YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [f](x0) = x0 + 2,
   
   [a](x0) = x0 + 3,
   
   [c](x0) = x0,
   
   [b](x0) = 3x0 + 8,
   
   [d](x0) = x0 + 1
  orientation:
   a(b(x1)) = 3x1 + 11 >= 3x1 + 11 = b(d(x1))
   
   a(c(x1)) = x1 + 3 >= x1 + 3 = d(d(d(x1)))
   
   b(d(x1)) = 3x1 + 11 >= 3x1 + 11 = a(c(b(x1)))
   
   c(f(x1)) = x1 + 2 >= x1 + 2 = d(d(c(x1)))
   
   d(d(x1)) = x1 + 2 >= x1 + 2 = f(x1)
   
   f(f(x1)) = x1 + 4 >= x1 + 3 = a(x1)
  problem:
   a(b(x1)) -> b(d(x1))
   a(c(x1)) -> d(d(d(x1)))
   b(d(x1)) -> a(c(b(x1)))
   c(f(x1)) -> d(d(c(x1)))
   d(d(x1)) -> f(x1)
  String Reversal Processor:
   b(a(x1)) -> d(b(x1))
   c(a(x1)) -> d(d(d(x1)))
   d(b(x1)) -> b(c(a(x1)))
   f(c(x1)) -> c(d(d(x1)))
   d(d(x1)) -> f(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]  
     [f](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]     [0]
     [a](x0) = [0 0 1]x0 + [0]
               [1 1 1]     [1],
     
               [1 0 0]  
     [c](x0) = [0 0 1]x0
               [0 1 0]  ,
     
               [1 0 1]  
     [b](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]  
     [d](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
                [2 1 1]     [1]    [1 0 1]             
     b(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = d(b(x1))
                [0 0 0]     [0]    [0 0 0]             
     
                [1 0 0]     [0]    [1 0 0]                
     c(a(x1)) = [1 1 1]x1 + [1] >= [0 0 0]x1 = d(d(d(x1)))
                [0 0 1]     [0]    [0 0 0]                
     
                [1 0 1]      [1 0 1]                
     d(b(x1)) = [0 0 0]x1 >= [0 0 0]x1 = b(c(a(x1)))
                [0 0 0]      [0 0 0]                
     
                [1 0 0]      [1 0 0]                
     f(c(x1)) = [0 0 0]x1 >= [0 0 0]x1 = c(d(d(x1)))
                [0 0 0]      [0 0 0]                
     
                [1 0 0]      [1 0 0]          
     d(d(x1)) = [0 0 0]x1 >= [0 0 0]x1 = f(x1)
                [0 0 0]      [0 0 0]          
    problem:
     c(a(x1)) -> d(d(d(x1)))
     d(b(x1)) -> b(c(a(x1)))
     f(c(x1)) -> c(d(d(x1)))
     d(d(x1)) -> f(x1)
    String Reversal Processor:
     a(c(x1)) -> d(d(d(x1)))
     b(d(x1)) -> a(c(b(x1)))
     c(f(x1)) -> d(d(c(x1)))
     d(d(x1)) -> f(x1)
     WPO Processor:
      algebra: Sum
      weight function:
       w0 = 0
       w(b) = 4
       w(f) = w(c) = w(d) = w(a) = 0
      status function:
       st(f) = st(c) = st(d) = st(a) = st(b) = [0]
      precedence:
       b > c > d > f ~ a
      problem:
       
      Qed