Termination Proof using AProVE

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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

Termination of R to be shown.

     ↳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, 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)
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_{_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)
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) -> 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(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)

Furthermore, R contains six SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))

eight new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))
SEL(s(X), cons(Y, n_₀)) -> SEL(X, 0)
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))
SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
SEL(s(X), cons(Y, n_₀)) -> SEL(X, 0)
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))
SEL(s(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(X1', X2'))) -> SEL(X, first(X1', X2'))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, n_{_first}(X1'', X2''))) -> SEL(X, n_{_first}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))
SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
SEL(s(X), cons(Y, n_₀)) -> SEL(X, 0)
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(X, from(X''))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))
SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
SEL(s(X), cons(Y, n_₀)) -> SEL(X, 0)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_₀)) -> SEL(X, 0)

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_₀)) -> SEL(X, n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 10

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))
SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, nil)
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, nil)

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_nil})) -> SEL(X, n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 11

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(X, s(X''))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_s}(X'''))) -> SEL(X, n_{_s}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 12

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(X, sel(X1', X2'))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_sel}(s(X''), cons(Y'', Z')))) -> SEL(X, sel(X'', activate(Z')))
SEL(s(X), cons(Y, n_{_sel}(0, cons(X'', Z')))) -> SEL(X, X'')
SEL(s(X), cons(Y, n_{_sel}(X1'', X2''))) -> SEL(X, n_{_sel}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_sel}(0, cons(X'', Z')))) -> SEL(X, X'')
SEL(s(X), cons(Y, n_{_sel}(s(X''), cons(Y'', Z')))) -> SEL(X, sel(X'', activate(Z')))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(0, X2''))) -> SEL(X, nil)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(X1, X2)

Additionally, the following usable rules using the Ce-refinement can be oriented:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> 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
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
0 -> n_₀

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(activate(x₁)) = x₁

POL(SEL(x₁, x₂)) = x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(ACTIVATE(x₁)) = x₁

POL(n__from(x₁)) = x₁

POL(n__cons(x₁, x₂)) = x₁ + x₂

POL(n__sel(x₁, x₂)) = x₂

POL(0) = 0

POL(first(x₁, x₂)) = 1 + x₂

POL(cons(x₁, x₂)) = x₁ + x₂

POL(FIRST(x₁, x₂)) = x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_sel}(0, cons(X'', Z')))) -> SEL(X, X'')
SEL(s(X), cons(Y, n_{_sel}(s(X''), cons(Y'', Z')))) -> SEL(X, sel(X'', activate(Z')))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 15

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_sel}(s(X''), cons(Y'', Z')))) -> SEL(X, sel(X'', activate(Z')))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, n_{_sel}(0, cons(X'', Z')))) -> SEL(X, X'')

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pairs can be strictly oriented:

ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
SEL(s(X), cons(Y, n_{_sel}(0, cons(X'', Z')))) -> SEL(X, X'')

Additionally, the following usable rules using the Ce-refinement can be oriented:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> 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
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
0 -> n_₀
nil -> n_{_nil}

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(activate(x₁)) = x₁

POL(SEL(x₁, x₂)) = x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = 1 + x₂

POL(n__nil) = 0

POL(ACTIVATE(x₁)) = x₁

POL(n__from(x₁)) = x₁

POL(n__cons(x₁, x₂)) = x₁ + x₂

POL(n__sel(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(first(x₁, x₂)) = x₂

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 17

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))
SEL(s(X), cons(Y, n_{_sel}(s(X''), cons(Y'', Z')))) -> SEL(X, sel(X'', activate(Z')))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, n_{_sel}(s(X''), cons(Y'', Z')))) -> SEL(X, sel(X'', activate(Z')))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
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
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
0 -> n_₀
nil -> n_{_nil}

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 18

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, n_{_cons}(X1', X2'))) -> SEL(X, cons(X1', X2'))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
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
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
0 -> n_₀
nil -> n_{_nil}

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = 1 + x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 19

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z')))) -> SEL(X, cons(Y'', n_{_first}(X'', activate(Z'))))

nine new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', Z'')))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', activate(Z''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', 0)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 20

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', 0)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', first(X1', X2'))))

four new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_first}(X1'', X2''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', first(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 21

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', 0)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', from(X'''))))

three new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''''), cons(Y''', n_{_from}(X'''''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X'''', from(X'''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 22

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', 0)))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', 0)))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_₀)))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', 0)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 23

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_nil})))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 24

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X'''))))) -> SEL(X, cons(Y'', n_{_first}(X'', s(X'''))))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''''), cons(Y''', n_{_s}(X'''''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X'''', s(X'''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 25

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X1', X2'))))

four new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_sel}(X1'', X2''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', sel(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 26

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', nil)))

two new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_first}(0, X2''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', nil)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 27

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_first}(X1'', X2''))))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_first}(X1''', X2'''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', n_{_first}(X1''', X2'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 28

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_from}(X''''))))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_from}(X'''''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', n_{_from}(X'''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 29

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_₀)))) -> SEL(X, cons(Y'', n_{_first}(X'', n_₀)))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_₀)))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', n_₀)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 30

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_nil})))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_nil})))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', n_{_nil})))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 31

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_s}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_s}(X''''))))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_s}(X'''''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', n_{_s}(X'''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 32

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', activate(Z')))))

10 new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''''), cons(Y'''', n_{_sel}(s(X'''''), cons(Y''', Z'')))))) -> SEL(X, n_{_cons}(Y'''', n_{_first}(X'''', sel(X''''', activate(Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X''''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X'''', activate(Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', first(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', from(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', 0))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', cons(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', nil))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', s(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', sel(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', Z''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 33

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', Z''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', sel(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', s(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', nil))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', cons(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', 0))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', from(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', first(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X''''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X'''', activate(Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(X1'', X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X1'', X2''))))

one new Dependency Pair is created:

SEL(s(X), cons(Y, n_{_first}(s(X'''), cons(Y''', n_{_sel}(X1''', X2'''))))) -> SEL(X, n_{_cons}(Y''', n_{_first}(X''', n_{_sel}(X1''', X2'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 34

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', sel(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', s(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', nil))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', cons(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', 0))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', from(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', first(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X''''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X'''', activate(Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', Z''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(0, X2''))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_nil})))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(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
s(X) -> n_{_s}(X)
nil -> n_{_nil}
0 -> n_₀
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = x₁

POL(activate(x₁)) = x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(0) = 0

POL(first(x₁, x₂)) = 1 + x₂

POL(cons(x₁, x₂)) = x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 35

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', sel(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', s(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', nil))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', cons(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', 0))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', from(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', first(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X''''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X'''', activate(Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', Z''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', sel(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', s(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', nil))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', cons(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', 0))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', from(X'0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', first(X1', X2')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X''''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_sel}(X'''', activate(Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(0, cons(X''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', X''')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', Z'')))) -> SEL(X, cons(Y'', n_{_first}(X'', Z'')))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_sel}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', sel(X''', Z''))))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(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
s(X) -> n_{_s}(X)
nil -> n_{_nil}
0 -> n_₀
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(0) = 0

POL(first(x₁, x₂)) = 1 + x₂

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 36

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_cons}(X1', X2'))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X1', X2'))))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
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
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
0 -> n_₀
nil -> n_{_nil}

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = 1 + x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 37

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

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

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', activate(Z'))))))

10 new Dependency Pairs are created:

SEL(s(X), cons(Y, n_{_first}(s(X''''), cons(Y''0, n_{_first}(s(X'''''), cons(Y'''', Z'')))))) -> SEL(X, n_{_cons}(Y''0, n_{_first}(X'''', cons(Y'''', n_{_first}(X''''', activate(Z''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X''''), cons(Y'''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_cons}(Y'''', n_{_first}(X'''', activate(Z''))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', 0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', cons(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', nil)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', s(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', sel(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', Z'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 38

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', sel(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', s(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', nil)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', cons(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', 0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X''''), cons(Y'''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_cons}(Y'''', n_{_first}(X'''', activate(Z''))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', Z'')))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_sel}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', sel(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_s}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', s(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_nil})))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', nil)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_₀)))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', 0)))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X''''), cons(Y'''', Z'')))))) -> SEL(X, cons(Y'', n_{_first}(X'', n_{_cons}(Y'''', n_{_first}(X'''', activate(Z''))))))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(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
s(X) -> n_{_s}(X)
nil -> n_{_nil}
0 -> n_₀
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(0) = 0

POL(first(x₁, x₂)) = 1 + x₂

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 7

             ↳Nar

...

               →DP Problem 39

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

       →DP Problem 6

         ↳Remaining

Dependency Pairs:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', cons(X1', X2'))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_cons}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', cons(X1', X2'))))))

Additionally, the following usable rules using the Ce-refinement can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> 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
s(X) -> n_{_s}(X)
nil -> n_{_nil}
0 -> n_₀

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(from(x₁)) = 1 + x₁

POL(activate(x₁)) = 1 + x₁

POL(SEL(x₁, x₂)) = 1 + x₂

POL(n__s(x₁)) = 0

POL(sel(x₁, x₂)) = x₂

POL(n__nil) = 0

POL(n__cons(x₁, x₂)) = 1 + x₁ + x₂

POL(n__from(x₁)) = x₁

POL(n__sel(x₁, x₂)) = x₂

POL(first(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(nil) = 0

POL(n__0) = 0

POL(s(x₁)) = 0

POL(n__first(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pairs:
QUOTE(n_{_sel}(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
SEL1(0, cons(X, Z)) -> QUOTE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
QUOTE1(n_{_cons}(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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pairs:
QUOTE(n_{_sel}(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
SEL1(0, cons(X, Z)) -> QUOTE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
QUOTE1(n_{_cons}(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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pairs:
QUOTE(n_{_sel}(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
SEL1(0, cons(X, Z)) -> QUOTE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
QUOTE1(n_{_cons}(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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pairs:
QUOTE(n_{_sel}(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
SEL1(0, cons(X, Z)) -> QUOTE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
QUOTE1(n_{_cons}(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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pairs:
QUOTE(n_{_sel}(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
SEL1(0, cons(X, Z)) -> QUOTE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
QUOTE1(n_{_cons}(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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_from}(X'0))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', from(X'0))))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_first}(s(X'''), cons(Y''', n_{_first}(X1', X2'))))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(Y''', n_{_first}(X''', first(X1', X2'))))))
SEL(s(X), cons(Y, n_{_from}(X'''))) -> SEL(X, cons(X''', n_{_from}(s(X'''))))
SEL(s(X), cons(Y, n_{_first}(s(X''), cons(Y'', n_{_from}(X''''))))) -> SEL(X, cons(Y'', n_{_first}(X'', cons(X'''', n_{_from}(s(X''''))))))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pairs:
QUOTE(n_{_sel}(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(n_{_s}(X)) -> QUOTE(activate(X))
SEL1(0, cons(X, Z)) -> QUOTE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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
Dependency Pair:
QUOTE1(n_{_cons}(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) -> n_{_sel}(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, n_{_first}(X, activate(Z)))
first(X1, X2) -> n_{_first}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(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
quote(n_{_s}(X)) -> s1(quote(activate(X)))
quote(n_{_sel}(X, Z)) -> sel1(activate(X), activate(Z))
quote1(n_{_cons}(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n_{_nil}) -> nil1
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)
0 -> n_₀
cons(X1, X2) -> n_{_cons}(X1, X2)
nil -> n_{_nil}
s(X) -> n_{_s}(X)
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

Termination of R could not be shown.
Duration:
0:09 minutes

POL(from(x₁))	= x₁
POL(activate(x₁))	= x₁
POL(SEL(x₁, x₂))	= x₂
POL(n__s(x₁))	= 0
POL(sel(x₁, x₂))	= x₂
POL(n__nil)	= 0
POL(ACTIVATE(x₁))	= x₁
POL(n__from(x₁))	= x₁
POL(n__cons(x₁, x₂))	= x₁ + x₂
POL(n__sel(x₁, x₂))	= x₂
POL(0)	= 0
POL(first(x₁, x₂))	= 1 + x₂
POL(cons(x₁, x₂))	= x₁ + x₂
POL(FIRST(x₁, x₂))	= x₂
POL(nil)	= 0
POL(n__0)	= 0
POL(s(x₁))	= 0
POL(n__first(x₁, x₂))	= 1 + x₂

POL(from(x₁))	= 1 + x₁
POL(activate(x₁))	= 1 + x₁
POL(SEL(x₁, x₂))	= 1 + x₂
POL(n__s(x₁))	= 0
POL(sel(x₁, x₂))	= x₂
POL(n__nil)	= 0
POL(n__cons(x₁, x₂))	= x₁ + x₂
POL(n__from(x₁))	= x₁
POL(n__sel(x₁, x₂))	= x₂
POL(first(x₁, x₂))	= 1 + x₂
POL(0)	= 0
POL(cons(x₁, x₂))	= 1 + x₁ + x₂
POL(nil)	= 0
POL(n__0)	= 0
POL(s(x₁))	= 0
POL(n__first(x₁, x₂))	= x₂