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:
 DP Processor:
  DPs:
   b#(a(c(x1))) -> b#(c(x1))
   a#(d(x1)) -> d#(c(x1))
   b#(b(b(x1))) -> b#(c(x1))
   b#(b(b(x1))) -> a#(b(c(x1)))
   d#(c(x1)) -> d#(x1)
   d#(c(x1)) -> b#(d(x1))
   d#(c(x1)) -> d#(b(d(x1)))
   d#(a(c(x1))) -> b#(x1)
   d#(a(c(x1))) -> b#(b(x1))
  TRS:
   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))
  Usable Rule Processor:
   DPs:
    b#(a(c(x1))) -> b#(c(x1))
    a#(d(x1)) -> d#(c(x1))
    b#(b(b(x1))) -> b#(c(x1))
    b#(b(b(x1))) -> a#(b(c(x1)))
    d#(c(x1)) -> d#(x1)
    d#(c(x1)) -> b#(d(x1))
    d#(c(x1)) -> d#(b(d(x1)))
    d#(a(c(x1))) -> b#(x1)
    d#(a(c(x1))) -> b#(b(x1))
   TRS:
    d(c(x1)) -> b(d(x1))
    d(c(x1)) -> d(b(d(x1)))
    d(a(c(x1))) -> b(b(x1))
    b(d(b(x1))) -> c(d(b(x1)))
    b(a(c(x1))) -> b(c(x1))
    b(b(b(x1))) -> a(b(c(x1)))
   Matrix Interpretation Processor: dim=5
    
    interpretation:
     [d#](x0) = [0 0 0 0 1]x0,
     
     [a#](x0) = [1 0 0 0 1]x0 + [1],
     
     [b#](x0) = [0 1 0 1 0]x0,
     
               [0 0 0 0 1]     [0]
               [0 0 0 0 0]     [0]
     [a](x0) = [1 0 1 0 1]x0 + [0]
               [0 0 0 0 0]     [1]
               [1 0 1 0 1]     [0],
     
               [0 1 0 1 0]     [0]
               [0 0 0 0 0]     [0]
     [c](x0) = [0 0 1 0 1]x0 + [1]
               [0 0 0 0 0]     [0]
               [0 0 1 0 1]     [1],
     
               [0 0 1 0 0]     [1]
               [0 0 0 0 1]     [0]
     [d](x0) = [0 0 0 0 1]x0 + [0]
               [0 0 1 0 0]     [0]
               [0 0 0 0 1]     [0],
     
               [0 1 0 1 0]     [0]
               [0 0 1 1 0]     [0]
     [b](x0) = [0 1 0 1 0]x0 + [1]
               [0 0 1 0 0]     [0]
               [0 1 0 1 0]     [0]
    orientation:
     b#(a(c(x1))) = [1] >= [0] = b#(c(x1))
     
     a#(d(x1)) = [0 0 1 0 1]x1 + [2] >= [0 0 1 0 1]x1 + [1] = d#(c(x1))
     
     b#(b(b(x1))) = [0 2 1 2 0]x1 + [2] >= [0] = b#(c(x1))
     
     b#(b(b(x1))) = [0 2 1 2 0]x1 + [2] >= [1] = a#(b(c(x1)))
     
     d#(c(x1)) = [0 0 1 0 1]x1 + [1] >= [0 0 0 0 1]x1 = d#(x1)
     
     d#(c(x1)) = [0 0 1 0 1]x1 + [1] >= [0 0 1 0 1]x1 = b#(d(x1))
     
     d#(c(x1)) = [0 0 1 0 1]x1 + [1] >= [0 0 1 0 1]x1 = d#(b(d(x1)))
     
     d#(a(c(x1))) = [0 1 2 1 2]x1 + [2] >= [0 1 0 1 0]x1 = b#(x1)
     
     d#(a(c(x1))) = [0 1 2 1 2]x1 + [2] >= [0 0 2 1 0]x1 = b#(b(x1))
     
                [0 0 1 0 1]     [2]    [0 0 1 0 1]     [0]           
                [0 0 1 0 1]     [1]    [0 0 1 0 1]     [0]           
     d(c(x1)) = [0 0 1 0 1]x1 + [1] >= [0 0 1 0 1]x1 + [1] = b(d(x1))
                [0 0 1 0 1]     [1]    [0 0 0 0 1]     [0]           
                [0 0 1 0 1]     [1]    [0 0 1 0 1]     [0]           
     
                [0 0 1 0 1]     [2]    [0 0 1 0 1]     [2]              
                [0 0 1 0 1]     [1]    [0 0 1 0 1]     [0]              
     d(c(x1)) = [0 0 1 0 1]x1 + [1] >= [0 0 1 0 1]x1 + [0] = d(b(d(x1)))
                [0 0 1 0 1]     [1]    [0 0 1 0 1]     [1]              
                [0 0 1 0 1]     [1]    [0 0 1 0 1]     [0]              
     
                   [0 1 2 1 2]     [3]    [0 0 2 1 0]     [0]           
                   [0 1 2 1 2]     [2]    [0 1 1 1 0]     [1]           
     d(a(c(x1))) = [0 1 2 1 2]x1 + [2] >= [0 0 2 1 0]x1 + [1] = b(b(x1))
                   [0 1 2 1 2]     [2]    [0 1 0 1 0]     [1]           
                   [0 1 2 1 2]     [2]    [0 0 2 1 0]     [0]           
     
                   [0 2 0 2 0]     [1]    [0 2 0 2 0]     [1]              
                   [0 2 0 2 0]     [1]    [0 0 0 0 0]     [0]              
     b(d(b(x1))) = [0 2 0 2 0]x1 + [2] >= [0 2 0 2 0]x1 + [1] = c(d(b(x1)))
                   [0 1 0 1 0]     [0]    [0 0 0 0 0]     [0]              
                   [0 2 0 2 0]     [1]    [0 2 0 2 0]     [1]              
     
                   [0 0 0 0 0]     [1]    [0 0 0 0 0]     [0]           
                   [0 1 2 1 2]     [3]    [0 0 1 0 1]     [1]           
     b(a(c(x1))) = [0 0 0 0 0]x1 + [2] >= [0 0 0 0 0]x1 + [1] = b(c(x1))
                   [0 1 2 1 2]     [2]    [0 0 1 0 1]     [1]           
                   [0 0 0 0 0]     [1]    [0 0 0 0 0]     [0]           
     
                   [0 2 1 2 0]     [2]    [0]              
                   [0 1 2 2 0]     [2]    [0]              
     b(b(b(x1))) = [0 2 1 2 0]x1 + [3] >= [1] = a(b(c(x1)))
                   [0 0 2 1 0]     [1]    [1]              
                   [0 2 1 2 0]     [2]    [1]              
    problem:
     DPs:
      
     TRS:
      d(c(x1)) -> b(d(x1))
      d(c(x1)) -> d(b(d(x1)))
      d(a(c(x1))) -> b(b(x1))
      b(d(b(x1))) -> c(d(b(x1)))
      b(a(c(x1))) -> b(c(x1))
      b(b(b(x1))) -> a(b(c(x1)))
    Qed