YES

Problem:
 d(x) -> e(u(x))
 d(u(x)) -> c(x)
 c(u(x)) -> b(x)
 v(e(x)) -> x
 b(u(x)) -> a(e(x))

Proof:
 DP Processor:
  DPs:
   d#(u(x)) -> c#(x)
   c#(u(x)) -> b#(x)
  TRS:
   d(x) -> e(u(x))
   d(u(x)) -> c(x)
   c(u(x)) -> b(x)
   v(e(x)) -> x
   b(u(x)) -> a(e(x))
  Usable Rule Processor:
   DPs:
    d#(u(x)) -> c#(x)
    c#(u(x)) -> b#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 4
    usable rules:
     
    interpretation:
     [b#](x0) = [0  -1 -2 -2]x0,
     
     [c#](x0) = [-& -& 0  0 ]x0 + [0],
     
     [d#](x0) = [-& 0  0  -&]x0 + [1],
     
               [-2 0  0  0 ]     [0 ]
               [-1 0  -2 -2]     [-2]
     [u](x0) = [1  0  1  1 ]x0 + [0 ]
               [0  0  1  1 ]     [0 ]
    orientation:
     d#(u(x)) = [1 0 1 1]x + [1] >= [-& -& 0  0 ]x + [0] = c#(x)
     
     c#(u(x)) = [1 0 1 1]x + [0] >= [0  -1 -2 -2]x = b#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed