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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(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) 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(activate(X))
ACTIVATE(n__natsFrom(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 12 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__natsFrom(X)) → ACTIVATE(X)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
ACTIVATE1 > cons2
[splitAt2, sel2, afterNth2, take2] > nil > cons2
[splitAt2, sel2, afterNth2, take2] > u4 > pair2 > cons2
[splitAt2, sel2, afterNth2, take2] > u4 > [activate1, tail1] > natsFrom1 > [ns1, s1] > cons2
[splitAt2, sel2, afterNth2, take2] > u4 > [activate1, tail1] > natsFrom1 > nnatsFrom1 > cons2
0 > cons2

Status:
sel2: [1,2]
tail1: [1]
afterNth2: [1,2]
nnatsFrom1: [1]
activate1: [1]
ns1: [1]
take2: [1,2]
0: []
splitAt2: [1,2]
cons2: [2,1]
u4: [1,4,2,3]
pair2: [1,2]
s1: [1]
nil: []
natsFrom1: [1]
ACTIVATE1: [1]


The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(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.
We have to consider all minimal (P,Q,R)-chains.

(11) 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)
n__s(x1)  =  n__s(x1)
fst(x1)  =  fst(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)
head(x1)  =  head(x1)
tail(x1)  =  tail(x1)
sel(x1, x2)  =  sel(x1, x2)
afterNth(x1, x2)  =  afterNth(x1, x2)
take(x1, x2)  =  take(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
SPLITAT2 > activate1 > natsFrom1 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]
0 > pair2 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]
tail1 > activate1 > natsFrom1 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]
sel2 > [splitAt2, u4, afterNth2] > activate1 > natsFrom1 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]
sel2 > [splitAt2, u4, afterNth2] > pair2 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]
take2 > [splitAt2, u4, afterNth2] > activate1 > natsFrom1 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]
take2 > [splitAt2, u4, afterNth2] > pair2 > [s1, cons2, nnatsFrom1, ns1, fst1, nil, head1]

Status:
sel2: [2,1]
tail1: [1]
afterNth2: [1,2]
head1: [1]
nnatsFrom1: [1]
activate1: [1]
ns1: [1]
take2: [1,2]
splitAt2: [1,2]
0: []
cons2: [1,2]
SPLITAT2: [1,2]
u4: [2,3,4,1]
fst1: [1]
pair2: [2,1]
s1: [1]
natsFrom1: [1]
nil: []


The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE