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:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [d](x0) = x0,
   
   [a](x0) = x0 + 8,
   
   [b](x0) = x0 + 8,
   
   [c](x0) = x0 + 8
  orientation:
   b(d(b(x1))) = x1 + 16 >= x1 + 16 = c(d(b(x1)))
   
   b(a(c(x1))) = x1 + 24 >= x1 + 16 = b(c(x1))
   
   a(d(x1)) = x1 + 8 >= x1 + 8 = d(c(x1))
   
   b(b(b(x1))) = x1 + 24 >= x1 + 24 = a(b(c(x1)))
   
   d(c(x1)) = x1 + 8 >= x1 + 8 = b(d(x1))
   
   d(c(x1)) = x1 + 8 >= x1 + 8 = d(b(d(x1)))
   
   d(a(c(x1))) = x1 + 16 >= x1 + 16 = 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))
  String Reversal Processor:
   b(d(b(x1))) -> b(d(c(x1)))
   d(a(x1)) -> c(d(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))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 1 0]  
     [d](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]     [0]
     [a](x0) = [0 1 0]x0 + [1]
               [0 0 0]     [1],
     
               [1 0 0]  
     [b](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]  
     [c](x0) = [0 0 0]x0
               [0 0 0]  
    orientation:
                   [1 0 0]      [1 0 0]                
     b(d(b(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(d(c(x1)))
                   [0 0 0]      [0 0 0]                
     
                [1 1 0]     [1]    [1 1 0]             
     d(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(d(x1))
                [0 0 0]     [0]    [0 0 0]             
     
                   [1 0 0]      [1 0 0]                
     b(b(b(x1))) = [0 0 0]x1 >= [0 0 0]x1 = c(b(a(x1)))
                   [0 0 0]      [0 0 0]                
     
                [1 1 0]      [1 0 0]             
     c(d(x1)) = [0 0 0]x1 >= [0 0 0]x1 = d(b(x1))
                [0 0 0]      [0 0 0]             
     
                [1 1 0]      [1 1 0]                
     c(d(x1)) = [0 0 0]x1 >= [0 0 0]x1 = d(b(d(x1)))
                [0 0 0]      [0 0 0]                
     
                   [1 1 0]      [1 0 0]             
     c(a(d(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(b(x1))
                   [0 0 0]      [0 0 0]             
    problem:
     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:
       b0(4) -> 1*
       b0(11) -> 12*
       b0(9) -> 13*
       b0(2) -> 9*
       b0(6) -> 7*
       c1(14) -> 15*
       c0(7) -> 5*
       c0(2) -> 3*
       d1(15) -> 16*
       b1(16) -> 17*
       a0(2) -> 6*
       f40() -> 2*
       d0(12) -> 10*
       d0(2) -> 11*
       d0(3) -> 4*
       d0(9) -> 8*
       17 -> 1,12,9
       13 -> 3*
       8 -> 3*
       1 -> 9,12
       5 -> 9,13
       9 -> 14*
       10 -> 3*
     problem:
      
     Qed