Termination w.r.t. Q of the following Term Rewriting System could be proven:

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(n__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)
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.


↳ QTRS
  ↳ DependencyPairsProof

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(n__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)
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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

U111(tt, N, X, XS) → ACTIVATE(XS)
ACTIVATE(n__natsFrom(X)) → NATSFROM(activate(X))
AFTERNTH(N, XS) → SPLITAT(N, XS)
U111(tt, N, X, XS) → ACTIVATE(N)
SEL(N, XS) → AFTERNTH(N, XS)
SPLITAT(s(N), cons(X, XS)) → U111(tt, N, X, activate(XS))
ACTIVATE(n__s(X)) → S(activate(X))
SPLITAT(s(N), cons(X, XS)) → ACTIVATE(XS)
U111(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))
AND(tt, X) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
TAKE(N, XS) → SPLITAT(N, XS)
U111(tt, N, X, XS) → ACTIVATE(X)
ACTIVATE(n__natsFrom(X)) → ACTIVATE(X)
U121(pair(YS, ZS), X) → ACTIVATE(X)
AFTERNTH(N, XS) → SND(splitAt(N, XS))
U111(tt, N, X, XS) → U121(splitAt(activate(N), activate(XS)), activate(X))
TAKE(N, XS) → FST(splitAt(N, XS))
TAIL(cons(N, XS)) → ACTIVATE(XS)
SEL(N, XS) → HEAD(afterNth(N, 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(n__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)
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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 16 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

ACTIVATE(n__natsFrom(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(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(n__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)
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__natsFrom(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__natsFrom(x1)) = 4 + (4)x_1   
POL(n__s(x1)) = 4 + x_1   
POL(ACTIVATE(x1)) = (4)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

SPLITAT(s(N), cons(X, XS)) → U111(tt, N, X, activate(XS))
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(n__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)
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


U111(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))
The remaining pairs can at least be oriented weakly.

SPLITAT(s(N), cons(X, XS)) → U111(tt, N, X, activate(XS))
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = (1/4)x_2   
POL(tt) = 2   
POL(SPLITAT(x1, x2)) = 1/4 + (4)x_1   
POL(U111(x1, x2, x3, x4)) = 1/4 + (4)x_1 + (4)x_2   
POL(n__natsFrom(x1)) = 0   
POL(s(x1)) = 2 + (2)x_1   
POL(activate(x1)) = x_1   
POL(n__s(x1)) = 2 + (2)x_1   
POL(natsFrom(x1)) = 0   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof

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

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(n__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)
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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.