(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, x3)
activate(x1)  =  activate(x1)
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 [LPO].
Precedence:
tail1 > activate1 > natsFrom1 > ns1
tail1 > activate1 > natsFrom1 > nnatsFrom1
tail1 > activate1 > natsFrom1 > cons2
tail1 > activate1 > s1 > ns1
sel2 > head1
sel2 > afterNth2 > splitAt2 > nil
sel2 > afterNth2 > splitAt2 > u2 > pair2 > cons2
sel2 > afterNth2 > splitAt2 > u2 > activate1 > natsFrom1 > ns1
sel2 > afterNth2 > splitAt2 > u2 > activate1 > natsFrom1 > nnatsFrom1
sel2 > afterNth2 > splitAt2 > u2 > activate1 > natsFrom1 > cons2
sel2 > afterNth2 > splitAt2 > u2 > activate1 > s1 > ns1
take2 > splitAt2 > nil
take2 > splitAt2 > u2 > pair2 > cons2
take2 > splitAt2 > u2 > activate1 > natsFrom1 > ns1
take2 > splitAt2 > u2 > activate1 > natsFrom1 > nnatsFrom1
take2 > splitAt2 > u2 > activate1 > natsFrom1 > cons2
take2 > splitAt2 > u2 > activate1 > s1 > ns1

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)
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)  =  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, 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 [LPO].
Precedence:
tail1 > activate1 > s1 > ns1
tail1 > activate1 > natsFrom1 > cons2
tail1 > activate1 > natsFrom1 > nnatsFrom1
tail1 > activate1 > natsFrom1 > ns1
sel2 > head1
sel2 > afterNth2 > splitAt2 > nil
sel2 > afterNth2 > splitAt2 > u3 > activate1 > s1 > ns1
sel2 > afterNth2 > splitAt2 > u3 > activate1 > natsFrom1 > cons2
sel2 > afterNth2 > splitAt2 > u3 > activate1 > natsFrom1 > nnatsFrom1
sel2 > afterNth2 > splitAt2 > u3 > activate1 > natsFrom1 > ns1
sel2 > afterNth2 > splitAt2 > u3 > pair2
take2 > splitAt2 > nil
take2 > splitAt2 > u3 > activate1 > s1 > ns1
take2 > splitAt2 > u3 > activate1 > natsFrom1 > cons2
take2 > splitAt2 > u3 > activate1 > natsFrom1 > nnatsFrom1
take2 > splitAt2 > u3 > activate1 > natsFrom1 > ns1
take2 > splitAt2 > u3 > pair2

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