   f(a1(x2),x3) -> a2(x2,x3)
   f(a0(),x2) -> a1(x2)
  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 0]     [0 0 1]  ,
    
                  [1 0 1]     [1 0 1]  
    [f](x0, x1) = [1 0 0]x0 + [0 0 0]x1
                  [0 1 0]     [0 0 1]  ,
    
               [1 0 1]     [0]
    [a1](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
    
                       [1 0 1]     [1 0 0]     [1 0 1]  
    [a3](x0, x1, x2) = [1 0 1]x0 + [0 0 0]x1 + [0 0 0]x2
                       [0 0 0]     [0 0 0]     [0 0 1]  ,
    
           [0]
    [a0] = [1]
           [0]
   orientation:
                 [1 0 1]     [1 0 1]                           
    a2(x,a0()) = [0 0 0]x >= [0 0 0]x = a3(a0(),f(x,a0()),a0())
                 [0 0 0]     [0 0 0]                           
    
                    [1 0 1]    [1 0 1]      [1 0 1]    [1 0 1]                                  
    a3(x,a0(),x1) = [1 0 1]x + [0 0 0]x1 >= [1 0 1]x + [0 0 0]x1 = f(a3(a0(),f(x,a0()),a0()),x1)
                    [0 0 0]    [0 0 1]      [0 0 0]    [0 0 1]                                  
    
                      [1 0 1]     [1 0 1]     [1 0 1]      [1 0 1]     [1 0 0]     [1 0 1]                 
    f(a2(x2,x3),x4) = [1 0 1]x2 + [1 0 0]x3 + [0 0 0]x4 >= [1 0 1]x2 + [0 0 0]x3 + [0 0 0]x4 = a3(x2,x3,x4)
                      [0 0 0]     [0 0 0]     [0 0 1]      [0 0 0]     [0 0 0]     [0 0 1]                 
    
                   [1 0 1]     [1 0 1]     [1]    [1 0 1]     [1 0 0]              
    f(a1(x2),x3) = [1 0 1]x2 + [0 0 0]x3 + [0] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3)
                   [0 0 0]     [0 0 1]     [0]    [0 0 0]     [0 0 1]              
    
                 [1 0 1]     [0]    [1 0 1]     [0]         
    f(a0(),x2) = [0 0 0]x2 + [0] >= [0 0 0]x2 + [0] = a1(x2)
                 [0 0 1]     [1]    [0 0 0]     [1]         
   problem:
    a2(x,a0()) -> a3(a0(),f(x,a0()),a0())
    a3(x,a0(),x1) -> f(a3(a0(),f(x,a0()),a0()),x1)
    f(a2(x2,x3),x4) -> a3(x2,x3,x4)
    f(a0(),x2) -> a1(x2)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1 1 0]     [1 1 1]     [1]
     [a2](x0, x1) = [1 0 0]x0 + [0 0 0]x1 + [1]
                    [0 0 0]     [0 0 0]     [1],
     
                   [1 0 0]     [1 0 0]  
     [f](x0, x1) = [1 0 0]x0 + [0 0 0]x1
                   [0 1 0]     [0 0 0]  ,
     
                [1 0 0]  
     [a1](x0) = [0 0 0]x0
                [0 0 0]  ,
     
                        [1 1 0]     [1 0 1]     [1 0 0]     [0]
     [a3](x0, x1, x2) = [1 1 0]x0 + [0 1 0]x1 + [0 0 0]x2 + [1]
                        [1 0 0]     [0 0 0]     [0 0 0]     [1],
     
            [0]
     [a0] = [0]
            [0]
    orientation:
                  [1 1 0]    [1]    [1 1 0]    [0]                          
     a2(x,a0()) = [1 0 0]x + [1] >= [1 0 0]x + [1] = a3(a0(),f(x,a0()),a0())
                  [0 0 0]    [1]    [0 0 0]    [1]                          
     
                     [1 1 0]    [1 0 0]     [0]    [1 1 0]    [1 0 0]     [0]                                
     a3(x,a0(),x1) = [1 1 0]x + [0 0 0]x1 + [1] >= [1 1 0]x + [0 0 0]x1 + [0] = f(a3(a0(),f(x,a0()),a0()),x1)
                     [1 0 0]    [0 0 0]     [1]    [1 0 0]    [0 0 0]     [1]                                
     
                       [1 1 0]     [1 1 1]     [1 0 0]     [1]    [1 1 0]     [1 0 1]     [1 0 0]     [0]               
     f(a2(x2,x3),x4) = [1 1 0]x2 + [1 1 1]x3 + [0 0 0]x4 + [1] >= [1 1 0]x2 + [0 1 0]x3 + [0 0 0]x4 + [1] = a3(x2,x3,x4)
                       [1 0 0]     [0 0 0]     [0 0 0]     [1]    [1 0 0]     [0 0 0]     [0 0 0]     [1]               
     
                  [1 0 0]      [1 0 0]           
     f(a0(),x2) = [0 0 0]x2 >= [0 0 0]x2 = a1(x2)
                  [0 0 0]      [0 0 0]           
    problem:
     a3(x,a0(),x1) -> f(a3(a0(),f(x,a0()),a0()),x1)
     f(a0(),x2) -> a1(x2)
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                    [1 0 0]     [1 0 0]  
      [f](x0, x1) = [0 0 0]x0 + [0 0 1]x1
                    [0 0 0]     [0 0 0]  ,
      
                 [1 0 0]  
      [a1](x0) = [0 0 0]x0
                 [0 0 0]  ,
      
                         [1 1 0]     [1 0 1]     [1 0 0]  
      [a3](x0, x1, x2) = [0 0 0]x0 + [0 0 0]x1 + [1 0 1]x2
                         [0 0 0]     [0 0 0]     [0 0 0]  ,
      
             [0]
      [a0] = [0]
             [1]
     orientation:
                      [1 1 0]    [1 0 0]     [1]    [1 0 0]    [1 0 0]                                  
      a3(x,a0(),x1) = [0 0 0]x + [1 0 1]x1 + [0] >= [0 0 0]x + [0 0 1]x1 = f(a3(a0(),f(x,a0()),a0()),x1)
                      [0 0 0]    [0 0 0]     [0]    [0 0 0]    [0 0 0]                                  
      
                   [1 0 0]      [1 0 0]           
      f(a0(),x2) = [0 0 1]x2 >= [0 0 0]x2 = a1(x2)
                   [0 0 0]      [0 0 0]           
     problem:
      f(a0(),x2) -> a1(x2)
     Matrix Interpretation Processor: dim=3
      
      interpretation:
                     [1 0 0]     [1 0 0]     [1]
       [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                     [0 0 0]     [0 0 0]     [1],
       
                  [1 0 0]     [0]
       [a1](x0) = [0 0 0]x0 + [0]
                  [0 0 0]     [1],
       
              [0]
       [a0] = [0]
              [0]
      orientation:
                    [1 0 0]     [1]    [1 0 0]     [0]         
       f(a0(),x2) = [0 0 0]x2 + [0] >= [0 0 0]x2 + [0] = a1(x2)
                    [0 0 0]     [1]    [0 0 0]     [1]         
      problem:
       
      Qed