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))
  Usable Rule Processor:
   DPs:
    sum#(s(x)) -> sum#(x)
    sum#(s(x)) -> sqr#(s(x))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [sqr#](x0) = -4x0 + 0,
     
     [sum#](x0) = x0,
     
     [s](x0) = 1x0 + 4
    orientation:
     sum#(s(x)) = 1x + 4 >= x = sum#(x)
     
     sum#(s(x)) = 1x + 4 >= -3x + 0 = sqr#(s(x))
    problem:
     DPs:
      
     TRS:
      
    Qed