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))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [sum#](x0) = 1x0,
      
      [*](x0, x1) = x0 + x1 + 0,
      
      [+](x0, x1) = 1x1,
      
      [sqr](x0) = 1x0 + 4,
      
      [s](x0) = 2x0,
      
      [sum](x0) = 2x0,
      
      [0] = 3
     orientation:
      sum#(s(x)) = 3x >= 1x = sum#(x)
      
      sum(0()) = 5 >= 3 = 0()
      
      sum(s(x)) = 4x >= 3x = +(sqr(s(x)),sum(x))
      
      sqr(x) = 1x + 4 >= x + 0 = *(x,x)
      
      sum(s(x)) = 4x >= 3x = +(*(s(x),s(x)),sum(x))
     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