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

Innermost 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 four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOTE(n_{_s}(X)) -> QUOTE(activate(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

Strategy:

innermost

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

QUOTE(n_{_s}(X)) -> QUOTE(activate(X))

eight new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rewriting Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

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

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

QUOTE(n_{_s}(n_₀)) -> QUOTE(0)

one new Dependency Pair is created:

QUOTE(n_{_s}(n_₀)) -> QUOTE(n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 6

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_cons}(X1', X2'))) -> QUOTE(cons(X1', X2'))

one new Dependency Pair is created:

QUOTE(n_{_s}(n_{_cons}(X1', X2'))) -> QUOTE(n_{_cons}(X1', X2'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 7

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(X'')) -> QUOTE(X'')
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(s(X''))
QUOTE(n_{_s}(n_{_nil})) -> QUOTE(nil)
QUOTE(n_{_s}(n_{_from}(X''))) -> QUOTE(from(X''))
QUOTE(n_{_s}(n_{_first}(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_nil})) -> QUOTE(nil)

one new Dependency Pair is created:

QUOTE(n_{_s}(n_{_nil})) -> QUOTE(n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 8

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(s(X''))
QUOTE(n_{_s}(n_{_from}(X''))) -> QUOTE(from(X''))
QUOTE(n_{_s}(n_{_first}(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(n_{_s}(X'')) -> QUOTE(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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(s(X''))

one new Dependency Pair is created:

QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(n_{_s}(X''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 9

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(n_{_s}(X''))
QUOTE(n_{_s}(X'')) -> QUOTE(X'')
QUOTE(n_{_s}(n_{_from}(X''))) -> QUOTE(from(X''))
QUOTE(n_{_s}(n_{_first}(X1', X2'))) -> QUOTE(first(X1', X2'))
QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_first}(X1', X2'))) -> QUOTE(first(X1', X2'))

one new Dependency Pair is created:

QUOTE(n_{_s}(n_{_first}(X1'', X2''))) -> QUOTE(n_{_first}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 10

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(X'')) -> QUOTE(X'')
QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(n_{_s}(n_{_from}(X''))) -> QUOTE(from(X''))
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_from}(X''))) -> QUOTE(from(X''))

two new Dependency Pairs are created:

QUOTE(n_{_s}(n_{_from}(X'''))) -> QUOTE(cons(X''', n_{_from}(s(X'''))))
QUOTE(n_{_s}(n_{_from}(X'''))) -> QUOTE(n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 11

                 ↳Rewriting Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(n_{_from}(X'''))) -> QUOTE(cons(X''', n_{_from}(s(X'''))))
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(n_{_s}(X''))
QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(n_{_s}(X'')) -> QUOTE(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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_from}(X'''))) -> QUOTE(cons(X''', n_{_from}(s(X'''))))

one new Dependency Pair is created:

QUOTE(n_{_s}(n_{_from}(X'''))) -> QUOTE(n_{_cons}(X''', n_{_from}(s(X'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 12

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(X'')) -> QUOTE(X'')
QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(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

Strategy:

innermost

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

QUOTE(n_{_s}(n_{_sel}(X1', X2'))) -> QUOTE(sel(X1', X2'))

one new Dependency Pair is created:

QUOTE(n_{_s}(n_{_sel}(X1'', X2''))) -> QUOTE(n_{_sel}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(n_{_s}(X''))
QUOTE(n_{_s}(X'')) -> QUOTE(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

Strategy:

innermost

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

QUOTE(n_{_s}(X'')) -> QUOTE(X'')

two new Dependency Pairs are created:

QUOTE(n_{_s}(n_{_s}(X''''))) -> QUOTE(n_{_s}(X''''))
QUOTE(n_{_s}(n_{_s}(n_{_s}(X'''')))) -> QUOTE(n_{_s}(n_{_s}(X'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 14

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE(n_{_s}(n_{_s}(n_{_s}(X'''')))) -> QUOTE(n_{_s}(n_{_s}(X'''')))
QUOTE(n_{_s}(n_{_s}(X''''))) -> QUOTE(n_{_s}(X''''))
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

QUOTE(n_{_s}(n_{_s}(n_{_s}(X'''')))) -> QUOTE(n_{_s}(n_{_s}(X'''')))
QUOTE(n_{_s}(n_{_s}(X''''))) -> QUOTE(n_{_s}(X''''))
QUOTE(n_{_s}(n_{_s}(X''))) -> QUOTE(n_{_s}(X''))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOTE(x₁)) = 1 + x₁

POL(n__s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 5

             ↳Rw

...

               →DP Problem 15

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

           →DP Problem 16

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

           →DP Problem 16

             ↳FwdInst

...

               →DP Problem 17

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s1(x₁)) = 1 + x₁

POL(UNQUOTE(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

           →DP Problem 16

             ↳FwdInst

...

               →DP Problem 18

                 ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, Z)) -> QUOTE1(activate(Z))

eight new Dependency Pairs are created:

QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_₀)) -> QUOTE1(0)
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(cons(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_nil})) -> QUOTE1(nil)
QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(s(X''))
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(s(X''))
QUOTE1(n_{_cons}(X, n_{_nil})) -> QUOTE1(nil)
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(cons(X1', X2'))
QUOTE1(n_{_cons}(X, n_₀)) -> QUOTE1(0)
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_₀)) -> QUOTE1(0)

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_₀)) -> QUOTE1(n_₀)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 20

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(s(X''))
QUOTE1(n_{_cons}(X, n_{_nil})) -> QUOTE1(nil)
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(cons(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(cons(X1', X2'))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 21

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')
QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(s(X''))
QUOTE1(n_{_cons}(X, n_{_nil})) -> QUOTE1(nil)
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_nil})) -> QUOTE1(nil)

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_nil})) -> QUOTE1(n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 22

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(s(X''))
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(s(X''))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_s}(X''))) -> QUOTE1(n_{_s}(X''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 23

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_first}(X1', X2'))) -> QUOTE1(first(X1', X2'))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_first}(X1'', X2''))) -> QUOTE1(n_{_first}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 24

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_from}(X''))) -> QUOTE1(from(X''))

two new Dependency Pairs are created:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(cons(X''', n_{_from}(s(X'''))))
QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 25

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(cons(X''', n_{_from}(s(X'''))))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(cons(X''', n_{_from}(s(X'''))))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(s(X'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 26

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(s(X'''))))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(s(X'''))))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(n_{_s}(X'''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 27

                 ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(n_{_s}(X'''))))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))
QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_sel}(X1', X2'))) -> QUOTE1(sel(X1', X2'))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X, n_{_sel}(X1'', X2''))) -> QUOTE1(n_{_sel}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 28

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))
QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')
QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, Z')) -> QUOTE1(Z')

three new Dependency Pairs are created:

QUOTE1(n_{_cons}(X, n_{_cons}(X'', Z'''))) -> QUOTE1(n_{_cons}(X'', Z'''))
QUOTE1(n_{_cons}(X, n_{_cons}(X'', n_{_cons}(X1''', X2''')))) -> QUOTE1(n_{_cons}(X'', n_{_cons}(X1''', X2''')))
QUOTE1(n_{_cons}(X, n_{_cons}(X'', n_{_from}(X''''')))) -> QUOTE1(n_{_cons}(X'', n_{_from}(X''''')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 29

                 ↳Instantiation Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pair:

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(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

Strategy:

innermost

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

QUOTE1(n_{_cons}(X, n_{_from}(X'''))) -> QUOTE1(n_{_cons}(X''', n_{_from}(n_{_s}(X'''))))

one new Dependency Pair is created:

QUOTE1(n_{_cons}(X', n_{_from}(n_{_s}(X')))) -> QUOTE1(n_{_cons}(n_{_s}(X'), n_{_from}(n_{_s}(n_{_s}(X')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOTE1(n_{_cons}(X', n_{_from}(n_{_s}(X')))) -> QUOTE1(n_{_cons}(n_{_s}(X'), n_{_from}(n_{_s}(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

Strategy:
innermost
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

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 30

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

QUOTE1(n_{_cons}(X, n_{_cons}(X'', n_{_cons}(X1''', X2''')))) -> QUOTE1(n_{_cons}(X'', n_{_cons}(X1''', X2''')))
QUOTE1(n_{_cons}(X, n_{_cons}(X'', Z'''))) -> QUOTE1(n_{_cons}(X'', Z'''))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

QUOTE1(n_{_cons}(X, n_{_cons}(X'', n_{_cons}(X1''', X2''')))) -> QUOTE1(n_{_cons}(X'', n_{_cons}(X1''', X2''')))
QUOTE1(n_{_cons}(X, n_{_cons}(X'', Z'''))) -> QUOTE1(n_{_cons}(X'', Z'''))
QUOTE1(n_{_cons}(X, n_{_cons}(X1', X2'))) -> QUOTE1(n_{_cons}(X1', X2'))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__cons(x₁, x₂)) = 1 + x₂

POL(QUOTE1(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 19

             ↳Rw

...

               →DP Problem 31

                 ↳Dependency Graph

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> 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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOTE1(n_{_cons}(X', n_{_from}(n_{_s}(X')))) -> QUOTE1(n_{_cons}(n_{_s}(X'), n_{_from}(n_{_s}(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

Strategy:
innermost
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

Strategy:
innermost

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

POL(n__cons(x₁, x₂))	= 1 + x₂
POL(QUOTE1(x₁))	= 1 + x₁