YES

Problem:
 f(0()) -> cons(0(),n__f(s(0())))
 f(s(0())) -> f(p(s(0())))
 p(s(X)) -> X
 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(X)) -> X
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [activate#](x0) = 4x0 + -16,
    
    [p#](x0) = -15x0 + 0,
    
    [f#](x0) = -7x0 + 0,
    
    [activate](x0) = 6x0 + 1,
    
    [p](x0) = -4x0 + 0,
    
    [cons](x0, x1) = x0 + x1 + -16,
    
    [n__f](x0) = -8x0 + 0,
    
    [s](x0) = 5x0 + 8,
    
    [f](x0) = -3x0 + 1,
    
    [0] = 0
   orientation:
    f#(s(0())) = 1 >= 0 = p#(s(0()))
    
    f#(s(0())) = 1 >= 0 = f#(p(s(0())))
    
    activate#(n__f(X)) = -4X + 4 >= -7X + 0 = f#(X)
    
    f(0()) = 1 >= 0 = cons(0(),n__f(s(0())))
    
    f(s(0())) = 5 >= 1 = f(p(s(0())))
    
    p(s(X)) = 1X + 4 >= X = X
    
    f(X) = -3X + 1 >= -8X + 0 = n__f(X)
    
    activate(n__f(X)) = -2X + 6 >= -3X + 1 = f(X)
    
    activate(X) = 6X + 1 >= X = X
   problem:
    DPs:
     
    TRS:
     f(0()) -> cons(0(),n__f(s(0())))
     f(s(0())) -> f(p(s(0())))
     p(s(X)) -> X
     f(X) -> n__f(X)
     activate(n__f(X)) -> f(X)
     activate(X) -> X
   Qed