YES

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

Proof:
 DP Processor:
  DPs:
   a#(d(x1)) -> b#(x1)
   a#(x1) -> b#(x1)
   a#(x1) -> b#(b(x1))
   a#(x1) -> b#(b(b(x1)))
   b#(d(b(x1))) -> c#(x1)
   b#(d(b(x1))) -> a#(c(x1))
  TRS:
   a(d(x1)) -> d(b(x1))
   a(x1) -> b(b(b(x1)))
   b(d(b(x1))) -> a(c(x1))
   c(x1) -> d(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c#](x0) = 5x0 + 5/2,
    
    [b#](x0) = 2x0,
    
    [a#](x0) = 2x0 + 5/2,
    
    [c](x0) = 4x0,
    
    [b](x0) = x0 + 1/2,
    
    [a](x0) = x0 + 2,
    
    [d](x0) = 4x0
   orientation:
    a#(d(x1)) = 8x1 + 5/2 >= 2x1 = b#(x1)
    
    a#(x1) = 2x1 + 5/2 >= 2x1 = b#(x1)
    
    a#(x1) = 2x1 + 5/2 >= 2x1 + 1 = b#(b(x1))
    
    a#(x1) = 2x1 + 5/2 >= 2x1 + 2 = b#(b(b(x1)))
    
    b#(d(b(x1))) = 8x1 + 4 >= 5x1 + 5/2 = c#(x1)
    
    b#(d(b(x1))) = 8x1 + 4 >= 8x1 + 5/2 = a#(c(x1))
    
    a(d(x1)) = 4x1 + 2 >= 4x1 + 2 = d(b(x1))
    
    a(x1) = x1 + 2 >= x1 + 3/2 = b(b(b(x1)))
    
    b(d(b(x1))) = 4x1 + 5/2 >= 4x1 + 2 = a(c(x1))
    
    c(x1) = 4x1 >= 4x1 = d(x1)
   problem:
    DPs:
     
    TRS:
     a(d(x1)) -> d(b(x1))
     a(x1) -> b(b(b(x1)))
     b(d(b(x1))) -> a(c(x1))
     c(x1) -> d(x1)
   Qed