Termination w.r.t. Q proof of /tmp/tmpOuq2sL/LISTUTILITIES_nosorts

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

            ↳ UsableRulesProof

          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

ACTIVATE(n__natsFrom(X)) → ACTIVATE(X)
The graph contains the following edges 1 > 1
ACTIVATE(n__s(X)) → ACTIVATE(X)
The graph contains the following edges 1 > 1

↳ 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.

SPLITAT(s(N), cons(X, XS)) → U11¹(tt, N, X, activate(XS))

The remaining pairs can at least be oriented weakly.

U11¹(tt, N, X, XS) → SPLITAT(activate(N), activate(XS))

Used ordering: Polynomial interpretation [25]:

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

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

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