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

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}(n_{_s}(X)))
SEL1(s(X), cons(Y, Z)) -> SEL1(X, activate(Z))
SEL1(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL1(0, cons(X, Z)) -> QUOTE(X)
FIRST1(s(X), cons(Y, Z)) -> QUOTE(Y)
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, activate(Z))
FIRST1(s(X), cons(Y, Z)) -> ACTIVATE(Z)
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(activate(X1), activate(X2))
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_₀) -> 0'
ACTIVATE(n_{_cons}(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_nil}) -> NIL
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(activate(X1), activate(X2))
ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X2)

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_from}(X)) -> ACTIVATE(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(n__cons(x₁, x₂)) = x₁

POL(n__from(x₁)) = 1 + x₁

POL(n__sel(x₁, x₂)) = x₁ + x₂

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(n__first(x₁, x₂)) = x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)

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)) = x₁

POL(n__sel(x₁, x₂)) = x₁ + x₂

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(n__first(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_sel}(X1, X2)) -> ACTIVATE(X1)

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)) = x₁

POL(n__sel(x₁, x₂)) = 1 + x₁ + x₂

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)

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 + x₁

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 9

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(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(n__s(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 6

             ↳Polo

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

UNQUOTE(s1(X)) -> UNQUOTE(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)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

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

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

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

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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Rw

...

               →DP Problem 13

                 ↳Rewriting Transformation

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

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

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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 12

             ↳Rw

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pairs:

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

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'')

six new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:38 minutes

POL(n__cons(x₁, x₂))	= x₁
POL(n__from(x₁))	= 1 + x₁
POL(n__sel(x₁, x₂))	= x₁ + x₂
POL(n__s(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁
POL(n__first(x₁, x₂))	= x₁ + x₂