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))
  TDG 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))
   graph:
    d#(u(x)) -> c#(x) -> c#(u(x)) -> b#(x)
   CDG 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))
    graph:
     
    Qed