(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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(s(X)))
FROM(X) → 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(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__0) → 01
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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 27 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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)

Lexicographic Path Order [LPO].
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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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)

Lexicographic Path Order [LPO].
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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
SEL(s(X), cons(Y, Z)) → SEL(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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
[SEL2, nsel2, sel2] > [activate1, first1, n0, 0] > from1 > s1 > [cons2, ncons2, nnil, nil, ns1]
[SEL2, nsel2, sel2] > [activate1, first1, n0, 0] > from1 > nfrom1 > [cons2, ncons2, nnil, nil, ns1]


The following usable rules [FROCOS05] were oriented:

activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X
first(0, Z) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
cons(X1, X2) → n__cons(X1, X2)
sel(0, cons(X, Z)) → X
sel(X1, X2) → n__sel(X1, X2)
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
niln__nil
from(X) → cons(X, n__from(s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
0n__0

(17) Obligation:

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

ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(19) TRUE

(20) 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(21) 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(22) 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(x1)
n__first(x1, x2)  =  n__first
first(x1, x2)  =  first(x2)
n__from(x1)  =  n__from(x1)
from(x1)  =  from(x1)
n__0  =  n__0
0  =  0
n__cons(x1, x2)  =  n__cons
n__nil  =  n__nil
nil  =  nil
n__s(x1)  =  x1
n__sel(x1, x2)  =  n__sel
sel(x1, x2)  =  sel(x2)

Lexicographic Path Order [LPO].
Precedence:
FIRST11 > [s1, activate1, first1] > [cons1, nfirst, from1, nnil, nil, sel1]
nfrom1 > [cons1, nfirst, from1, nnil, nil, sel1]
0 > n0 > [cons1, nfirst, from1, nnil, nil, sel1]
ncons > [cons1, nfirst, from1, nnil, nil, sel1]
nsel > [cons1, nfirst, from1, nnil, nil, sel1]


The following usable rules [FROCOS05] were oriented: none

(23) 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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