YES(?,O(n^1))

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

Strict Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { natsFrom^#(N) -> c_1()
  , natsFrom^#(X) -> c_2()
  , fst^#(pair(XS, YS)) -> c_3()
  , snd^#(pair(XS, YS)) -> c_4()
  , splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , splitAt^#(0(), XS) -> c_6()
  , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
  , activate^#(X) -> c_8()
  , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
  , head^#(cons(N, XS)) -> c_10()
  , tail^#(cons(N, XS)) -> c_11(activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }

and mark the set of starting terms.

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

Strict DPs:
  { natsFrom^#(N) -> c_1()
  , natsFrom^#(X) -> c_2()
  , fst^#(pair(XS, YS)) -> c_3()
  , snd^#(pair(XS, YS)) -> c_4()
  , splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , splitAt^#(0(), XS) -> c_6()
  , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
  , activate^#(X) -> c_8()
  , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
  , head^#(cons(N, XS)) -> c_10()
  , tail^#(cons(N, XS)) -> c_11(activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {1,2,3,4,6,8,10} by
applications of Pre({1,2,3,4,6,8,10}) = {5,7,9,11,12,13,14}. Here
rules are labeled as follows:

  DPs:
    { 1: natsFrom^#(N) -> c_1()
    , 2: natsFrom^#(X) -> c_2()
    , 3: fst^#(pair(XS, YS)) -> c_3()
    , 4: snd^#(pair(XS, YS)) -> c_4()
    , 5: splitAt^#(s(N), cons(X, XS)) ->
         c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
             splitAt^#(N, activate(XS)),
             activate^#(XS),
             activate^#(XS))
    , 6: splitAt^#(0(), XS) -> c_6()
    , 7: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
    , 8: activate^#(X) -> c_8()
    , 9: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
    , 10: head^#(cons(N, XS)) -> c_10()
    , 11: tail^#(cons(N, XS)) -> c_11(activate^#(XS))
    , 12: sel^#(N, XS) ->
          c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
    , 13: afterNth^#(N, XS) ->
          c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
    , 14: take^#(N, XS) ->
          c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }

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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
  , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
  , tail^#(cons(N, XS)) -> c_11(activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }
Weak DPs:
  { natsFrom^#(N) -> c_1()
  , natsFrom^#(X) -> c_2()
  , fst^#(pair(XS, YS)) -> c_3()
  , snd^#(pair(XS, YS)) -> c_4()
  , splitAt^#(0(), XS) -> c_6()
  , activate^#(X) -> c_8()
  , head^#(cons(N, XS)) -> c_10() }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {3} by applications of
Pre({3}) = {1,2,4}. Here rules are labeled as follows:

  DPs:
    { 1: splitAt^#(s(N), cons(X, XS)) ->
         c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
             splitAt^#(N, activate(XS)),
             activate^#(XS),
             activate^#(XS))
    , 2: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
    , 3: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
    , 4: tail^#(cons(N, XS)) -> c_11(activate^#(XS))
    , 5: sel^#(N, XS) ->
         c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
    , 6: afterNth^#(N, XS) ->
         c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
    , 7: take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS))
    , 8: natsFrom^#(N) -> c_1()
    , 9: natsFrom^#(X) -> c_2()
    , 10: fst^#(pair(XS, YS)) -> c_3()
    , 11: snd^#(pair(XS, YS)) -> c_4()
    , 12: splitAt^#(0(), XS) -> c_6()
    , 13: activate^#(X) -> c_8()
    , 14: head^#(cons(N, XS)) -> c_10() }

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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
  , tail^#(cons(N, XS)) -> c_11(activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }
Weak DPs:
  { natsFrom^#(N) -> c_1()
  , natsFrom^#(X) -> c_2()
  , fst^#(pair(XS, YS)) -> c_3()
  , snd^#(pair(XS, YS)) -> c_4()
  , splitAt^#(0(), XS) -> c_6()
  , activate^#(X) -> c_8()
  , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
  , head^#(cons(N, XS)) -> c_10() }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {2,3} by applications of
Pre({2,3}) = {1}. Here rules are labeled as follows:

  DPs:
    { 1: splitAt^#(s(N), cons(X, XS)) ->
         c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
             splitAt^#(N, activate(XS)),
             activate^#(XS),
             activate^#(XS))
    , 2: u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
    , 3: tail^#(cons(N, XS)) -> c_11(activate^#(XS))
    , 4: sel^#(N, XS) ->
         c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
    , 5: afterNth^#(N, XS) ->
         c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
    , 6: take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS))
    , 7: natsFrom^#(N) -> c_1()
    , 8: natsFrom^#(X) -> c_2()
    , 9: fst^#(pair(XS, YS)) -> c_3()
    , 10: snd^#(pair(XS, YS)) -> c_4()
    , 11: splitAt^#(0(), XS) -> c_6()
    , 12: activate^#(X) -> c_8()
    , 13: activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
    , 14: head^#(cons(N, XS)) -> c_10() }

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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }
Weak DPs:
  { natsFrom^#(N) -> c_1()
  , natsFrom^#(X) -> c_2()
  , fst^#(pair(XS, YS)) -> c_3()
  , snd^#(pair(XS, YS)) -> c_4()
  , splitAt^#(0(), XS) -> c_6()
  , u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
  , activate^#(X) -> c_8()
  , activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
  , head^#(cons(N, XS)) -> c_10()
  , tail^#(cons(N, XS)) -> c_11(activate^#(XS)) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
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.

{ natsFrom^#(N) -> c_1()
, natsFrom^#(X) -> c_2()
, fst^#(pair(XS, YS)) -> c_3()
, snd^#(pair(XS, YS)) -> c_4()
, splitAt^#(0(), XS) -> c_6()
, u^#(pair(YS, ZS), N, X, XS) -> c_7(activate^#(X))
, activate^#(X) -> c_8()
, activate^#(n__natsFrom(X)) -> c_9(natsFrom^#(X))
, head^#(cons(N, XS)) -> c_10()
, tail^#(cons(N, XS)) -> c_11(activate^#(XS)) }

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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { splitAt^#(s(N), cons(X, XS)) ->
    c_5(u^#(splitAt(N, activate(XS)), N, X, activate(XS)),
        splitAt^#(N, activate(XS)),
        activate^#(XS),
        activate^#(XS))
  , sel^#(N, XS) -> c_12(head^#(afterNth(N, XS)), afterNth^#(N, XS))
  , afterNth^#(N, XS) ->
    c_13(snd^#(splitAt(N, XS)), splitAt^#(N, XS))
  , take^#(N, XS) -> c_14(fst^#(splitAt(N, XS)), splitAt^#(N, XS)) }

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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) -> c_1(splitAt^#(N, activate(XS)))
  , sel^#(N, XS) -> c_2(afterNth^#(N, XS))
  , afterNth^#(N, XS) -> c_3(splitAt^#(N, XS))
  , take^#(N, XS) -> c_4(splitAt^#(N, XS)) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , fst(pair(XS, YS)) -> XS
  , snd(pair(XS, YS)) -> YS
  , splitAt(s(N), cons(X, XS)) ->
    u(splitAt(N, activate(XS)), N, X, activate(XS))
  , splitAt(0(), XS) -> pair(nil(), XS)
  , u(pair(YS, ZS), N, X, XS) -> pair(cons(activate(X), YS), ZS)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X)
  , head(cons(N, XS)) -> N
  , tail(cons(N, XS)) -> activate(XS)
  , sel(N, XS) -> head(afterNth(N, XS))
  , afterNth(N, XS) -> snd(splitAt(N, XS))
  , take(N, XS) -> fst(splitAt(N, XS)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
    , natsFrom(X) -> n__natsFrom(X)
    , activate(X) -> X
    , activate(n__natsFrom(X)) -> natsFrom(X) }

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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) -> c_1(splitAt^#(N, activate(XS)))
  , sel^#(N, XS) -> c_2(afterNth^#(N, XS))
  , afterNth^#(N, XS) -> c_3(splitAt^#(N, XS))
  , take^#(N, XS) -> c_4(splitAt^#(N, XS)) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Consider the dependency graph

  1: splitAt^#(s(N), cons(X, XS)) -> c_1(splitAt^#(N, activate(XS)))
     -->_1 splitAt^#(s(N), cons(X, XS)) ->
           c_1(splitAt^#(N, activate(XS))) :1
  
  2: sel^#(N, XS) -> c_2(afterNth^#(N, XS))
     -->_1 afterNth^#(N, XS) -> c_3(splitAt^#(N, XS)) :3
  
  3: afterNth^#(N, XS) -> c_3(splitAt^#(N, XS))
     -->_1 splitAt^#(s(N), cons(X, XS)) ->
           c_1(splitAt^#(N, activate(XS))) :1
  
  4: take^#(N, XS) -> c_4(splitAt^#(N, XS))
     -->_1 splitAt^#(s(N), cons(X, XS)) ->
           c_1(splitAt^#(N, activate(XS))) :1
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { sel^#(N, XS) -> c_2(afterNth^#(N, XS))
  , take^#(N, XS) -> c_4(splitAt^#(N, XS)) }


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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) -> c_1(splitAt^#(N, activate(XS)))
  , afterNth^#(N, XS) -> c_3(splitAt^#(N, XS)) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(X) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Consider the dependency graph

  1: splitAt^#(s(N), cons(X, XS)) -> c_1(splitAt^#(N, activate(XS)))
     -->_1 splitAt^#(s(N), cons(X, XS)) ->
           c_1(splitAt^#(N, activate(XS))) :1
  
  2: afterNth^#(N, XS) -> c_3(splitAt^#(N, XS))
     -->_1 splitAt^#(s(N), cons(X, XS)) ->
           c_1(splitAt^#(N, activate(XS))) :1
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { afterNth^#(N, XS) -> c_3(splitAt^#(N, XS)) }


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

Strict DPs:
  { splitAt^#(s(N), cons(X, XS)) -> c_1(splitAt^#(N, activate(XS))) }
Weak Trs:
  { natsFrom(N) -> cons(N, n__natsFrom(s(N)))
  , natsFrom(X) -> n__natsFrom(X)
  , activate(X) -> X
  , activate(n__natsFrom(X)) -> natsFrom(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(natsFrom) = {}, safe(cons) = {1, 2}, safe(n__natsFrom) = {1},
 safe(s) = {1}, safe(activate) = {}, safe(splitAt^#) = {2},
 safe(c_1) = {}

and precedence

 natsFrom ~ activate, natsFrom ~ splitAt^#, activate ~ splitAt^# .

Following symbols are considered recursive:

 {natsFrom, activate, splitAt^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(natsFrom) = [], pi(cons) = [], pi(n__natsFrom) = [1],
 pi(s) = [1], pi(activate) = [], pi(splitAt^#) = [1], pi(c_1) = [1]

Usable defined function symbols are a subset of:

 {splitAt^#}

For your convenience, here are the satisfied ordering constraints:

  pi(splitAt^#(s(N), cons(X, XS))) = splitAt^#(s(; N);)                 
                                   > c_1(splitAt^#(N;);)                
                                   = pi(c_1(splitAt^#(N, activate(XS))))
                                                                        

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