YES

Problem:
 sum(0()) -> 0()
 sum(s(x)) -> +(sum(x),s(x))
 sum1(0()) -> 0()
 sum1(s(x)) -> s(+(sum1(x),+(x,x)))

Proof:
 DP Processor:
  DPs:
   sum#(s(x)) -> sum#(x)
   sum1#(s(x)) -> sum1#(x)
  TRS:
   sum(0()) -> 0()
   sum(s(x)) -> +(sum(x),s(x))
   sum1(0()) -> 0()
   sum1(s(x)) -> s(+(sum1(x),+(x,x)))
  Usable Rule Processor:
   DPs:
    sum#(s(x)) -> sum#(x)
    sum1#(s(x)) -> sum1#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [sum1#](x0) = x0,
     
     [sum#](x0) = x0,
     
     [s](x0) = 1x0 + -2
    orientation:
     sum#(s(x)) = 1x + -2 >= x = sum#(x)
     
     sum1#(s(x)) = 1x + -2 >= x = sum1#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed