YES

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

Proof:
 DP Processor:
  DPs:
   sum#(s(x)) -> sum#(x)
   sum#(s(x)) -> sqr#(s(x))
  TRS:
   sum(0()) -> 0()
   sum(s(x)) -> +(sqr(s(x)),sum(x))
   sqr(x) -> *(x,x)
   sum(s(x)) -> +(*(s(x),s(x)),sum(x))
  TDG Processor:
   DPs:
    sum#(s(x)) -> sum#(x)
    sum#(s(x)) -> sqr#(s(x))
   TRS:
    sum(0()) -> 0()
    sum(s(x)) -> +(sqr(s(x)),sum(x))
    sqr(x) -> *(x,x)
    sum(s(x)) -> +(*(s(x),s(x)),sum(x))
   graph:
    sum#(s(x)) -> sum#(x) -> sum#(s(x)) -> sqr#(s(x))
    sum#(s(x)) -> sum#(x) -> sum#(s(x)) -> sum#(x)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     sum#(s(x)) -> sum#(x)
    TRS:
     sum(0()) -> 0()
     sum(s(x)) -> +(sqr(s(x)),sum(x))
     sqr(x) -> *(x,x)
     sum(s(x)) -> +(*(s(x),s(x)),sum(x))
    Subterm Criterion Processor:
     simple projection:
      pi(sum#) = 0
     problem:
      DPs:
       
      TRS:
       sum(0()) -> 0()
       sum(s(x)) -> +(sqr(s(x)),sum(x))
       sqr(x) -> *(x,x)
       sum(s(x)) -> +(*(s(x),s(x)),sum(x))
     Qed