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)
  , head(cons(X, XS)) -> X
  , 2nd(cons(X, XS)) -> head(activate(XS))
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__take(X1, X2)) -> take(X1, X2)
  , take(X1, X2) -> n__take(X1, X2)
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , take(0(), XS) -> nil()
  , sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
  , sel(0(), cons(X, XS)) -> 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()
  , head^#(cons(X, XS)) -> c_3()
  , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X))
  , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
  , take^#(X1, X2) -> c_8()
  , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
  , take^#(0(), XS) -> c_10()
  , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS)))
  , sel^#(0(), cons(X, XS)) -> c_12() }

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()
  , head^#(cons(X, XS)) -> c_3()
  , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X))
  , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
  , take^#(X1, X2) -> c_8()
  , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
  , take^#(0(), XS) -> c_10()
  , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS)))
  , sel^#(0(), cons(X, XS)) -> c_12() }
Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , head(cons(X, XS)) -> X
  , 2nd(cons(X, XS)) -> head(activate(XS))
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__take(X1, X2)) -> take(X1, X2)
  , take(X1, X2) -> n__take(X1, X2)
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , take(0(), XS) -> nil()
  , sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
  , sel(0(), cons(X, XS)) -> 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)
    , activate(n__take(X1, X2)) -> take(X1, X2)
    , take(X1, X2) -> n__take(X1, X2)
    , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
    , take(0(), XS) -> nil() }

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()
  , head^#(cons(X, XS)) -> c_3()
  , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X))
  , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
  , take^#(X1, X2) -> c_8()
  , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
  , take^#(0(), XS) -> c_10()
  , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS)))
  , sel^#(0(), cons(X, XS)) -> c_12() }
Strict Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__take(X1, X2)) -> take(X1, X2)
  , take(X1, X2) -> n__take(X1, X2)
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , take(0(), XS) -> nil() }
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(cons) = {2}, Uargs(n__take) = {2}, Uargs(head^#) = {1},
  Uargs(c_4) = {1}, Uargs(c_6) = {1}, Uargs(c_7) = {1},
  Uargs(c_9) = {1}, Uargs(sel^#) = {2}, Uargs(c_11) = {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 + [2]         
                                           
     [activate](x1) = [1] x1 + [2]         
                                           
     [take](x1, x2) = [1] x1 + [1] x2 + [2]
                                           
                [0] = [1]                  
                                           
              [nil] = [2]                  
                                           
  [n__take](x1, x2) = [1] x1 + [1] x2 + [1]
                                           
       [from^#](x1) = [1]                  
                                           
              [c_1] = [0]                  
                                           
              [c_2] = [0]                  
                                           
       [head^#](x1) = [1] x1 + [2]         
                                           
              [c_3] = [1]                  
                                           
        [2nd^#](x1) = [1] x1 + [2]         
                                           
          [c_4](x1) = [1] x1 + [2]         
                                           
   [activate^#](x1) = [2] x1 + [2]         
                                           
              [c_5] = [1]                  
                                           
          [c_6](x1) = [1] x1 + [2]         
                                           
          [c_7](x1) = [1] x1 + [1]         
                                           
   [take^#](x1, x2) = [1] x1 + [2] x2 + [2]
                                           
              [c_8] = [1]                  
                                           
          [c_9](x1) = [1] x1 + [1]         
                                           
             [c_10] = [2]                  
                                           
    [sel^#](x1, x2) = [2] x1 + [2] x2 + [2]
                                           
         [c_11](x1) = [1] x1 + [1]         
                                           
             [c_12] = [1]                  

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)]                          
                                                                   
  [activate(n__take(X1, X2))] = [1] X1 + [1] X2 + [3]              
                              > [1] X1 + [1] X2 + [2]              
                              = [take(X1, X2)]                     
                                                                   
               [take(X1, X2)] = [1] X1 + [1] X2 + [2]              
                              > [1] X1 + [1] X2 + [1]              
                              = [n__take(X1, X2)]                  
                                                                   
    [take(s(N), cons(X, XS))] = [1] XS + [1] N + [4]               
                              > [1] XS + [1] N + [3]               
                              = [cons(X, n__take(N, activate(XS)))]
                                                                   
              [take(0(), XS)] = [1] XS + [3]                       
                              > [2]                                
                              = [nil()]                            
                                                                   

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:
  { 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
  , sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) }
Weak DPs:
  { from^#(X) -> c_1()
  , from^#(X) -> c_2()
  , head^#(cons(X, XS)) -> c_3()
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X))
  , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
  , take^#(X1, X2) -> c_8()
  , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
  , take^#(0(), XS) -> c_10()
  , sel^#(0(), cons(X, XS)) -> c_12() }
Weak Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__take(X1, X2)) -> take(X1, X2)
  , take(X1, X2) -> n__take(X1, X2)
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , take(0(), XS) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {1} by applications of
Pre({1}) = {}. Here rules are labeled as follows:

  DPs:
    { 1: 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
    , 2: sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS)))
    , 3: from^#(X) -> c_1()
    , 4: from^#(X) -> c_2()
    , 5: head^#(cons(X, XS)) -> c_3()
    , 6: activate^#(X) -> c_5()
    , 7: activate^#(n__from(X)) -> c_6(from^#(X))
    , 8: activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
    , 9: take^#(X1, X2) -> c_8()
    , 10: take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
    , 11: take^#(0(), XS) -> c_10()
    , 12: sel^#(0(), cons(X, XS)) -> c_12() }

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

Strict DPs:
  { sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) }
Weak DPs:
  { from^#(X) -> c_1()
  , from^#(X) -> c_2()
  , head^#(cons(X, XS)) -> c_3()
  , 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
  , activate^#(X) -> c_5()
  , activate^#(n__from(X)) -> c_6(from^#(X))
  , activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
  , take^#(X1, X2) -> c_8()
  , take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
  , take^#(0(), XS) -> c_10()
  , sel^#(0(), cons(X, XS)) -> c_12() }
Weak Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__take(X1, X2)) -> take(X1, X2)
  , take(X1, X2) -> n__take(X1, X2)
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , take(0(), XS) -> nil() }
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()
, head^#(cons(X, XS)) -> c_3()
, 2nd^#(cons(X, XS)) -> c_4(head^#(activate(XS)))
, activate^#(X) -> c_5()
, activate^#(n__from(X)) -> c_6(from^#(X))
, activate^#(n__take(X1, X2)) -> c_7(take^#(X1, X2))
, take^#(X1, X2) -> c_8()
, take^#(s(N), cons(X, XS)) -> c_9(activate^#(XS))
, take^#(0(), XS) -> c_10()
, sel^#(0(), cons(X, XS)) -> c_12() }

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

Strict DPs:
  { sel^#(s(N), cons(X, XS)) -> c_11(sel^#(N, activate(XS))) }
Weak Trs:
  { from(X) -> cons(X, n__from(s(X)))
  , from(X) -> n__from(X)
  , activate(X) -> X
  , activate(n__from(X)) -> from(X)
  , activate(n__take(X1, X2)) -> take(X1, X2)
  , take(X1, X2) -> n__take(X1, X2)
  , take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
  , take(0(), XS) -> nil() }
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(take) = {}, safe(0) = {},
 safe(nil) = {}, safe(n__take) = {1, 2}, safe(sel^#) = {2},
 safe(c_11) = {}

and precedence

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

Following symbols are considered recursive:

 {from, activate, take, 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(take) = [], pi(0) = [], pi(nil) = [],
 pi(n__take) = [1], pi(sel^#) = [1], pi(c_11) = [1]

Usable defined function symbols are a subset of:

 {sel^#}

For your convenience, here are the satisfied ordering constraints:

  pi(sel^#(s(N), cons(X, XS))) = sel^#(s(; N);)                  
                               > c_11(sel^#(N;);)                
                               = pi(c_11(sel^#(N, activate(XS))))
                                                                 

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