YES(O(1),O(n^1))

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

Strict Trs:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , after(0(), XS) -> XS
  , after(s(N), cons(X, XS)) -> after(N, activate(XS))
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

Arguments of following rules are not normal-forms:

{ after(s(N), cons(X, XS)) -> after(N, activate(XS)) }

All above mentioned rules can be savely removed.

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

Strict Trs:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , after(0(), XS) -> XS
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

Trs:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , after(0(), XS) -> XS
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
   safe(n__s) = {1}, safe(after) = {}, safe(0) = {}, safe(s) = {1},
   safe(activate) = {}
  
  and precedence
  
   from > s, after > from, after > s, activate > from, activate > s,
   after ~ activate .
  
  Following symbols are considered recursive:
  
   {activate}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
                  from(; X) > cons(; X,  n__from(; n__s(; X)))
                                                              
                  from(; X) > n__from(; X)                    
                                                              
           after(0(),  XS;) > XS                              
                                                              
                     s(; X) > n__s(; X)                       
                                                              
               activate(X;) > X                               
                                                              
    activate(n__from(; X);) > from(; activate(X;))            
                                                              
       activate(n__s(; X);) > s(; activate(X;))               
                                                              

We return to the main proof.

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

Weak Trs:
  { from(X) -> cons(X, n__from(n__s(X)))
  , from(X) -> n__from(X)
  , after(0(), XS) -> XS
  , s(X) -> n__s(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(activate(X))
  , activate(n__s(X)) -> s(activate(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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