Term Rewriting System R:
[X, Y, Z, X1, X2]
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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, nfirst(X, activate(Z)))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FROM(X) -> CONS(X, nfrom(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)
QUOTE(ns(X)) -> QUOTE(activate(X))
QUOTE(ns(X)) -> ACTIVATE(X)
QUOTE(nsel(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(nsel(X, Z)) -> ACTIVATE(X)
QUOTE(nsel(X, Z)) -> ACTIVATE(Z)
QUOTE1(ncons(X, Z)) -> QUOTE(activate(X))
QUOTE1(ncons(X, Z)) -> ACTIVATE(X)
QUOTE1(ncons(X, Z)) -> QUOTE1(activate(Z))
QUOTE1(ncons(X, Z)) -> ACTIVATE(Z)
QUOTE1(nfirst(X, Z)) -> FIRST1(activate(X), activate(Z))
QUOTE1(nfirst(X, Z)) -> ACTIVATE(X)
QUOTE1(nfirst(X, Z)) -> ACTIVATE(Z)
UNQUOTE(01) -> 0'
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(nfirst(X1, X2)) -> FIRST(X1, X2)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(n0) -> 0'
ACTIVATE(ncons(X1, X2)) -> CONS(X1, X2)
ACTIVATE(nnil) -> NIL
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pair:

QUOTE(ns(X)) -> QUOTE(activate(X))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(X)) -> QUOTE(activate(X))
eight new Dependency Pairs are created:

QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(n0)) -> QUOTE(0)
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(X1', X2'))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(ns(X''))) -> QUOTE(s(X''))
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(X'')) -> QUOTE(X'')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rewriting Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(ns(X''))) -> QUOTE(s(X''))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(X1', X2'))
QUOTE(ns(n0)) -> QUOTE(0)
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(n0)) -> QUOTE(0)
one new Dependency Pair is created:

QUOTE(ns(n0)) -> QUOTE(n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 6
Rewriting Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(ns(X''))) -> QUOTE(s(X''))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(X1', X2'))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(ns(X'')) -> QUOTE(X'')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(X1', X2'))
one new Dependency Pair is created:

QUOTE(ns(ncons(X1', X2'))) -> QUOTE(ncons(X1', X2'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 7
Rewriting Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(ns(X''))) -> QUOTE(s(X''))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(nnil)) -> QUOTE(nil)
one new Dependency Pair is created:

QUOTE(ns(nnil)) -> QUOTE(nnil)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 8
Rewriting Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(ns(X''))) -> QUOTE(s(X''))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(ns(X'')) -> QUOTE(X'')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(ns(X''))) -> QUOTE(s(X''))
one new Dependency Pair is created:

QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 9
Narrowing Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))
QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(X1', X2'))
one new Dependency Pair is created:

QUOTE(ns(nfirst(X1'', X2''))) -> QUOTE(nfirst(X1'', X2''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 10
Narrowing Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(nfrom(X''))) -> QUOTE(from(X''))
two new Dependency Pairs are created:

QUOTE(ns(nfrom(X'''))) -> QUOTE(cons(X''', nfrom(s(X'''))))
QUOTE(ns(nfrom(X'''))) -> QUOTE(nfrom(X'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 11
Rewriting Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(nfrom(X'''))) -> QUOTE(cons(X''', nfrom(s(X'''))))
QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(X'')) -> QUOTE(X'')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(nfrom(X'''))) -> QUOTE(cons(X''', nfrom(s(X'''))))
one new Dependency Pair is created:

QUOTE(ns(nfrom(X'''))) -> QUOTE(ncons(X''', nfrom(s(X'''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 12
Narrowing Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(X1', X2'))
one new Dependency Pair is created:

QUOTE(ns(nsel(X1'', X2''))) -> QUOTE(nsel(X1'', X2''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 13
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))
QUOTE(ns(X'')) -> QUOTE(X'')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(X'')) -> QUOTE(X'')
two new Dependency Pairs are created:

QUOTE(ns(ns(X''''))) -> QUOTE(ns(X''''))
QUOTE(ns(ns(ns(X'''')))) -> QUOTE(ns(ns(X'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 14
Polynomial Ordering
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE(ns(ns(ns(X'''')))) -> QUOTE(ns(ns(X'''')))
QUOTE(ns(ns(X''''))) -> QUOTE(ns(X''''))
QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

QUOTE(ns(ns(ns(X'''')))) -> QUOTE(ns(ns(X'''')))
QUOTE(ns(ns(X''''))) -> QUOTE(ns(X''''))
QUOTE(ns(ns(X''))) -> QUOTE(ns(X''))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOTE(x1))=  1 + x1  
  POL(n__s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 5
Rw
             ...
               →DP Problem 15
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Forward Instantiation Transformation
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pair:

UNQUOTE(s1(X)) -> UNQUOTE(X)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

UNQUOTE(s1(X)) -> UNQUOTE(X)
one new Dependency Pair is created:

UNQUOTE(s1(s1(X''))) -> UNQUOTE(s1(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
           →DP Problem 16
Forward Instantiation Transformation
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pair:

UNQUOTE(s1(s1(X''))) -> UNQUOTE(s1(X''))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

UNQUOTE(s1(s1(X''))) -> UNQUOTE(s1(X''))
one new Dependency Pair is created:

UNQUOTE(s1(s1(s1(X'''')))) -> UNQUOTE(s1(s1(X'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
           →DP Problem 16
FwdInst
             ...
               →DP Problem 17
Polynomial Ordering
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pair:

UNQUOTE(s1(s1(s1(X'''')))) -> UNQUOTE(s1(s1(X'''')))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

UNQUOTE(s1(s1(s1(X'''')))) -> UNQUOTE(s1(s1(X'''')))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s1(x1))=  1 + x1  
  POL(UNQUOTE(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
           →DP Problem 16
FwdInst
             ...
               →DP Problem 18
Dependency Graph
       →DP Problem 3
Nar
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Narrowing Transformation
       →DP Problem 4
Remaining


Dependency Pair:

QUOTE1(ncons(X, Z)) -> QUOTE1(activate(Z))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, Z)) -> QUOTE1(activate(Z))
eight new Dependency Pairs are created:

QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, n0)) -> QUOTE1(0)
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(cons(X1', X2'))
QUOTE1(ncons(X, nnil)) -> QUOTE1(nil)
QUOTE1(ncons(X, ns(X''))) -> QUOTE1(s(X''))
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rewriting Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, ns(X''))) -> QUOTE1(s(X''))
QUOTE1(ncons(X, nnil)) -> QUOTE1(nil)
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(cons(X1', X2'))
QUOTE1(ncons(X, n0)) -> QUOTE1(0)
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, n0)) -> QUOTE1(0)
one new Dependency Pair is created:

QUOTE1(ncons(X, n0)) -> QUOTE1(n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 20
Rewriting Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, ns(X''))) -> QUOTE1(s(X''))
QUOTE1(ncons(X, nnil)) -> QUOTE1(nil)
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(cons(X1', X2'))
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(cons(X1', X2'))
one new Dependency Pair is created:

QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 21
Rewriting Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
QUOTE1(ncons(X, ns(X''))) -> QUOTE1(s(X''))
QUOTE1(ncons(X, nnil)) -> QUOTE1(nil)
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nnil)) -> QUOTE1(nil)
one new Dependency Pair is created:

QUOTE1(ncons(X, nnil)) -> QUOTE1(nnil)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 22
Rewriting Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, ns(X''))) -> QUOTE1(s(X''))
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, ns(X''))) -> QUOTE1(s(X''))
one new Dependency Pair is created:

QUOTE1(ncons(X, ns(X''))) -> QUOTE1(ns(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 23
Narrowing Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nfirst(X1', X2'))) -> QUOTE1(first(X1', X2'))
one new Dependency Pair is created:

QUOTE1(ncons(X, nfirst(X1'', X2''))) -> QUOTE1(nfirst(X1'', X2''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 24
Narrowing Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nfrom(X''))) -> QUOTE1(from(X''))
two new Dependency Pairs are created:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(cons(X''', nfrom(s(X'''))))
QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(nfrom(X'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 25
Rewriting Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(cons(X''', nfrom(s(X'''))))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(cons(X''', nfrom(s(X'''))))
one new Dependency Pair is created:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(s(X'''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 26
Rewriting Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(s(X'''))))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(s(X'''))))
one new Dependency Pair is created:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(ns(X'''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 27
Narrowing Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(ns(X'''))))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))
QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nsel(X1', X2'))) -> QUOTE1(sel(X1', X2'))
one new Dependency Pair is created:

QUOTE1(ncons(X, nsel(X1'', X2''))) -> QUOTE1(nsel(X1'', X2''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 28
Forward Instantiation Transformation
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))
QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(ns(X'''))))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, Z')) -> QUOTE1(Z')
three new Dependency Pairs are created:

QUOTE1(ncons(X, ncons(X'', Z'''))) -> QUOTE1(ncons(X'', Z'''))
QUOTE1(ncons(X, ncons(X'', ncons(X1''', X2''')))) -> QUOTE1(ncons(X'', ncons(X1''', X2''')))
QUOTE1(ncons(X, ncons(X'', nfrom(X''''')))) -> QUOTE1(ncons(X'', nfrom(X''''')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 29
Instantiation Transformation
       →DP Problem 4
Remaining


Dependency Pair:

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(ns(X'''))))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

QUOTE1(ncons(X, nfrom(X'''))) -> QUOTE1(ncons(X''', nfrom(ns(X'''))))
one new Dependency Pair is created:

QUOTE1(ncons(X', nfrom(ns(X')))) -> QUOTE1(ncons(ns(X'), nfrom(ns(ns(X')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 30
Polynomial Ordering
       →DP Problem 4
Remaining


Dependency Pairs:

QUOTE1(ncons(X, ncons(X'', ncons(X1''', X2''')))) -> QUOTE1(ncons(X'', ncons(X1''', X2''')))
QUOTE1(ncons(X, ncons(X'', Z'''))) -> QUOTE1(ncons(X'', Z'''))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

QUOTE1(ncons(X, ncons(X'', ncons(X1''', X2''')))) -> QUOTE1(ncons(X'', ncons(X1''', X2''')))
QUOTE1(ncons(X, ncons(X'', Z'''))) -> QUOTE1(ncons(X'', Z'''))
QUOTE1(ncons(X, ncons(X1', X2'))) -> QUOTE1(ncons(X1', X2'))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__cons(x1, x2))=  1 + x2  
  POL(QUOTE1(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 19
Rw
             ...
               →DP Problem 31
Dependency Graph
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
s(X) -> ns(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(nfrom(X)) -> from(X)
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(nnil) -> nil
activate(ns(X)) -> s(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:04 minutes