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) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SPLITAT(s(N), cons(X, XS)) → U(splitAt(N, activate(XS)), N, X, activate(XS))
SPLITAT(s(N), cons(X, XS)) → SPLITAT(N, activate(XS))
SPLITAT(s(N), cons(X, XS)) → ACTIVATE(XS)
U(pair(YS, ZS), N, X, XS) → ACTIVATE(X)
TAIL(cons(N, XS)) → ACTIVATE(XS)
SEL(N, XS) → HEAD(afterNth(N, XS))
SEL(N, XS) → AFTERNTH(N, XS)
TAKE(N, XS) → FST(splitAt(N, XS))
TAKE(N, XS) → SPLITAT(N, XS)
AFTERNTH(N, XS) → SND(splitAt(N, XS))
AFTERNTH(N, XS) → SPLITAT(N, XS)
ACTIVATE(n__natsFrom(X)) → NATSFROM(X)

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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 11 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SPLITAT(s(N), cons(X, XS)) → SPLITAT(N, activate(XS))

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.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SPLITAT(s(N), cons(X, XS)) → SPLITAT(N, activate(XS))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SPLITAT(x1, x2)  =  SPLITAT(x1, x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
activate(x1)  =  activate(x1)
natsFrom(x1)  =  natsFrom(x1)
n__natsFrom(x1)  =  n__natsFrom(x1)
fst(x1)  =  x1
pair(x1, x2)  =  pair(x1, x2)
snd(x1)  =  snd(x1)
splitAt(x1, x2)  =  splitAt(x1, x2)
0  =  0
nil  =  nil
u(x1, x2, x3, x4)  =  u(x1, x2, x3, x4)
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:

SPLITAT₂ > [activate₁, natsFrom₁] > s₁
SPLITAT₂ > [activate₁, natsFrom₁] > cons₂
SPLITAT₂ > [activate₁, natsFrom₁] > n_natsFrom1
tail₁ > [activate₁, natsFrom₁] > s₁
tail₁ > [activate₁, natsFrom₁] > cons₂
tail₁ > [activate₁, natsFrom₁] > n_natsFrom1
[sel₂, afterNth₂] > snd₁
[sel₂, afterNth₂] > [splitAt₂, u₄, take₂] > [activate₁, natsFrom₁] > s₁
[sel₂, afterNth₂] > [splitAt₂, u₄, take₂] > [activate₁, natsFrom₁] > cons₂
[sel₂, afterNth₂] > [splitAt₂, u₄, take₂] > [activate₁, natsFrom₁] > n_natsFrom1
[sel₂, afterNth₂] > [splitAt₂, u₄, take₂] > [pair₂, 0] > nil

Status:

SPLITAT₂: [1,2]
s₁: multiset
cons₂: multiset
activate₁: [1]
natsFrom₁: [1]
n_natsFrom1: multiset
pair₂: multiset
snd₁: multiset
splitAt₂: [1,2]
0: multiset
nil: multiset
u₄: [2,3,1,4]
tail₁: multiset
sel₂: multiset
afterNth₂: multiset
take₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE