YES(?,O(n^1))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { f(X) -> n__f(X)
  , f(0()) -> cons(0(), n__f(n__s(n__0())))
  , f(s(0())) -> f(p(s(0())))
  , 0() -> n__0()
  , s(X) -> n__s(X)
  , p(s(0())) -> 0()
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , activate(n__0()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Arguments of following rules are not normal-forms:

{ f(0()) -> cons(0(), n__f(n__s(n__0())))
, f(s(0())) -> f(p(s(0())))
, p(s(0())) -> 0() }

All above mentioned rules can be savely removed.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { f(X) -> n__f(X)
  , 0() -> n__0()
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__f(X)) -> f(activate(X))
  , activate(n__s(X)) -> s(activate(X))
  , activate(n__0()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping

 safe(f) = {1}, safe(0) = {}, safe(cons) = {1, 2}, safe(n__f) = {1},
 safe(n__s) = {1}, safe(n__0) = {}, safe(s) = {1}, safe(p) = {},
 safe(activate) = {}

and precedence

 activate > f, activate > 0, activate > s, f ~ 0, f ~ s, 0 ~ s .

Following symbols are considered recursive:

 {activate}

The recursion depth is 1.

For your convenience, here are the satisfied ordering constraints:

                f(; X) > n__f(; X)        
                                          
                   0() > n__0()           
                                          
                s(; X) > n__s(; X)        
                                          
          activate(X;) > X                
                                          
  activate(n__f(; X);) > f(; activate(X;))
                                          
  activate(n__s(; X);) > s(; activate(X;))
                                          
     activate(n__0();) > 0()              
                                          

Hurray, we answered YES(?,O(n^1))