YES

Problem:
 f(0()) -> cons(0(),n__f(s(0())))
 f(s(0())) -> f(p(s(0())))
 p(s(0())) -> 0()
 f(X) -> n__f(X)
 activate(n__f(X)) -> f(X)
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   f#(s(0())) -> p#(s(0()))
   f#(s(0())) -> f#(p(s(0())))
   activate#(n__f(X)) -> f#(X)
  TRS:
   f(0()) -> cons(0(),n__f(s(0())))
   f(s(0())) -> f(p(s(0())))
   p(s(0())) -> 0()
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [activate#](x0) = 6x0 + -16,
    
    [p#](x0) = -5x0 + 0,
    
    [f#](x0) = -4x0 + 0,
    
    [activate](x0) = 6x0 + 12,
    
    [p](x0) = -1x0 + 1,
    
    [cons](x0, x1) = -3x0 + -1x1 + 0,
    
    [n__f](x0) = -6x0 + 1,
    
    [s](x0) = 5,
    
    [f](x0) = -1x0 + 2,
    
    [0] = 3
   orientation:
    f#(s(0())) = 1 >= 0 = p#(s(0()))
    
    f#(s(0())) = 1 >= 0 = f#(p(s(0())))
    
    activate#(n__f(X)) = X + 7 >= -4X + 0 = f#(X)
    
    f(0()) = 2 >= 0 = cons(0(),n__f(s(0())))
    
    f(s(0())) = 4 >= 3 = f(p(s(0())))
    
    p(s(0())) = 4 >= 3 = 0()
    
    f(X) = -1X + 2 >= -6X + 1 = n__f(X)
    
    activate(n__f(X)) = X + 12 >= -1X + 2 = f(X)
    
    activate(X) = 6X + 12 >= X = X
   problem:
    DPs:
     
    TRS:
     f(0()) -> cons(0(),n__f(s(0())))
     f(s(0())) -> f(p(s(0())))
     p(s(0())) -> 0()
     f(X) -> n__f(X)
     activate(n__f(X)) -> f(X)
     activate(X) -> X
   Qed