Termination w.r.t. Q proof of /tmp/tmpJZhp5D/Ex6_15_AEL02

Termination w.r.t. Q of the given QTRS could not be shown:

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

  ↳15 QDP

  ↳16 QDP

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
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:

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
FIRST(0, Z) → NIL
FIRST(s(X), cons(Y, Z)) → CONS(Y, n__first(X, activate(Z)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
FROM(X) → CONS(X, n__from(n__s(X)))
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL1(0, cons(X, Z)) → QUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
FIRST1(s(X), cons(Y, Z)) → ACTIVATE(Z)
QUOTE1(n__cons(X, Z)) → QUOTE(activate(X))
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
QUOTE1(n__cons(X, Z)) → ACTIVATE(Z)
QUOTE(n__s(X)) → QUOTE(activate(X))
QUOTE(n__s(X)) → ACTIVATE(X)
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
QUOTE(n__sel(X, Z)) → ACTIVATE(X)
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
QUOTE1(n__first(X, Z)) → FIRST1(activate(X), activate(Z))
QUOTE1(n__first(X, Z)) → ACTIVATE(X)
QUOTE1(n__first(X, Z)) → ACTIVATE(Z)
UNQUOTE(01) → 0¹
UNQUOTE(s1(X)) → S(unquote(X))
UNQUOTE(s1(X)) → UNQUOTE(X)
UNQUOTE1(nil1) → NIL
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
FCONS(X, Z) → CONS(X, Z)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
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 6 SCCs with 26 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

UNQUOTE(s1(X)) → UNQUOTE(X)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
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.

UNQUOTE(s1(X)) → UNQUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
UNQUOTE(x1) = x1
s1(x1) = s1(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
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:

UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
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.

UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
UNQUOTE1(x1) = x1
cons1(x1, x2) = cons1(x1, x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
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

(15) Obligation:

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

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

(16) Obligation:

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

SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE(n__s(X)) → QUOTE(activate(X))
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

(17) Obligation:

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

FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST1(x1, x2) = FIRST1(x1)
s(x1) = s(x1)
cons(x1, x2) = cons(x2)
activate(x1) = activate
n__first(x1, x2) = n__first(x2)
first(x1, x2) = first(x1, x2)
n__from(x1) = n__from(x1)
from(x1) = from(x1)
n__s(x1) = n__s(x1)
n__0 = n__0
0 = 0
n__cons(x1, x2) = n__cons(x2)
n__nil = n__nil
nil = nil
n__sel(x1, x2) = n__sel(x2)
sel(x1, x2) = sel(x2)

Recursive Path Order [RPO].
Precedence:

FIRST1₁ > [cons₁, activate, n_{_from₁}, from₁, n_{_s₁}, n_{_nil}, nil, sel₁]
[s₁, n_{_first₁}, first₂] > [cons₁, activate, n_{_from₁}, from₁, n_{_s₁}, n_{_nil}, nil, sel₁]
0 > n_₀ > [cons₁, activate, n_{_from₁}, from₁, n_{_s₁}, n_{_nil}, nil, sel₁]
n_{_cons₁} > [cons₁, activate, n_{_from₁}, from₁, n_{_s₁}, n_{_nil}, nil, sel₁]
n_{_sel₁} > [cons₁, activate, n_{_from₁}, from₁, n_{_s₁}, n_{_nil}, nil, sel₁]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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