YES

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

Proof:
 Uncurry Processor:
  a1(b1(x)) -> b1(a1(x))
  b1(c1(x)) -> c1(b1(x))
  c1(a1(x)) -> a1(c1(x))
  f(a(),x1) -> a1(x1)
  f(b(),x1) -> b1(x1)
  f(c(),x1) -> c1(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
               [1 0 1]     [1]
    [c1](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [0],
    
               [1 0 0]     [0]
    [b1](x0) = [0 1 0]x0 + [1]
               [0 0 0]     [0],
    
               [1 1 0]  
    [a1](x0) = [0 1 0]x0
               [0 0 1]  ,
    
          [0]
    [c] = [0]
          [1],
    
                  [1 0 1]     [1 1 1]  
    [f](x0, x1) = [0 1 1]x0 + [1 1 1]x1
                  [0 0 0]     [0 0 1]  ,
    
          [0]
    [b] = [1]
          [0],
    
          [0]
    [a] = [0]
          [0]
   orientation:
                [1 1 0]    [1]    [1 1 0]    [0]            
    a1(b1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = b1(a1(x))
                [0 0 0]    [0]    [0 0 0]    [0]            
    
                [1 0 1]    [1]    [1 0 0]    [1]            
    b1(c1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [0] = c1(b1(x))
                [0 0 0]    [0]    [0 0 0]    [0]            
    
                [1 1 1]    [1]    [1 0 1]    [1]            
    c1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(c1(x))
                [0 0 0]    [0]    [0 0 0]    [0]            
    
                [1 1 1]      [1 1 0]           
    f(a(),x1) = [1 1 1]x1 >= [0 1 0]x1 = a1(x1)
                [0 0 1]      [0 0 1]           
    
                [1 1 1]     [0]    [1 0 0]     [0]         
    f(b(),x1) = [1 1 1]x1 + [1] >= [0 1 0]x1 + [1] = b1(x1)
                [0 0 1]     [0]    [0 0 0]     [0]         
    
                [1 1 1]     [1]    [1 0 1]     [1]         
    f(c(),x1) = [1 1 1]x1 + [1] >= [0 0 0]x1 + [0] = c1(x1)
                [0 0 1]     [0]    [0 0 0]     [0]         
   problem:
    b1(c1(x)) -> c1(b1(x))
    c1(a1(x)) -> a1(c1(x))
    f(a(),x1) -> a1(x1)
    f(b(),x1) -> b1(x1)
    f(c(),x1) -> c1(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                [1 1 0]     [0]
     [c1](x0) = [0 1 0]x0 + [0]
                [0 0 1]     [1],
     
                [1 0 1]  
     [b1](x0) = [0 1 0]x0
                [0 0 1]  ,
     
                     [0]
     [a1](x0) = x0 + [1]
                     [0],
     
           [0]
     [c] = [0]
           [0],
     
                   [1 0 0]     [1 1 1]     [0]
     [f](x0, x1) = [1 0 0]x0 + [0 1 0]x1 + [0]
                   [0 0 0]     [0 1 1]     [1],
     
           [0]
     [b] = [0]
           [0],
     
           [1]
     [a] = [0]
           [0]
    orientation:
                 [1 1 1]    [1]    [1 1 1]    [0]            
     b1(c1(x)) = [0 1 0]x + [0] >= [0 1 0]x + [0] = c1(b1(x))
                 [0 0 1]    [1]    [0 0 1]    [1]            
     
                 [1 1 0]    [1]    [1 1 0]    [0]            
     c1(a1(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = a1(c1(x))
                 [0 0 1]    [1]    [0 0 1]    [1]            
     
                 [1 1 1]     [1]         [0]         
     f(a(),x1) = [0 1 0]x1 + [1] >= x1 + [1] = a1(x1)
                 [0 1 1]     [1]         [0]         
     
                 [1 1 1]     [0]    [1 0 1]           
     f(b(),x1) = [0 1 0]x1 + [0] >= [0 1 0]x1 = b1(x1)
                 [0 1 1]     [1]    [0 0 1]           
     
                 [1 1 1]     [0]    [1 1 0]     [0]         
     f(c(),x1) = [0 1 0]x1 + [0] >= [0 1 0]x1 + [0] = c1(x1)
                 [0 1 1]     [1]    [0 0 1]     [1]         
    problem:
     f(b(),x1) -> b1(x1)
     f(c(),x1) -> c1(x1)
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                 [1 0 0]  
      [c1](x0) = [0 0 0]x0
                 [0 0 0]  ,
      
                 [1 0 0]  
      [b1](x0) = [0 0 0]x0
                 [0 0 0]  ,
      
            [0]
      [c] = [0]
            [0],
      
                    [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]     [0],
      
            [0]
      [b] = [0]
            [0]
     orientation:
                  [1 0 0]     [1]    [1 0 0]           
      f(b(),x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 = b1(x1)
                  [0 0 0]     [0]    [0 0 0]           
      
                  [1 0 0]     [1]    [1 0 0]           
      f(c(),x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c1(x1)
                  [0 0 0]     [0]    [0 0 0]           
     problem:
      
     Qed