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:

U11(tt, N, X, XS) → U12(splitAt(activate(N), activate(XS)), activate(X))
U12(pair(YS, ZS), X) → pair(cons(activate(X), YS), ZS)
afterNth(N, XS) → snd(splitAt(N, XS))
and(tt, X) → activate(X)
fst(pair(X, Y)) → X
head(cons(N, XS)) → N
natsFrom(N) → cons(N, n__natsFrom(s(N)))
sel(N, XS) → head(afterNth(N, XS))
snd(pair(X, Y)) → Y
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → U11(tt, N, X, activate(XS))
tail(cons(N, XS)) → activate(XS)
take(N, XS) → fst(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:

U11¹(tt, N, X, XS) → U12¹(splitAt(activate(N), activate(XS)), activate(X))
U11¹(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))
U11¹(tt, N, X, XS) → ACTIVATE(N)
U11¹(tt, N, X, XS) → ACTIVATE(XS)
U11¹(tt, N, X, XS) → ACTIVATE(X)
U12¹(pair(YS, ZS), X) → ACTIVATE(X)
AFTERNTH(N, XS) → SND(splitAt(N, XS))
AFTERNTH(N, XS) → SPLITAT(N, XS)
AND(tt, X) → ACTIVATE(X)
SEL(N, XS) → HEAD(afterNth(N, XS))
SEL(N, XS) → AFTERNTH(N, XS)
SPLITAT(s(N), cons(X, XS)) → U11¹(tt, N, X, activate(XS))
SPLITAT(s(N), cons(X, XS)) → ACTIVATE(XS)
TAIL(cons(N, XS)) → ACTIVATE(XS)
TAKE(N, XS) → FST(splitAt(N, XS))
TAKE(N, XS) → SPLITAT(N, XS)
ACTIVATE(n__natsFrom(X)) → NATSFROM(X)

The TRS R consists of the following rules:

U11(tt, N, X, XS) → U12(splitAt(activate(N), activate(XS)), activate(X))
U12(pair(YS, ZS), X) → pair(cons(activate(X), YS), ZS)
afterNth(N, XS) → snd(splitAt(N, XS))
and(tt, X) → activate(X)
fst(pair(X, Y)) → X
head(cons(N, XS)) → N
natsFrom(N) → cons(N, n__natsFrom(s(N)))
sel(N, XS) → head(afterNth(N, XS))
snd(pair(X, Y)) → Y
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → U11(tt, N, X, activate(XS))
tail(cons(N, XS)) → activate(XS)
take(N, XS) → fst(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 15 less nodes.

(4) Obligation:

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

U11¹(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))
SPLITAT(s(N), cons(X, XS)) → U11¹(tt, N, X, activate(XS))

The TRS R consists of the following rules:

U11(tt, N, X, XS) → U12(splitAt(activate(N), activate(XS)), activate(X))
U12(pair(YS, ZS), X) → pair(cons(activate(X), YS), ZS)
afterNth(N, XS) → snd(splitAt(N, XS))
and(tt, X) → activate(X)
fst(pair(X, Y)) → X
head(cons(N, XS)) → N
natsFrom(N) → cons(N, n__natsFrom(s(N)))
sel(N, XS) → head(afterNth(N, XS))
snd(pair(X, Y)) → Y
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → U11(tt, N, X, activate(XS))
tail(cons(N, XS)) → activate(XS)
take(N, XS) → fst(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)) → U11¹(tt, N, X, activate(XS))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2, x3, x4)  =  U11¹(x2)
tt  =  tt
SPLITAT(x1, x2)  =  x1
activate(x1)  =  activate(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  x2
n__natsFrom(x1)  =  n__natsFrom
natsFrom(x1)  =  natsFrom

Recursive Path Order [RPO].
Precedence:

s₁ > [U11^1₁, tt, activate₁, n_natsFrom, natsFrom]

The following usable rules [FROCOS05] were oriented:

activate(n__natsFrom(X)) → natsFrom(X)
activate(X) → X
natsFrom(N) → cons(N, n__natsFrom(s(N)))
natsFrom(X) → n__natsFrom(X)

(6) Obligation:

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

U11¹(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))

The TRS R consists of the following rules:

U11(tt, N, X, XS) → U12(splitAt(activate(N), activate(XS)), activate(X))
U12(pair(YS, ZS), X) → pair(cons(activate(X), YS), ZS)
afterNth(N, XS) → snd(splitAt(N, XS))
and(tt, X) → activate(X)
fst(pair(X, Y)) → X
head(cons(N, XS)) → N
natsFrom(N) → cons(N, n__natsFrom(s(N)))
sel(N, XS) → head(afterNth(N, XS))
snd(pair(X, Y)) → Y
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → U11(tt, N, X, activate(XS))
tail(cons(N, XS)) → activate(XS)
take(N, XS) → fst(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) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(8) TRUE