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(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)
activate(n__natsFrom(X)) → natsFrom(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
natsFrom(x1)  =  natsFrom(x1)
cons(x1, x2)  =  cons(x1, x2)
n__natsFrom(x1)  =  x1
s(x1)  =  s(x1)
fst(x1)  =  x1
pair(x1, x2)  =  pair(x1, x2)
snd(x1)  =  x1
splitAt(x1, x2)  =  splitAt(x1, x2)
0  =  0
nil  =  nil
u(x1, x2, x3, x4)  =  u(x1, x2, x3, x4)
activate(x1)  =  activate(x1)
head(x1)  =  x1
tail(x1)  =  tail(x1)
sel(x1, x2)  =  sel(x1, x2)
afterNth(x1, x2)  =  afterNth(x1, x2)
take(x1, x2)  =  take(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

[0, nil] > pair₂ > cons₂
tail₁ > [natsFrom₁, activate₁] > cons₂
tail₁ > [natsFrom₁, activate₁] > s₁
[sel₂, afterNth₂] > [splitAt₂, take₂] > u₄ > [natsFrom₁, activate₁] > cons₂
[sel₂, afterNth₂] > [splitAt₂, take₂] > u₄ > [natsFrom₁, activate₁] > s₁
[sel₂, afterNth₂] > [splitAt₂, take₂] > u₄ > pair₂ > cons₂

Status:

sel₂: multiset
afterNth₂: multiset
tail₁: multiset
activate₁: [1]
take₂: [1,2]
0: multiset
splitAt₂: [1,2]
cons₂: multiset
u₄: [1,2,4,3]
pair₂: multiset
s₁: multiset
nil: multiset
natsFrom₁: [1]

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(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)
afterNth(N, XS) → snd(splitAt(N, XS))
natsFrom(X) → n__natsFrom(X)
activate(X) → X

(2) Obligation:

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

sel(N, XS) → head(afterNth(N, XS))
take(N, XS) → fst(splitAt(N, XS))
activate(n__natsFrom(X)) → natsFrom(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x₁)) = 1 + x₁   
POL(afterNth(x₁, x₂)) = x₁ + x₂   
POL(fst(x₁)) = x₁   
POL(head(x₁)) = x₁   
POL(n__natsFrom(x₁)) = x₁   
POL(natsFrom(x₁)) = x₁   
POL(sel(x₁, x₂)) = 1 + x₁ + x₂   
POL(splitAt(x₁, x₂)) = x₁ + x₂   
POL(take(x₁, x₂)) = 1 + x₁ + x₂   

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

sel(N, XS) → head(afterNth(N, XS))
take(N, XS) → fst(splitAt(N, XS))
activate(n__natsFrom(X)) → natsFrom(X)

(4) Obligation:

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

(5) RisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE

(7) RisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE

(9) RisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE