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)
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 4/9
    DPs:
     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))
    Subterm Criterion Processor:
     simple projection:
      pi(e#) = 0
     problem:
      DPs:
       
      TRS:
       d(0()) -> 0()
       d(s(x)) -> s(s(d(x)))
       e(0(),x) -> x
       e(s(x),y) -> e(x,d(y))
     Qed
    
    DPs:
     d#(s(x)) -> d#(x)
    TRS:
     d(0()) -> 0()
     d(s(x)) -> s(s(d(x)))
     e(0(),x) -> x
     e(s(x),y) -> e(x,d(y))
    Subterm Criterion Processor:
     simple projection:
      pi(d#) = 0
     problem:
      DPs:
       
      TRS:
       d(0()) -> 0()
       d(s(x)) -> s(s(d(x)))
       e(0(),x) -> x
       e(s(x),y) -> e(x,d(y))
     Qed