YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> c(b(x1))
  b(b(x1)) -> d(c(x1))
  c(c(x1)) -> d(d(d(x1)))
  c(d(x1)) -> f(b(x1))
  d(d(d(x1))) -> c(a(x1))
  f(f(x1)) -> b(f(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0) = x0 + 5,
    
    [c](x0) = x0 + 6,
    
    [d](x0) = x0 + 4,
    
    [a](x0) = x0 + 6,
    
    [b](x0) = x0 + 5
   orientation:
    a(a(x1)) = x1 + 12 >= x1 + 11 = c(b(x1))
    
    b(b(x1)) = x1 + 10 >= x1 + 10 = d(c(x1))
    
    c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1)))
    
    c(d(x1)) = x1 + 10 >= x1 + 10 = f(b(x1))
    
    d(d(d(x1))) = x1 + 12 >= x1 + 12 = c(a(x1))
    
    f(f(x1)) = x1 + 10 >= x1 + 10 = b(f(x1))
   problem:
    b(b(x1)) -> d(c(x1))
    c(c(x1)) -> d(d(d(x1)))
    c(d(x1)) -> f(b(x1))
    d(d(d(x1))) -> c(a(x1))
    f(f(x1)) -> b(f(x1))
   String Reversal Processor:
    b(b(x1)) -> c(d(x1))
    c(c(x1)) -> d(d(d(x1)))
    d(c(x1)) -> b(f(x1))
    d(d(d(x1))) -> a(c(x1))
    f(f(x1)) -> f(b(x1))
    WPO Processor:
     algebra: Sum
     weight function:
      w0 = 0
      w(c) = 3
      w(f) = w(b) = 2
      w(d) = 1
      w(a) = 0
     status function:
      st(f) = st(d) = st(b) = st(c) = st(a) = [0]
     precedence:
      f > d > b > c ~ a
     problem:
      
     Qed