Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

natsFrom(N) → cons(N, n__natsFrom(n__s(N)))
fst(pair(XS, YS)) → XS
snd(pair(XS, YS)) → YS
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → u(splitAt(N, activate(XS)), N, X, activate(XS))
u(pair(YS, ZS), N, X, XS) → pair(cons(activate(X), YS), ZS)
head(cons(N, XS)) → N
tail(cons(N, XS)) → activate(XS)
sel(N, XS) → head(afterNth(N, XS))
take(N, XS) → fst(splitAt(N, XS))
afterNth(N, XS) → snd(splitAt(N, XS))
natsFrom(X) → n__natsFrom(X)
s(X) → n__s(X)
activate(n__natsFrom(X)) → natsFrom(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic Path Order [LPO].
Precedence:

0 > [cons₂, n_natsFrom1, n_s1, fst₁]
sel₂ > head₁ > [cons₂, n_natsFrom1, n_s1, fst₁]
sel₂ > afterNth₂ > snd₁ > [cons₂, n_natsFrom1, n_s1, fst₁]
sel₂ > afterNth₂ > splitAt₂ > nil > [cons₂, n_natsFrom1, n_s1, fst₁]
sel₂ > afterNth₂ > splitAt₂ > u₄ > [pair₂, activate₁, tail₁] > natsFrom₁ > [cons₂, n_natsFrom1, n_s1, fst₁]
sel₂ > afterNth₂ > splitAt₂ > u₄ > [pair₂, activate₁, tail₁] > s₁ > [cons₂, n_natsFrom1, n_s1, fst₁]
take₂ > splitAt₂ > nil > [cons₂, n_natsFrom1, n_s1, fst₁]
take₂ > splitAt₂ > u₄ > [pair₂, activate₁, tail₁] > natsFrom₁ > [cons₂, n_natsFrom1, n_s1, fst₁]
take₂ > splitAt₂ > u₄ > [pair₂, activate₁, tail₁] > s₁ > [cons₂, n_natsFrom1, n_s1, fst₁]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

natsFrom(N) → cons(N, n__natsFrom(n__s(N)))
fst(pair(XS, YS)) → XS
snd(pair(XS, YS)) → YS
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → u(splitAt(N, activate(XS)), N, X, activate(XS))
u(pair(YS, ZS), N, X, XS) → pair(cons(activate(X), YS), ZS)
head(cons(N, XS)) → N
tail(cons(N, XS)) → activate(XS)
sel(N, XS) → head(afterNth(N, XS))
take(N, XS) → fst(splitAt(N, XS))
afterNth(N, XS) → snd(splitAt(N, XS))
natsFrom(X) → n__natsFrom(X)
s(X) → n__s(X)
activate(n__natsFrom(X)) → natsFrom(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE