Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

DBL(s(X)) -> S(n_{_s}(n_{_dbl}(activate(X))))
DBL(s(X)) -> ACTIVATE(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)
DBLS(cons(X, Y)) -> ACTIVATE(Y)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
SEL(s(X), cons(Y, Z)) -> SEL(activate(X), activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(X)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(Y)
FROM(X) -> ACTIVATE(X)
DBL1(s(X)) -> DBL1(activate(X))
DBL1(s(X)) -> ACTIVATE(X)
SEL1(0, cons(X, Y)) -> ACTIVATE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(activate(X), activate(Z))
SEL1(s(X), cons(Y, Z)) -> ACTIVATE(X)
SEL1(s(X), cons(Y, Z)) -> ACTIVATE(Z)
QUOTE(s(X)) -> QUOTE(activate(X))
QUOTE(s(X)) -> ACTIVATE(X)
QUOTE(dbl(X)) -> DBL1(X)
QUOTE(sel(X, Y)) -> SEL1(X, Y)
ACTIVATE(n_{_s}(X)) -> S(X)
ACTIVATE(n_{_dbl}(X)) -> DBL(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
ACTIVATE(n_{_indx}(X1, X2)) -> INDX(X1, X2)
ACTIVATE(n_{_from}(X)) -> FROM(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

Dependency Pairs:

INDX(cons(X, Y), Z) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
FROM(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(X)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
ACTIVATE(n_{_indx}(X1, X2)) -> INDX(X1, X2)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(Y)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_from}(X)) -> FROM(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FROM(x₁)) = x₁

POL(n__from(x₁)) = 1 + x₁

POL(n__indx(x₁, x₂)) = x₁ + x₂

POL(n__sel(x₁, x₂)) = x₂

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(SEL(x₁, x₂)) = x₂

POL(n__dbls(x₁)) = x₁

POL(INDX(x₁, x₂)) = x₁ + x₂

POL(ACTIVATE(x₁)) = x₁

POL(DBLS(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳Dependency Graph

Dependency Pairs:

INDX(cons(X, Y), Z) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
FROM(X) -> ACTIVATE(X)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
ACTIVATE(n_{_indx}(X1, X2)) -> INDX(X1, X2)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(Y)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 3

                 ↳Polynomial Ordering

Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
ACTIVATE(n_{_indx}(X1, X2)) -> INDX(X1, X2)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
INDX(cons(X, Y), Z) -> ACTIVATE(Y)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_indx}(X1, X2)) -> INDX(X1, X2)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__indx(x₁, x₂)) = 1 + x₁ + x₂

POL(n__sel(x₁, x₂)) = x₂

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(SEL(x₁, x₂)) = x₂

POL(n__dbls(x₁)) = x₁

POL(INDX(x₁, x₂)) = x₁ + x₂

POL(ACTIVATE(x₁)) = x₁

POL(DBLS(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 4

                 ↳Dependency Graph

Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
INDX(cons(X, Y), Z) -> ACTIVATE(Y)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 5

                 ↳Polynomial Ordering

Dependency Pairs:

SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_sel}(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(Y)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__sel(x₁, x₂)) = 1 + x₂

POL(0) = 0

POL(SEL(x₁, x₂)) = x₂

POL(cons(x₁, x₂)) = x₁ + x₂

POL(n__dbls(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(DBLS(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pairs:

SEL(0, cons(X, Y)) -> ACTIVATE(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(Y)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pairs can be strictly oriented:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
DBLS(cons(X, Y)) -> ACTIVATE(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = 1 + x₁ + x₂

POL(n__dbls(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

POL(DBLS(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 2

             ↳DGraph

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

ACTIVATE(n_{_dbls}(X)) -> DBLS(X)

Rules:

dbl(0) -> 0
dbl(s(X)) -> s(n_{_s}(n_{_dbl}(activate(X))))
dbl(X) -> n_{_dbl}(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(n_{_dbl}(activate(X)), n_{_dbls}(activate(Y)))
dbls(X) -> n_{_dbls}(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> n_{_sel}(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(n_{_sel}(activate(X), activate(Z)), n_{_indx}(activate(Y), activate(Z)))
indx(X1, X2) -> n_{_indx}(X1, X2)
from(X) -> cons(activate(X), n_{_from}(n_{_s}(activate(X))))
from(X) -> n_{_from}(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> n_{_s}(X)
activate(n_{_s}(X)) -> s(X)
activate(n_{_dbl}(X)) -> dbl(X)
activate(n_{_dbls}(X)) -> dbls(X)
activate(n_{_sel}(X1, X2)) -> sel(X1, X2)
activate(n_{_indx}(X1, X2)) -> indx(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes

POL(FROM(x₁))	= x₁
POL(n__from(x₁))	= 1 + x₁
POL(n__indx(x₁, x₂))	= x₁ + x₂
POL(n__sel(x₁, x₂))	= x₂
POL(0)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(SEL(x₁, x₂))	= x₂
POL(n__dbls(x₁))	= x₁
POL(INDX(x₁, x₂))	= x₁ + x₂
POL(ACTIVATE(x₁))	= x₁
POL(DBLS(x₁))	= x₁

POL(n__indx(x₁, x₂))	= 1 + x₁ + x₂
POL(n__sel(x₁, x₂))	= x₂
POL(0)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(SEL(x₁, x₂))	= x₂
POL(n__dbls(x₁))	= x₁
POL(INDX(x₁, x₂))	= x₁ + x₂
POL(ACTIVATE(x₁))	= x₁
POL(DBLS(x₁))	= x₁

POL(n__sel(x₁, x₂))	= 1 + x₂
POL(0)	= 0
POL(SEL(x₁, x₂))	= x₂
POL(cons(x₁, x₂))	= x₁ + x₂
POL(n__dbls(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁
POL(DBLS(x₁))	= x₁