YES

Problem:
 f(0()) -> s(0())
 f(s(0())) -> s(s(0()))
 f(s(0())) -> *(s(s(0())),f(0()))
 f(+(x,s(0()))) -> +(s(s(0())),f(x))
 f(+(x,y)) -> *(f(x),f(y))

Proof:
 DP Processor:
  DPs:
   f#(s(0())) -> f#(0())
   f#(+(x,s(0()))) -> f#(x)
   f#(+(x,y)) -> f#(y)
   f#(+(x,y)) -> f#(x)
  TRS:
   f(0()) -> s(0())
   f(s(0())) -> s(s(0()))
   f(s(0())) -> *(s(s(0())),f(0()))
   f(+(x,s(0()))) -> +(s(s(0())),f(x))
   f(+(x,y)) -> *(f(x),f(y))
  Usable Rule Processor:
   DPs:
    f#(s(0())) -> f#(0())
    f#(+(x,s(0()))) -> f#(x)
    f#(+(x,y)) -> f#(y)
    f#(+(x,y)) -> f#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0) = x0 + 0,
     
     [+](x0, x1) = 1x0 + 4x1 + 4,
     
     [s](x0) = 2x0 + 0,
     
     [0] = 1
    orientation:
     f#(s(0())) = 3 >= 1 = f#(0())
     
     f#(+(x,s(0()))) = 1x + 7 >= x + 0 = f#(x)
     
     f#(+(x,y)) = 1x + 4y + 4 >= y + 0 = f#(y)
     
     f#(+(x,y)) = 1x + 4y + 4 >= x + 0 = f#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed