YES

Problem:
 d(0()) -> 0()
 d(s(x)) -> s(s(d(x)))
 e(0()) -> s(0())
 e(r(x)) -> d(e(x))

Proof:
 DP Processor:
  DPs:
   d#(s(x)) -> d#(x)
   e#(r(x)) -> e#(x)
   e#(r(x)) -> d#(e(x))
  TRS:
   d(0()) -> 0()
   d(s(x)) -> s(s(d(x)))
   e(0()) -> s(0())
   e(r(x)) -> d(e(x))
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    d(0()) -> 0()
    d(s(x)) -> s(s(d(x)))
    e(0()) -> s(0())
    e(r(x)) -> d(e(x))
   interpretation:
    [e#](x0) = x0 + 6,
    
    [d#](x0) = x0,
    
    [r](x0) = 4x0 + 4,
    
    [e](x0) = 4x0,
    
    [s](x0) = x0 + 1,
    
    [d](x0) = 2x0 + 1,
    
    [0] = 1
   orientation:
    d#(s(x)) = x + 1 >= x = d#(x)
    
    e#(r(x)) = 4x + 10 >= x + 6 = e#(x)
    
    e#(r(x)) = 4x + 10 >= 4x = d#(e(x))
    
    d(0()) = 3 >= 1 = 0()
    
    d(s(x)) = 2x + 3 >= 2x + 3 = s(s(d(x)))
    
    e(0()) = 4 >= 2 = s(0())
    
    e(r(x)) = 16x + 16 >= 8x + 1 = d(e(x))
   problem:
    DPs:
     
    TRS:
     d(0()) -> 0()
     d(s(x)) -> s(s(d(x)))
     e(0()) -> s(0())
     e(r(x)) -> d(e(x))
   Qed