YES

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

Proof:
 String Reversal Processor:
  a(d(b(d(b(a(a(x1))))))) -> d(b(a(a(c(a(a(x1)))))))
  c(a(a(x1))) -> a(a(c(c(x1))))
  c(c(c(x1))) -> d(b(c(d(b(x1)))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [d](x0) = [0 0 0]x0
              [0 1 1]  ,
    
              [1 0 0]  
    [c](x0) = [0 1 0]x0
              [0 0 0]  ,
    
              [1 0 1]     [0]
    [a](x0) = [0 0 0]x0 + [1]
              [0 0 1]     [0],
    
              [1 0 0]  
    [b](x0) = [0 0 1]x0
              [0 1 0]  
   orientation:
                              [1 0 3]     [1]    [1 0 2]     [0]                          
    a(d(b(d(b(a(a(x1))))))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = d(b(a(a(c(a(a(x1)))))))
                              [0 0 1]     [1]    [0 0 0]     [1]                          
    
                  [1 0 2]     [0]    [1 0 0]     [0]                 
    c(a(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(a(c(c(x1))))
                  [0 0 0]     [0]    [0 0 0]     [0]                 
    
                  [1 0 0]      [1 0 0]                      
    c(c(c(x1))) = [0 1 0]x1 >= [0 0 0]x1 = d(b(c(d(b(x1)))))
                  [0 0 0]      [0 0 0]                      
   problem:
    c(a(a(x1))) -> a(a(c(c(x1))))
    c(c(c(x1))) -> d(b(c(d(b(x1)))))
   String Reversal Processor:
    a(a(c(x1))) -> c(c(a(a(x1))))
    c(c(c(x1))) -> b(d(c(b(d(x1)))))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {6,1}
      transitions:
       c0(4) -> 5*
       c0(5) -> 1*
       c0(8) -> 9*
       c1(31) -> 32*
       c1(13) -> 14*
       b0(7) -> 8*
       b0(10) -> 6*
       d1(29) -> 30*
       d1(14) -> 15*
       d1(11) -> 12*
       d1(32) -> 33*
       b1(12) -> 13*
       b1(15) -> 16*
       b1(33) -> 34*
       b1(30) -> 31*
       d0(2) -> 7*
       d0(9) -> 10*
       f40() -> 2*
       a0(2) -> 3*
       a0(3) -> 4*
       16 -> 1*
       4 -> 29*
       1 -> 3,4
       34 -> 5*
       5 -> 11*
     problem:
      
     Qed