YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   b ==> b0/0 b1/1
   a ==> a0/0 a1/1
   
  uncurry-rules:
   f(a0(),x1) -> a1(x1)
   f(b0(),x3) -> b1(x3)
  eta-rules:
   
  problem:
   a1(b1(a1(x))) -> a1(b1(b1(a1(x))))
   b1(b1(b1(x))) -> b1(b1(x))
   f(a0(),x1) -> a1(x1)
   f(b0(),x3) -> b1(x3)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
               [1 0 1]     [0]
    [a1](x0) = [0 1 0]x0 + [1]
               [1 0 0]     [0],
    
           [0]
    [a0] = [0]
           [0],
    
           [0]
    [b0] = [0]
           [0],
    
               [1 0 0]  
    [b1](x0) = [0 0 0]x0
               [0 1 0]  ,
    
                  [1 0 0]     [1 1 1]     [1]
    [f](x0, x1) = [0 0 0]x0 + [0 1 1]x1 + [1]
                  [0 0 0]     [1 1 0]     [1]
   orientation:
                    [1 1 1]    [1]    [1 0 1]    [0]                    
    a1(b1(a1(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a1(b1(b1(a1(x))))
                    [1 0 1]    [0]    [1 0 1]    [0]                    
    
                    [1 0 0]     [1 0 0]             
    b1(b1(b1(x))) = [0 0 0]x >= [0 0 0]x = b1(b1(x))
                    [0 0 0]     [0 0 0]             
    
                 [1 1 1]     [1]    [1 0 1]     [0]         
    f(a0(),x1) = [0 1 1]x1 + [1] >= [0 1 0]x1 + [1] = a1(x1)
                 [1 1 0]     [1]    [1 0 0]     [0]         
    
                 [1 1 1]     [1]    [1 0 0]           
    f(b0(),x3) = [0 1 1]x3 + [1] >= [0 0 0]x3 = b1(x3)
                 [1 1 0]     [1]    [0 1 0]           
   problem:
    b1(b1(b1(x))) -> b1(b1(x))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                [1 1 0]     [0]
     [b1](x0) = [0 0 1]x0 + [0]
                [0 0 0]     [1]
    orientation:
                     [1 1 1]    [1]    [1 1 1]    [0]            
     b1(b1(b1(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = b1(b1(x))
                     [0 0 0]    [1]    [0 0 0]    [1]            
    problem:
     
    Qed