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=2
   
   interpretation:
    [c#](x0) = [0 4]x0,
    
    [b#](x0) = [0 4]x0 + [1],
    
    [a#](x0) = [0 4]x0 + [4],
    
              [4 0]     [2]
    [c](x0) = [1 1]x0 + [0],
    
                   [1]
    [b](x0) = x0 + [0],
    
                   [4]
    [a](x0) = x0 + [1],
    
              [4 0]     [2]
    [d](x0) = [1 1]x0 + [0]
   orientation:
    a#(d(x1)) = [4 4]x1 + [4] >= [0 4]x1 + [1] = b#(x1)
    
    a#(x1) = [0 4]x1 + [4] >= [0 4]x1 + [1] = b#(x1)
    
    a#(x1) = [0 4]x1 + [4] >= [0 4]x1 + [1] = b#(b(x1))
    
    a#(x1) = [0 4]x1 + [4] >= [0 4]x1 + [1] = b#(b(b(x1)))
    
    b#(d(b(x1))) = [4 4]x1 + [5] >= [0 4]x1 = c#(x1)
    
    b#(d(b(x1))) = [4 4]x1 + [5] >= [4 4]x1 + [4] = a#(c(x1))
    
               [4 0]     [6]    [4 0]     [6]           
    a(d(x1)) = [1 1]x1 + [1] >= [1 1]x1 + [1] = d(b(x1))
    
                 [4]         [3]              
    a(x1) = x1 + [1] >= x1 + [0] = b(b(b(x1)))
    
                  [4 0]     [7]    [4 0]     [6]           
    b(d(b(x1))) = [1 1]x1 + [1] >= [1 1]x1 + [1] = a(c(x1))
    
            [4 0]     [2]    [4 0]     [2]        
    c(x1) = [1 1]x1 + [0] >= [1 1]x1 + [0] = 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