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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(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.


U111(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))
SPLITAT(s(N), cons(X, XS)) → U111(tt, N, X, activate(XS))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
U111(x0, x1, x2, x3, x4)  =  U111(x1, x2)
SPLITAT(x0, x1, x2)  =  SPLITAT(x1)

Tags:
U111 has argument tags [1,6,7,5,0] and root tag 0
SPLITAT has argument tags [0,6,5] and root tag 1

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(SPLITAT(x1, x2)) = 1 + x2   
POL(U111(x1, x2, x3, x4)) = x3   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = 0   
POL(n__natsFrom(x1)) = 0   
POL(natsFrom(x1)) = 0   
POL(s(x1)) = 1 + x1   
POL(tt) = 1   

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:
P is empty.
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) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE