YES

Problem:
 f(0()) -> cons(0(),n__f(n__s(n__0())))
 f(s(0())) -> f(p(s(0())))
 p(s(X)) -> X
 f(X) -> n__f(X)
 s(X) -> n__s(X)
 0() -> n__0()
 activate(n__f(X)) -> f(activate(X))
 activate(n__s(X)) -> s(activate(X))
 activate(n__0()) -> 0()
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   f#(s(0())) -> p#(s(0()))
   f#(s(0())) -> f#(p(s(0())))
   activate#(n__f(X)) -> activate#(X)
   activate#(n__f(X)) -> f#(activate(X))
   activate#(n__s(X)) -> activate#(X)
   activate#(n__s(X)) -> s#(activate(X))
   activate#(n__0()) -> 0#()
  TRS:
   f(0()) -> cons(0(),n__f(n__s(n__0())))
   f(s(0())) -> f(p(s(0())))
   p(s(X)) -> X
   f(X) -> n__f(X)
   s(X) -> n__s(X)
   0() -> n__0()
   activate(n__f(X)) -> f(activate(X))
   activate(n__s(X)) -> s(activate(X))
   activate(n__0()) -> 0()
   activate(X) -> X
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [activate#](x0) = 3x0,
    
    [0#] = 0,
    
    [s#](x0) = 0,
    
    [p#](x0) = 0,
    
    [f#](x0) = 2x0,
    
    [activate](x0) = 2x0,
    
    [p](x0) = 1/2x0 + 2,
    
    [s](x0) = 3x0 + 3,
    
    [cons](x0, x1) = 3x0 + 3,
    
    [n__f](x0) = 7/2x0 + 3/2,
    
    [n__s](x0) = 3x0 + 3/2,
    
    [n__0] = 5/2,
    
    [f](x0) = 7/2x0 + 3/2,
    
    [0] = 3
   orientation:
    f#(s(0())) = 24 >= 0 = p#(s(0()))
    
    f#(s(0())) = 24 >= 16 = f#(p(s(0())))
    
    activate#(n__f(X)) = 21/2X + 9/2 >= 3X = activate#(X)
    
    activate#(n__f(X)) = 21/2X + 9/2 >= 4X = f#(activate(X))
    
    activate#(n__s(X)) = 9X + 9/2 >= 3X = activate#(X)
    
    activate#(n__s(X)) = 9X + 9/2 >= 0 = s#(activate(X))
    
    activate#(n__0()) = 15/2 >= 0 = 0#()
    
    f(0()) = 12 >= 12 = cons(0(),n__f(n__s(n__0())))
    
    f(s(0())) = 87/2 >= 59/2 = f(p(s(0())))
    
    p(s(X)) = 3/2X + 7/2 >= X = X
    
    f(X) = 7/2X + 3/2 >= 7/2X + 3/2 = n__f(X)
    
    s(X) = 3X + 3 >= 3X + 3/2 = n__s(X)
    
    0() = 3 >= 5/2 = n__0()
    
    activate(n__f(X)) = 7X + 3 >= 7X + 3/2 = f(activate(X))
    
    activate(n__s(X)) = 6X + 3 >= 6X + 3 = s(activate(X))
    
    activate(n__0()) = 5 >= 3 = 0()
    
    activate(X) = 2X >= X = X
   problem:
    DPs:
     
    TRS:
     f(0()) -> cons(0(),n__f(n__s(n__0())))
     f(s(0())) -> f(p(s(0())))
     p(s(X)) -> X
     f(X) -> n__f(X)
     s(X) -> n__s(X)
     0() -> n__0()
     activate(n__f(X)) -> f(activate(X))
     activate(n__s(X)) -> s(activate(X))
     activate(n__0()) -> 0()
     activate(X) -> X
   Qed