YES

Problem:
 d(0()) -> 0()
 d(s(x)) -> s(s(d(x)))
 e(0(),x) -> x
 e(s(x),y) -> e(x,d(y))

Proof:
 DP Processor:
  DPs:
   d#(s(x)) -> d#(x)
   e#(s(x),y) -> d#(y)
   e#(s(x),y) -> e#(x,d(y))
  TRS:
   d(0()) -> 0()
   d(s(x)) -> s(s(d(x)))
   e(0(),x) -> x
   e(s(x),y) -> e(x,d(y))
  TDG Processor:
   DPs:
    d#(s(x)) -> d#(x)
    e#(s(x),y) -> d#(y)
    e#(s(x),y) -> e#(x,d(y))
   TRS:
    d(0()) -> 0()
    d(s(x)) -> s(s(d(x)))
    e(0(),x) -> x
    e(s(x),y) -> e(x,d(y))
   graph:
    e#(s(x),y) -> e#(x,d(y)) -> e#(s(x),y) -> e#(x,d(y))
    e#(s(x),y) -> e#(x,d(y)) -> e#(s(x),y) -> d#(y)
    e#(s(x),y) -> d#(y) -> d#(s(x)) -> d#(x)
    d#(s(x)) -> d#(x) -> d#(s(x)) -> d#(x)
   CDG Processor:
    DPs:
     d#(s(x)) -> d#(x)
     e#(s(x),y) -> d#(y)
     e#(s(x),y) -> e#(x,d(y))
    TRS:
     d(0()) -> 0()
     d(s(x)) -> s(s(d(x)))
     e(0(),x) -> x
     e(s(x),y) -> e(x,d(y))
    graph:
     
    Qed