MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { U11(tt(), N, XS) -> U12(tt(), N, XS)
  , U12(tt(), N, XS) -> snd(splitAt(N, XS))
  , snd(pair(X, Y)) -> U51(tt(), Y)
  , splitAt(s(N), cons(X, XS)) -> U61(tt(), N, X, XS)
  , splitAt(0(), XS) -> pair(nil(), XS)
  , U21(tt(), X) -> U22(tt(), X)
  , U22(tt(), X) -> X
  , U31(tt(), N) -> U32(tt(), N)
  , U32(tt(), N) -> N
  , U41(tt(), N, XS) -> U42(tt(), N, XS)
  , U42(tt(), N, XS) -> head(afterNth(N, XS))
  , head(cons(N, XS)) -> U31(tt(), N)
  , afterNth(N, XS) -> U11(tt(), N, XS)
  , U51(tt(), Y) -> U52(tt(), Y)
  , U52(tt(), Y) -> Y
  , U61(tt(), N, X, XS) -> U62(tt(), N, X, XS)
  , U62(tt(), N, X, XS) -> U63(tt(), N, X, XS)
  , U63(tt(), N, X, XS) -> U64(splitAt(N, XS), X)
  , U64(pair(YS, ZS), X) -> pair(cons(X, YS), ZS)
  , U71(tt(), XS) -> U72(tt(), XS)
  , U72(tt(), XS) -> XS
  , U81(tt(), N, XS) -> U82(tt(), N, XS)
  , U82(tt(), N, XS) -> fst(splitAt(N, XS))
  , fst(pair(X, Y)) -> U21(tt(), X)
  , natsFrom(N) -> cons(N, natsFrom(s(N)))
  , sel(N, XS) -> U41(tt(), N, XS)
  , tail(cons(N, XS)) -> U71(tt(), XS)
  , take(N, XS) -> U81(tt(), N, XS) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..