YES

Problem:
 a(b(x1)) -> d(x1)
 b(a(x1)) -> a(b(x1))
 d(c(x1)) -> f(a(b(b(c(x1)))))
 d(f(x1)) -> f(a(b(x1)))
 a(f(x1)) -> a(x1)

Proof:
 String Reversal Processor:
  b(a(x1)) -> d(x1)
  a(b(x1)) -> b(a(x1))
  c(d(x1)) -> c(b(b(a(f(x1)))))
  f(d(x1)) -> b(a(f(x1)))
  f(a(x1)) -> a(x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [0]
    [f](x0) = [1 1 0]x0 + [1]
              [0 0 1]     [0],
    
              [1 0 0]  
    [c](x0) = [0 0 0]x0
              [1 0 0]  ,
    
              [1 0 1]     [1]
    [d](x0) = [0 1 0]x0 + [0]
              [0 0 1]     [1],
    
              [1 0 1]     [1]
    [a](x0) = [0 1 0]x0 + [1]
              [0 0 1]     [0],
    
              [1 0 0]     [0]
    [b](x0) = [0 1 1]x0 + [0]
              [0 0 1]     [1]
   orientation:
               [1 0 1]     [1]    [1 0 1]     [1]        
    b(a(x1)) = [0 1 1]x1 + [1] >= [0 1 0]x1 + [0] = d(x1)
               [0 0 1]     [1]    [0 0 1]     [1]        
    
               [1 0 1]     [2]    [1 0 1]     [1]           
    a(b(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = b(a(x1))
               [0 0 1]     [1]    [0 0 1]     [1]           
    
               [1 0 1]     [1]    [1 0 1]     [1]                    
    c(d(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(b(b(a(f(x1)))))
               [1 0 1]     [1]    [1 0 1]     [1]                    
    
               [1 0 1]     [1]    [1 0 1]     [1]              
    f(d(x1)) = [1 1 1]x1 + [2] >= [1 1 1]x1 + [2] = b(a(f(x1)))
               [0 0 1]     [1]    [0 0 1]     [1]              
    
               [1 0 1]     [1]    [1 0 1]     [1]        
    f(a(x1)) = [1 1 1]x1 + [3] >= [0 1 0]x1 + [1] = a(x1)
               [0 0 1]     [0]    [0 0 1]     [0]        
   problem:
    b(a(x1)) -> d(x1)
    c(d(x1)) -> c(b(b(a(f(x1)))))
    f(d(x1)) -> b(a(f(x1)))
    f(a(x1)) -> a(x1)
   String Reversal Processor:
    a(b(x1)) -> d(x1)
    d(c(x1)) -> f(a(b(b(c(x1)))))
    d(f(x1)) -> f(a(b(x1)))
    a(f(x1)) -> a(x1)
    Matrix Interpretation Processor: dim=3
     
     interpretation:
                [1 1 0]     [0]
      [f](x0) = [0 0 0]x0 + [1]
                [0 0 1]     [0],
      
                [1 0 0]     [0]
      [c](x0) = [0 0 0]x0 + [1]
                [0 0 0]     [0],
      
                [1 0 0]  
      [d](x0) = [0 1 0]x0
                [0 0 0]  ,
      
                [1 1 0]  
      [a](x0) = [0 0 1]x0
                [0 0 0]  ,
      
                [1 0 0]  
      [b](x0) = [0 0 0]x0
                [0 1 0]  
     orientation:
                 [1 0 0]      [1 0 0]          
      a(b(x1)) = [0 1 0]x1 >= [0 1 0]x1 = d(x1)
                 [0 0 0]      [0 0 0]          
      
                 [1 0 0]     [0]    [1 0 0]     [0]                    
      d(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = f(a(b(b(c(x1)))))
                 [0 0 0]     [0]    [0 0 0]     [0]                    
      
                 [1 1 0]     [0]    [1 1 0]     [0]              
      d(f(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = f(a(b(x1)))
                 [0 0 0]     [0]    [0 0 0]     [0]              
      
                 [1 1 0]     [1]    [1 1 0]          
      a(f(x1)) = [0 0 1]x1 + [0] >= [0 0 1]x1 = a(x1)
                 [0 0 0]     [0]    [0 0 0]          
     problem:
      a(b(x1)) -> d(x1)
      d(c(x1)) -> f(a(b(b(c(x1)))))
      d(f(x1)) -> f(a(b(x1)))
     String Reversal Processor:
      b(a(x1)) -> d(x1)
      c(d(x1)) -> c(b(b(a(f(x1)))))
      f(d(x1)) -> b(a(f(x1)))
      Matrix Interpretation Processor: dim=3
       
       interpretation:
                  [1 0 0]     [0]
        [f](x0) = [0 1 0]x0 + [1]
                  [0 0 0]     [1],
        
                  [1 1 0]     [1]
        [c](x0) = [0 0 0]x0 + [0]
                  [0 1 0]     [1],
        
                  [1 0 0]     [1]
        [d](x0) = [0 0 1]x0 + [1]
                  [0 0 0]     [0],
        
                  [1 0 0]     [1]
        [a](x0) = [0 0 0]x0 + [0]
                  [0 0 1]     [1],
        
                  [1 0 0]  
        [b](x0) = [0 0 1]x0
                  [0 0 0]  
       orientation:
                   [1 0 0]     [1]    [1 0 0]     [1]        
        b(a(x1)) = [0 0 1]x1 + [1] >= [0 0 1]x1 + [1] = d(x1)
                   [0 0 0]     [0]    [0 0 0]     [0]        
        
                   [1 0 1]     [3]    [1 0 0]     [2]                    
        c(d(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(b(b(a(f(x1)))))
                   [0 0 1]     [2]    [0 0 0]     [1]                    
        
                   [1 0 0]     [1]    [1 0 0]     [1]              
        f(d(x1)) = [0 0 1]x1 + [2] >= [0 0 0]x1 + [2] = b(a(f(x1)))
                   [0 0 0]     [1]    [0 0 0]     [0]              
       problem:
        b(a(x1)) -> d(x1)
        f(d(x1)) -> b(a(f(x1)))
       KBO Processor:
        weight function:
         w0 = 1
         w(f) = w(d) = w(b) = 1
         w(a) = 0
        precedence:
         a > f > b > d
        problem:
         
        Qed