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)))
  TDG 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)))
   graph:
    sum1#(s(x)) -> sum1#(x) -> sum1#(s(x)) -> sum1#(x)
    sum#(s(x)) -> sum#(x) -> sum#(s(x)) -> sum#(x)
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 2/4
    DPs:
     sum#(s(x)) -> sum#(x)
    TRS:
     sum(0()) -> 0()
     sum(s(x)) -> +(sum(x),s(x))
     sum1(0()) -> 0()
     sum1(s(x)) -> s(+(sum1(x),+(x,x)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [sum#](x0) = 1x0,
      
      [sum1](x0) = 1x0,
      
      [+](x0, x1) = x1,
      
      [s](x0) = 1x0,
      
      [sum](x0) = 2x0 + 4,
      
      [0] = 4
     orientation:
      sum#(s(x)) = 2x >= 1x = sum#(x)
      
      sum(0()) = 6 >= 4 = 0()
      
      sum(s(x)) = 3x + 4 >= 1x = +(sum(x),s(x))
      
      sum1(0()) = 5 >= 4 = 0()
      
      sum1(s(x)) = 2x >= 1x = s(+(sum1(x),+(x,x)))
     problem:
      DPs:
       
      TRS:
       sum(0()) -> 0()
       sum(s(x)) -> +(sum(x),s(x))
       sum1(0()) -> 0()
       sum1(s(x)) -> s(+(sum1(x),+(x,x)))
     Qed
    
    DPs:
     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)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [sum1#](x0) = 1x0,
      
      [sum1](x0) = 1x0,
      
      [+](x0, x1) = x1,
      
      [s](x0) = 1x0,
      
      [sum](x0) = 2x0 + 4,
      
      [0] = 4
     orientation:
      sum1#(s(x)) = 2x >= 1x = sum1#(x)
      
      sum(0()) = 6 >= 4 = 0()
      
      sum(s(x)) = 3x + 4 >= 1x = +(sum(x),s(x))
      
      sum1(0()) = 5 >= 4 = 0()
      
      sum1(s(x)) = 2x >= 1x = s(+(sum1(x),+(x,x)))
     problem:
      DPs:
       
      TRS:
       sum(0()) -> 0()
       sum(s(x)) -> +(sum(x),s(x))
       sum1(0()) -> 0()
       sum1(s(x)) -> s(+(sum1(x),+(x,x)))
     Qed