YES

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

Proof:
 String Reversal Processor:
  strict:
   b(a(x1)) -> a(x1)
   b(d(x1)) -> b(x1)
   c(d(x1)) -> c(x1)
  weak:
   a(x1) -> a(d(x1))
   c(x1) -> b(c(x1))
  Bounds Processor:
   bound: 1
   enrichment: match-rt
   automaton:
    final states: {5}
    transitions:
     d0(5) -> 5*
     a1(8) -> 9*
     c0(5) -> 5*
     a0(5) -> 5*
     b0(5) -> 5*
     b1(50) -> 51*
     d1(33) -> 34*
     d1(27) -> 28*
     c1(49) -> 50*
     28 -> 8*
     8 -> 33*
     34 -> 8*
     5 -> 49,27,8
     9 -> 5*
     51 -> 50,5
   problem:
    strict:
     b(d(x1)) -> b(x1)
     c(d(x1)) -> c(x1)
    weak:
     a(x1) -> a(d(x1))
     c(x1) -> b(c(x1))
   String Reversal Processor:
    strict:
     d(b(x1)) -> b(x1)
     d(c(x1)) -> c(x1)
    weak:
     a(x1) -> d(a(x1))
     c(x1) -> c(b(x1))
    Matrix Interpretation Processor: dim=2
     
     interpretation:
                [1 2]  
      [a](x0) = [0 0]x0,
      
                [1 0]  
      [c](x0) = [1 0]x0,
      
                [1 0]     [0]
      [b](x0) = [0 0]x0 + [2],
      
                [1 1]  
      [d](x0) = [0 1]x0
     orientation:
                 [1 0]     [2]    [1 0]     [0]        
      d(b(x1)) = [0 0]x1 + [2] >= [0 0]x1 + [2] = b(x1)
      
                 [2 0]      [1 0]          
      d(c(x1)) = [1 0]x1 >= [1 0]x1 = c(x1)
      
              [1 2]      [1 2]             
      a(x1) = [0 0]x1 >= [0 0]x1 = d(a(x1))
      
              [1 0]      [1 0]             
      c(x1) = [1 0]x1 >= [1 0]x1 = c(b(x1))
     problem:
      strict:
       d(c(x1)) -> c(x1)
      weak:
       a(x1) -> d(a(x1))
       c(x1) -> c(b(x1))
     String Reversal Processor:
      strict:
       c(d(x1)) -> c(x1)
      weak:
       a(x1) -> a(d(x1))
       c(x1) -> b(c(x1))
      Arctic Interpretation Processor:
       dimension: 2
       interpretation:
                  [0  -&]  
        [a](x0) = [0  -&]x0,
        
                  [0 0]  
        [c](x0) = [0 0]x0,
        
                  [0 0]  
        [b](x0) = [0 0]x0,
        
                  [0  -&]  
        [d](x0) = [3  1 ]x0
       orientation:
                   [3 1]      [0 0]          
        c(d(x1)) = [3 1]x1 >= [0 0]x1 = c(x1)
        
                [0  -&]      [0  -&]             
        a(x1) = [0  -&]x1 >= [0  -&]x1 = a(d(x1))
        
                [0 0]      [0 0]             
        c(x1) = [0 0]x1 >= [0 0]x1 = b(c(x1))
       problem:
        strict:
         
        weak:
         a(x1) -> a(d(x1))
         c(x1) -> b(c(x1))
       Qed