YES

Problem:
 sum(0()) -> 0()
 sum(s(x)) -> +(sum(x),s(x))
 +(x,0()) -> x
 +(x,s(y)) -> s(+(x,y))

Proof:
 DP Processor:
  DPs:
   sum#(s(x)) -> sum#(x)
   sum#(s(x)) -> +#(sum(x),s(x))
   +#(x,s(y)) -> +#(x,y)
  TRS:
   sum(0()) -> 0()
   sum(s(x)) -> +(sum(x),s(x))
   +(x,0()) -> x
   +(x,s(y)) -> s(+(x,y))
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    
   interpretation:
    [+#](x0, x1) = 4x1,
    
    [sum#](x0) = 5x0 + 6,
    
    [+](x0, x1) = 0,
    
    [s](x0) = 4x0 + 2,
    
    [sum](x0) = 6,
    
    [0] = 0
   orientation:
    sum#(s(x)) = 20x + 16 >= 5x + 6 = sum#(x)
    
    sum#(s(x)) = 20x + 16 >= 16x + 8 = +#(sum(x),s(x))
    
    +#(x,s(y)) = 16y + 8 >= 4y = +#(x,y)
    
    sum(0()) = 6 >= 0 = 0()
    
    sum(s(x)) = 6 >= 0 = +(sum(x),s(x))
    
    +(x,0()) = 0 >= x = x
    
    +(x,s(y)) = 0 >= 2 = s(+(x,y))
   problem:
    DPs:
     
    TRS:
     sum(0()) -> 0()
     sum(s(x)) -> +(sum(x),s(x))
     +(x,0()) -> x
     +(x,s(y)) -> s(+(x,y))
   Qed