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))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0) = x0,
    
    [+](x0, x1) = 4x0 + 1x1 + 4,
    
    [*](x0, x1) = x0 + 4,
    
    [s](x0) = 1x0 + 0,
    
    [f](x0) = 3x0 + 5,
    
    [0] = 0
   orientation:
    f#(s(0())) = 1 >= 0 = f#(0())
    
    f#(+(x,s(0()))) = 4x + 4 >= x = f#(x)
    
    f#(+(x,y)) = 4x + 1y + 4 >= y = f#(y)
    
    f#(+(x,y)) = 4x + 1y + 4 >= x = f#(x)
    
    f(0()) = 5 >= 1 = s(0())
    
    f(s(0())) = 5 >= 2 = s(s(0()))
    
    f(s(0())) = 5 >= 4 = *(s(s(0())),f(0()))
    
    f(+(x,s(0()))) = 7x + 7 >= 4x + 6 = +(s(s(0())),f(x))
    
    f(+(x,y)) = 7x + 4y + 7 >= 3x + 5 = *(f(x),f(y))
   problem:
    DPs:
     
    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))
   Qed