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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 Narrowing (⇔)
↳6 QDP
↳7 Narrowing (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 QDP
↳11 Narrowing (⇔)
↳12 QDP
↳13 Narrowing (⇔)
↳14 QDP
↳15 DependencyGraphProof (⇔)
↳16 QDP
↳17 QDPSizeChangeProof (⇔)
↳18 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) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U111(tt, N, X, XS) → SPLITAT(activate(N), activate(XS)) at position [0] we obtained the following new rules [LPAR04]:

U111(tt, n__natsFrom(x0), y1, y2) → SPLITAT(natsFrom(x0), activate(y2))
U111(tt, x0, y1, y2) → SPLITAT(x0, activate(y2))

(6) Obligation:

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__natsFrom(x0), y1, y2) → SPLITAT(natsFrom(x0), activate(y2))
U111(tt, x0, y1, y2) → SPLITAT(x0, activate(y2))

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) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U111(tt, n__natsFrom(x0), y1, y2) → SPLITAT(natsFrom(x0), activate(y2)) at position [0] we obtained the following new rules [LPAR04]:

U111(tt, n__natsFrom(x0), y1, y2) → SPLITAT(cons(x0, n__natsFrom(s(x0))), activate(y2))
U111(tt, n__natsFrom(x0), y1, y2) → SPLITAT(n__natsFrom(x0), activate(y2))

(8) Obligation:

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, x0, y1, y2) → SPLITAT(x0, activate(y2))
U111(tt, n__natsFrom(x0), y1, y2) → SPLITAT(cons(x0, n__natsFrom(s(x0))), activate(y2))
U111(tt, n__natsFrom(x0), y1, y2) → SPLITAT(n__natsFrom(x0), activate(y2))

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.

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(10) Obligation:

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

U111(tt, x0, y1, y2) → SPLITAT(x0, activate(y2))
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.

(11) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U111(tt, x0, y1, y2) → SPLITAT(x0, activate(y2)) at position [1] we obtained the following new rules [LPAR04]:

U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, natsFrom(x0))
U111(tt, y0, y1, x0) → SPLITAT(y0, x0)

(12) Obligation:

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, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, natsFrom(x0))
U111(tt, y0, y1, x0) → SPLITAT(y0, x0)

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.

(13) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, natsFrom(x0)) at position [1] we obtained the following new rules [LPAR04]:

U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))
U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, n__natsFrom(x0))

(14) Obligation:

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, y0, y1, x0) → SPLITAT(y0, x0)
U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))
U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, n__natsFrom(x0))

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.

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

U111(tt, y0, y1, x0) → SPLITAT(y0, x0)
SPLITAT(s(N), cons(X, XS)) → U111(tt, N, X, activate(XS))
U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))

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.

(17) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • SPLITAT(s(N), cons(X, XS)) → U111(tt, N, X, activate(XS))
    The graph contains the following edges 1 > 2, 2 > 3

  • U111(tt, y0, y1, x0) → SPLITAT(y0, x0)
    The graph contains the following edges 2 >= 1, 4 >= 2

  • U111(tt, y0, y1, n__natsFrom(x0)) → SPLITAT(y0, cons(x0, n__natsFrom(s(x0))))
    The graph contains the following edges 2 >= 1

(18) TRUE