YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [a](x0) = 4x0,
   
   [c](x0) = 4x0,
   
   [d](x0) = x0,
   
   [b](x0) = 4x0
  orientation:
   b(d(b(x1))) = 8x1 >= 8x1 = c(d(b(x1)))
   
   b(a(c(x1))) = 12x1 >= 8x1 = b(c(x1))
   
   a(d(x1)) = 4x1 >= 4x1 = d(c(x1))
   
   b(b(b(x1))) = 12x1 >= 12x1 = a(b(c(x1)))
   
   d(c(x1)) = 4x1 >= 4x1 = b(d(x1))
   
   d(c(x1)) = 4x1 >= 4x1 = d(b(d(x1)))
   
   d(a(c(x1))) = 8x1 >= 8x1 = b(b(x1))
  problem:
   b(d(b(x1))) -> c(d(b(x1)))
   a(d(x1)) -> d(c(x1))
   b(b(b(x1))) -> a(b(c(x1)))
   d(c(x1)) -> b(d(x1))
   d(c(x1)) -> d(b(d(x1)))
   d(a(c(x1))) -> b(b(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  2 ]  
    [a](x0) = [-& 1 ]x0,
    
              [0  0 ]  
    [c](x0) = [-& -&]x0,
    
              [0 0]  
    [d](x0) = [0 0]x0,
    
              [0  0 ]  
    [b](x0) = [-& 0 ]x0
   orientation:
                  [0 0]      [0  0 ]                
    b(d(b(x1))) = [0 0]x1 >= [-& -&]x1 = c(d(b(x1)))
    
               [2 2]      [0 0]             
    a(d(x1)) = [1 1]x1 >= [0 0]x1 = d(c(x1))
    
                  [0  0 ]      [0  0 ]                
    b(b(b(x1))) = [-& 0 ]x1 >= [-& -&]x1 = a(b(c(x1)))
    
               [0 0]      [0 0]             
    d(c(x1)) = [0 0]x1 >= [0 0]x1 = b(d(x1))
    
               [0 0]      [0 0]                
    d(c(x1)) = [0 0]x1 >= [0 0]x1 = d(b(d(x1)))
    
                  [0 0]      [0  0 ]             
    d(a(c(x1))) = [0 0]x1 >= [-& 0 ]x1 = b(b(x1))
   problem:
    b(d(b(x1))) -> c(d(b(x1)))
    b(b(b(x1))) -> a(b(c(x1)))
    d(c(x1)) -> b(d(x1))
    d(c(x1)) -> d(b(d(x1)))
    d(a(c(x1))) -> b(b(x1))
   String Reversal Processor:
    b(d(b(x1))) -> b(d(c(x1)))
    b(b(b(x1))) -> c(b(a(x1)))
    c(d(x1)) -> d(b(x1))
    c(d(x1)) -> d(b(d(x1)))
    c(a(d(x1))) -> b(b(x1))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {13,10,8,5,1}
      transitions:
       f40() -> 2*
       b0(2) -> 9*
       b0(9) -> 13*
       b0(4) -> 1*
       b0(11) -> 12*
       b0(6) -> 7*
       d0(12) -> 10*
       d0(2) -> 11*
       d0(9) -> 8*
       d0(3) -> 4*
       c0(7) -> 5*
       c0(2) -> 3*
       a0(2) -> 6*
       b1(16) -> 17*
       d1(15) -> 16*
       c1(14) -> 15*
       1 -> 9,12
       5 -> 9,13
       8 -> 3*
       9 -> 14*
       10 -> 3*
       13 -> 3*
       17 -> 1,12,9
     problem:
      
     Qed