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(s(X)))
  , from(X) -> n__from(X)
  , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
  , sel(0(), cons(X, Y)) -> X
  , activate(X) -> X
  , activate(n__from(X)) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { from^#(X) -> c_1()
  , from^#(X) -> c_2()
  , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z)))
  , sel^#(0(), cons(X, Y)) -> c_4()
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X)) }

and mark the set of starting terms.

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

Strict DPs:
  { from^#(X) -> c_1()
  , from^#(X) -> c_2()
  , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z)))
  , sel^#(0(), cons(X, Y)) -> c_4()
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X)) }
Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
  , sel(0(), cons(X, Y)) -> X
  , activate(X) -> X
  , activate(n__from(X)) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We replace rewrite rules by usable rules:

  Strict Usable Rules:
    { from(X) -> cons(X, n__from(s(X)))
    , from(X) -> n__from(X)
    , activate(X) -> X
    , activate(n__from(X)) -> from(X) }

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

Strict DPs:
  { from^#(X) -> c_1()
  , from^#(X) -> c_2()
  , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z)))
  , sel^#(0(), cons(X, Y)) -> c_4()
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X)) }
Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following constant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(sel^#) = {2}, Uargs(c_3) = {1}, Uargs(c_6) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

        [from](x1) = [2]                  
                                          
    [cons](x1, x2) = [1] x2 + [0]         
                                          
     [n__from](x1) = [1]                  
                                          
           [s](x1) = [1] x1 + [1]         
                                          
               [0] = [1]                  
                                          
    [activate](x1) = [1] x1 + [2]         
                                          
      [from^#](x1) = [1]                  
                                          
             [c_1] = [0]                  
                                          
             [c_2] = [0]                  
                                          
   [sel^#](x1, x2) = [2] x1 + [1] x2 + [2]
                                          
         [c_3](x1) = [1] x1 + [2]         
                                          
             [c_4] = [1]                  
                                          
  [activate^#](x1) = [2] x1 + [2]         
                                          
             [c_5] = [1]                  
                                          
         [c_6](x1) = [1] x1 + [2]         

This order satisfies following ordering constraints:

               [from(X)] = [2]                     
                         > [1]                     
                         = [cons(X, n__from(s(X)))]
                                                   
               [from(X)] = [2]                     
                         > [1]                     
                         = [n__from(X)]            
                                                   
           [activate(X)] = [1] X + [2]             
                         > [1] X + [0]             
                         = [X]                     
                                                   
  [activate(n__from(X))] = [3]                     
                         > [2]                     
                         = [from(X)]               
                                                   

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

Strict DPs:
  { sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) }
Weak DPs:
  { from^#(X) -> c_1()
  , from^#(X) -> c_2()
  , sel^#(0(), cons(X, Y)) -> c_4()
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X)) }
Weak Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ from^#(X) -> c_1()
, from^#(X) -> c_2()
, sel^#(0(), cons(X, Y)) -> c_4()
, activate^#(X) -> c_5()
, activate^#(n__from(X)) -> c_6(from^#(X)) }

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

Strict DPs:
  { sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) }
Weak Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

 safe(from) = {}, safe(cons) = {1, 2}, safe(n__from) = {1},
 safe(s) = {1}, safe(activate) = {}, safe(sel^#) = {2},
 safe(c_3) = {}

and precedence

 from ~ activate, from ~ sel^#, activate ~ sel^# .

Following symbols are considered recursive:

 {from, activate, sel^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(from) = [], pi(cons) = [], pi(n__from) = [], pi(s) = [1],
 pi(activate) = [], pi(sel^#) = [1], pi(c_3) = [1]

Usable defined function symbols are a subset of:

 {sel^#}

For your convenience, here are the satisfied ordering constraints:

  pi(sel^#(s(X), cons(Y, Z))) = sel^#(s(; X);)                
                              > c_3(sel^#(X;);)               
                              = pi(c_3(sel^#(X, activate(Z))))
                                                              

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