YES

Problem:
 f(0()) -> cons(0())
 f(s(0())) -> f(p(s(0())))
 p(s(0())) -> 0()

Proof:
 DP Processor:
  DPs:
   f#(s(0())) -> p#(s(0()))
   f#(s(0())) -> f#(p(s(0())))
  TRS:
   f(0()) -> cons(0())
   f(s(0())) -> f(p(s(0())))
   p(s(0())) -> 0()
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [p#](x0) = -4x0 + 0,
    
    [f#](x0) = x0 + 0,
    
    [p](x0) = 1,
    
    [s](x0) = 4x0 + 2,
    
    [cons](x0) = -3x0 + 0,
    
    [f](x0) = x0 + 1,
    
    [0] = 0
   orientation:
    f#(s(0())) = 4 >= 0 = p#(s(0()))
    
    f#(s(0())) = 4 >= 1 = f#(p(s(0())))
    
    f(0()) = 1 >= 0 = cons(0())
    
    f(s(0())) = 4 >= 1 = f(p(s(0())))
    
    p(s(0())) = 1 >= 0 = 0()
   problem:
    DPs:
     
    TRS:
     f(0()) -> cons(0())
     f(s(0())) -> f(p(s(0())))
     p(s(0())) -> 0()
   Qed