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()
  TDG 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()
   graph:
    f#(s(0())) -> f#(p(s(0()))) -> f#(s(0())) -> f#(p(s(0())))
    f#(s(0())) -> f#(p(s(0()))) -> f#(s(0())) -> p#(s(0()))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(s(0())) -> f#(p(s(0())))
    TRS:
     f(0()) -> cons(0())
     f(s(0())) -> f(p(s(0())))
     p(s(0())) -> 0()
    Matrix Interpretation Processor:
     dimension: 1
     interpretation:
      [f#](x0) = x0 + 1,
      
      [p](x0) = 1,
      
      [s](x0) = x0 + 1,
      
      [cons](x0) = 0,
      
      [f](x0) = 0,
      
      [0] = 1
     orientation:
      f#(s(0())) = 3 >= 2 = f#(p(s(0())))
      
      f(0()) = 0 >= 0 = cons(0())
      
      f(s(0())) = 0 >= 0 = f(p(s(0())))
      
      p(s(0())) = 1 >= 1 = 0()
     problem:
      DPs:
       
      TRS:
       f(0()) -> cons(0())
       f(s(0())) -> f(p(s(0())))
       p(s(0())) -> 0()
     Qed