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)
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

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)
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

         ↳Narrowing Transformation

Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(X)
SEL(s(X), cons(Y, Z)) -> SEL(activate(X), activate(Z))
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(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
ACTIVATE(n_{_dbl}(X)) -> DBL(X)
DBL(s(X)) -> 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)
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

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

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

14 new Dependency Pairs are created:

SEL(s(n_{_s}(X'')), cons(Y, Z)) -> SEL(s(X''), activate(Z))
SEL(s(n_{_dbl}(X'')), cons(Y, Z)) -> SEL(dbl(X''), activate(Z))
SEL(s(n_{_dbls}(X'')), cons(Y, Z)) -> SEL(dbls(X''), activate(Z))
SEL(s(n_{_sel}(X1', X2')), cons(Y, Z)) -> SEL(sel(X1', X2'), activate(Z))
SEL(s(n_{_indx}(X1', X2')), cons(Y, Z)) -> SEL(indx(X1', X2'), activate(Z))
SEL(s(n_{_from}(X'')), cons(Y, Z)) -> SEL(from(X''), activate(Z))
SEL(s(X''), cons(Y, Z)) -> SEL(X'', activate(Z))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(activate(X), s(X''))
SEL(s(X), cons(Y, n_{_dbl}(X''))) -> SEL(activate(X), dbl(X''))
SEL(s(X), cons(Y, n_{_dbls}(X''))) -> SEL(activate(X), dbls(X''))
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(activate(X), sel(X1', X2'))
SEL(s(X), cons(Y, n_{_indx}(X1', X2'))) -> SEL(activate(X), indx(X1', X2'))
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(activate(X), from(X''))
SEL(s(X), cons(Y, Z')) -> SEL(activate(X), Z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

SEL(s(X), cons(Y, Z')) -> SEL(activate(X), Z')
SEL(s(X), cons(Y, n_{_from}(X''))) -> SEL(activate(X), from(X''))
SEL(s(X), cons(Y, n_{_indx}(X1', X2'))) -> SEL(activate(X), indx(X1', X2'))
SEL(s(X), cons(Y, n_{_sel}(X1', X2'))) -> SEL(activate(X), sel(X1', X2'))
SEL(s(X), cons(Y, n_{_dbls}(X''))) -> SEL(activate(X), dbls(X''))
SEL(s(X), cons(Y, n_{_dbl}(X''))) -> SEL(activate(X), dbl(X''))
SEL(s(X), cons(Y, n_{_s}(X''))) -> SEL(activate(X), s(X''))
SEL(s(X''), cons(Y, Z)) -> SEL(X'', activate(Z))
SEL(s(n_{_from}(X'')), cons(Y, Z)) -> SEL(from(X''), activate(Z))
SEL(s(n_{_indx}(X1', X2')), cons(Y, Z)) -> SEL(indx(X1', X2'), activate(Z))
SEL(s(n_{_sel}(X1', X2')), cons(Y, Z)) -> SEL(sel(X1', X2'), activate(Z))
SEL(s(n_{_dbls}(X'')), cons(Y, Z)) -> SEL(dbls(X''), activate(Z))
SEL(s(n_{_dbl}(X'')), cons(Y, Z)) -> SEL(dbl(X''), activate(Z))
SEL(s(n_{_s}(X'')), cons(Y, Z)) -> SEL(s(X''), activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(X)
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(X)
ACTIVATE(n_{_dbls}(X)) -> DBLS(X)
DBL(s(X)) -> ACTIVATE(X)
ACTIVATE(n_{_dbl}(X)) -> DBL(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)
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

Termination of R could not be shown.
Duration:
0:07 minutes