Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z, X1, X2]
a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> A_{_SEL}(mark(X), mark(Z))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(dbl(X)) -> A_{_DBL}(mark(X))
MARK(dbl(X)) -> MARK(X)
MARK(dbls(X)) -> A_{_DBLS}(mark(X))
MARK(dbls(X)) -> MARK(X)
MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))
MARK(sel(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(indx(X1, X2)) -> A_{_INDX}(mark(X1), X2)
MARK(indx(X1, X2)) -> MARK(X1)
MARK(from(X)) -> A_{_FROM}(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(indx(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> A_{_SEL}(mark(X), mark(Z))
MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))
MARK(dbls(X)) -> MARK(X)
MARK(dbl(X)) -> MARK(X)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)

Rules:

a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

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

A_{_SEL}(s(X), cons(Y, Z)) -> A_{_SEL}(mark(X), mark(Z))

18 new Dependency Pairs are created:

A_{_SEL}(s(dbl(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbl}(mark(X'')), mark(Z))
A_{_SEL}(s(dbls(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbls}(mark(X'')), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(indx(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_indx}(mark(X1'), X2'), mark(Z))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(X''), mark(Z))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(X''), mark(Z))
A_{_SEL}(s(nil), cons(Y, Z)) -> A_{_SEL}(nil, mark(Z))
A_{_SEL}(s(cons(X1', X2')), cons(Y, Z)) -> A_{_SEL}(cons(X1', X2'), mark(Z))
A_{_SEL}(s(X), cons(Y, dbl(X''))) -> A_{_SEL}(mark(X), a_{_dbl}(mark(X'')))
A_{_SEL}(s(X), cons(Y, dbls(X''))) -> A_{_SEL}(mark(X), a_{_dbls}(mark(X'')))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, indx(X1', X2'))) -> A_{_SEL}(mark(X), a_{_indx}(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(X''))
A_{_SEL}(s(X), cons(Y, 0)) -> A_{_SEL}(mark(X), 0)
A_{_SEL}(s(X), cons(Y, s(X''))) -> A_{_SEL}(mark(X), s(X''))
A_{_SEL}(s(X), cons(Y, nil)) -> A_{_SEL}(mark(X), nil)
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(X1', X2'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(X1', X2'))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(X''))
A_{_SEL}(s(X), cons(Y, indx(X1', X2'))) -> A_{_SEL}(mark(X), a_{_indx}(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, dbls(X''))) -> A_{_SEL}(mark(X), a_{_dbls}(mark(X'')))
A_{_SEL}(s(X), cons(Y, dbl(X''))) -> A_{_SEL}(mark(X), a_{_dbl}(mark(X'')))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(X''), mark(Z))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(X''), mark(Z))
A_{_SEL}(s(indx(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_indx}(mark(X1'), X2'), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(dbls(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbls}(mark(X'')), mark(Z))
A_{_SEL}(s(dbl(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbl}(mark(X'')), mark(Z))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(indx(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))
MARK(dbls(X)) -> MARK(X)
MARK(dbl(X)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)

Rules:

a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

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

MARK(sel(X1, X2)) -> A_{_SEL}(mark(X1), mark(X2))

18 new Dependency Pairs are created:

MARK(sel(dbl(X'), X2)) -> A_{_SEL}(a_{_dbl}(mark(X')), mark(X2))
MARK(sel(dbls(X'), X2)) -> A_{_SEL}(a_{_dbls}(mark(X')), mark(X2))
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
MARK(sel(indx(X1'', X2''), X2)) -> A_{_SEL}(a_{_indx}(mark(X1''), X2''), mark(X2))
MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(X'), mark(X2))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(X'), mark(X2))
MARK(sel(nil, X2)) -> A_{_SEL}(nil, mark(X2))
MARK(sel(cons(X1'', X2''), X2)) -> A_{_SEL}(cons(X1'', X2''), mark(X2))
MARK(sel(X1, dbl(X'))) -> A_{_SEL}(mark(X1), a_{_dbl}(mark(X')))
MARK(sel(X1, dbls(X'))) -> A_{_SEL}(mark(X1), a_{_dbls}(mark(X')))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, indx(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_indx}(mark(X1''), X2''))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(X'))
MARK(sel(X1, 0)) -> A_{_SEL}(mark(X1), 0)
MARK(sel(X1, s(X'))) -> A_{_SEL}(mark(X1), s(X'))
MARK(sel(X1, nil)) -> A_{_SEL}(mark(X1), nil)
MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Negative Polynomial Order

Dependency Pairs:

MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(X1'', X2''))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(X'))
MARK(sel(X1, indx(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_indx}(mark(X1''), X2''))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, dbls(X'))) -> A_{_SEL}(mark(X1), a_{_dbls}(mark(X')))
MARK(sel(X1, dbl(X'))) -> A_{_SEL}(mark(X1), a_{_dbl}(mark(X')))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(X'), mark(X2))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(X'), mark(X2))
MARK(sel(indx(X1'', X2''), X2)) -> A_{_SEL}(a_{_indx}(mark(X1''), X2''), mark(X2))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(X''))
A_{_SEL}(s(X), cons(Y, indx(X1', X2'))) -> A_{_SEL}(mark(X), a_{_indx}(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, dbls(X''))) -> A_{_SEL}(mark(X), a_{_dbls}(mark(X'')))
A_{_SEL}(s(X), cons(Y, dbl(X''))) -> A_{_SEL}(mark(X), a_{_dbl}(mark(X'')))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(X''), mark(Z))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(X''), mark(Z))
A_{_SEL}(s(indx(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_indx}(mark(X1'), X2'), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(dbls(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbls}(mark(X'')), mark(Z))
A_{_SEL}(s(dbl(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbl}(mark(X'')), mark(Z))
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(sel(dbls(X'), X2)) -> A_{_SEL}(a_{_dbls}(mark(X')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(sel(dbl(X'), X2)) -> A_{_SEL}(a_{_dbl}(mark(X')), mark(X2))
MARK(indx(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
MARK(dbls(X)) -> MARK(X)
MARK(dbl(X)) -> MARK(X)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(X1', X2'))

Rules:

a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

The following Dependency Pairs can be strictly oriented using the given order.

MARK(sel(X1, dbl(X'))) -> A_{_SEL}(mark(X1), a_{_dbl}(mark(X')))
A_{_SEL}(s(X), cons(Y, dbl(X''))) -> A_{_SEL}(mark(X), a_{_dbl}(mark(X'')))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = 1

POL( A_{_SEL}(x₁, x₂) ) = x₂

POL( a_{_dbl}(x₁) ) = 0

POL( cons(x₁, x₂) ) = 1

POL( mark(x₁) ) = 1

POL( a_{_sel}(x₁, x₂) ) = 1

POL( a_{_from}(x₁) ) = 1

POL( a_{_indx}(x₁, x₂) ) = 1

POL( a_{_dbls}(x₁) ) = 1

POL( 0 ) = 0

POL( s(x₁) ) = 0

POL( nil ) = 0

POL( from(x₁) ) = 0

POL( indx(x₁, x₂) ) = 0

POL( sel(x₁, x₂) ) = 0

POL( dbls(x₁) ) = 0

POL( dbl(x₁) ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Negative Polynomial Order

Dependency Pairs:

MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(X1'', X2''))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(X'))
MARK(sel(X1, indx(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_indx}(mark(X1''), X2''))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, dbls(X'))) -> A_{_SEL}(mark(X1), a_{_dbls}(mark(X')))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(X'), mark(X2))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(X'), mark(X2))
MARK(sel(indx(X1'', X2''), X2)) -> A_{_SEL}(a_{_indx}(mark(X1''), X2''), mark(X2))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(X''))
A_{_SEL}(s(X), cons(Y, indx(X1', X2'))) -> A_{_SEL}(mark(X), a_{_indx}(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, dbls(X''))) -> A_{_SEL}(mark(X), a_{_dbls}(mark(X'')))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(X''), mark(Z))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(X''), mark(Z))
A_{_SEL}(s(indx(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_indx}(mark(X1'), X2'), mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(dbls(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbls}(mark(X'')), mark(Z))
A_{_SEL}(s(dbl(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbl}(mark(X'')), mark(Z))
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
MARK(sel(dbls(X'), X2)) -> A_{_SEL}(a_{_dbls}(mark(X')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(sel(dbl(X'), X2)) -> A_{_SEL}(a_{_dbl}(mark(X')), mark(X2))
MARK(indx(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
MARK(dbls(X)) -> MARK(X)
MARK(dbl(X)) -> MARK(X)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(X1', X2'))

Rules:

a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

The following Dependency Pairs can be strictly oriented using the given order.

MARK(sel(from(X'), X2)) -> A_{_SEL}(a_{_from}(X'), mark(X2))
MARK(sel(indx(X1'', X2''), X2)) -> A_{_SEL}(a_{_indx}(mark(X1''), X2''), mark(X2))
A_{_SEL}(s(from(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_from}(X''), mark(Z))
A_{_SEL}(s(indx(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_indx}(mark(X1'), X2'), mark(Z))
A_{_SEL}(s(dbls(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbls}(mark(X'')), mark(Z))
MARK(sel(dbls(X'), X2)) -> A_{_SEL}(a_{_dbls}(mark(X')), mark(X2))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)

Used ordering:
Polynomial Order with Interpretation:

POL( MARK(x₁) ) = 1

POL( A_{_SEL}(x₁, x₂) ) = x₁

POL( a_{_from}(x₁) ) = 0

POL( s(x₁) ) = 1

POL( a_{_sel}(x₁, x₂) ) = 1

POL( 0 ) = 1

POL( mark(x₁) ) = 1

POL( a_{_dbls}(x₁) ) = 0

POL( a_{_dbl}(x₁) ) = 1

POL( a_{_indx}(x₁, x₂) ) = 0

POL( nil ) = 0

POL( cons(x₁, x₂) ) = 0

POL( from(x₁) ) = 0

POL( indx(x₁, x₂) ) = 0

POL( sel(x₁, x₂) ) = 0

POL( dbls(x₁) ) = 0

POL( dbl(x₁) ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(sel(X1, cons(X1'', X2''))) -> A_{_SEL}(mark(X1), cons(X1'', X2''))
MARK(sel(X1, from(X'))) -> A_{_SEL}(mark(X1), a_{_from}(X'))
MARK(sel(X1, indx(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_indx}(mark(X1''), X2''))
MARK(sel(X1, sel(X1'', X2''))) -> A_{_SEL}(mark(X1), a_{_sel}(mark(X1''), mark(X2'')))
MARK(sel(X1, dbls(X'))) -> A_{_SEL}(mark(X1), a_{_dbls}(mark(X')))
MARK(sel(s(X'), X2)) -> A_{_SEL}(s(X'), mark(X2))
MARK(sel(0, X2)) -> A_{_SEL}(0, mark(X2))
A_{_SEL}(s(X), cons(Y, from(X''))) -> A_{_SEL}(mark(X), a_{_from}(X''))
A_{_SEL}(s(X), cons(Y, indx(X1', X2'))) -> A_{_SEL}(mark(X), a_{_indx}(mark(X1'), X2'))
A_{_SEL}(s(X), cons(Y, sel(X1', X2'))) -> A_{_SEL}(mark(X), a_{_sel}(mark(X1'), mark(X2')))
A_{_SEL}(s(X), cons(Y, dbls(X''))) -> A_{_SEL}(mark(X), a_{_dbls}(mark(X'')))
A_{_SEL}(s(s(X'')), cons(Y, Z)) -> A_{_SEL}(s(X''), mark(Z))
A_{_SEL}(s(0), cons(Y, Z)) -> A_{_SEL}(0, mark(Z))
A_{_SEL}(s(sel(X1', X2')), cons(Y, Z)) -> A_{_SEL}(a_{_sel}(mark(X1'), mark(X2')), mark(Z))
A_{_SEL}(s(dbl(X'')), cons(Y, Z)) -> A_{_SEL}(a_{_dbl}(mark(X'')), mark(Z))
MARK(sel(sel(X1'', X2''), X2)) -> A_{_SEL}(a_{_sel}(mark(X1''), mark(X2'')), mark(X2))
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(Z)
A_{_SEL}(s(X), cons(Y, Z)) -> MARK(X)
MARK(sel(dbl(X'), X2)) -> A_{_SEL}(a_{_dbl}(mark(X')), mark(X2))
MARK(indx(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
MARK(dbls(X)) -> MARK(X)
MARK(dbl(X)) -> MARK(X)
A_{_SEL}(0, cons(X, Y)) -> MARK(X)
A_{_SEL}(s(X), cons(Y, cons(X1', X2'))) -> A_{_SEL}(mark(X), cons(X1', X2'))

Rules:

a_{_dbl}(0) -> 0
a_{_dbl}(s(X)) -> s(s(dbl(X)))
a_{_dbl}(X) -> dbl(X)
a_{_dbls}(nil) -> nil
a_{_dbls}(cons(X, Y)) -> cons(dbl(X), dbls(Y))
a_{_dbls}(X) -> dbls(X)
a_{_sel}(0, cons(X, Y)) -> mark(X)
a_{_sel}(s(X), cons(Y, Z)) -> a_{_sel}(mark(X), mark(Z))
a_{_sel}(X1, X2) -> sel(X1, X2)
a_{_indx}(nil, X) -> nil
a_{_indx}(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
a_{_indx}(X1, X2) -> indx(X1, X2)
a_{_from}(X) -> cons(X, from(s(X)))
a_{_from}(X) -> from(X)
mark(dbl(X)) -> a_{_dbl}(mark(X))
mark(dbls(X)) -> a_{_dbls}(mark(X))
mark(sel(X1, X2)) -> a_{_sel}(mark(X1), mark(X2))
mark(indx(X1, X2)) -> a_{_indx}(mark(X1), X2)
mark(from(X)) -> a_{_from}(X)
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)

The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes