Term Rewriting System R:
[X, Y, X1, X2]
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, nfrom(ns(X)))
LENGTH(ncons(X, Y)) -> S(length1(activate(Y)))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))
LENGTH(ncons(X, Y)) -> ACTIVATE(Y)
LENGTH1(X) -> LENGTH(activate(X))
LENGTH1(X) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nnil) -> NIL
ACTIVATE(ncons(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 2`
`         ↳UsableRules`

Dependency Pairs:

ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

As we are in the innermost case, we can delete all 13 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 3`
`             ↳Size-Change Principle`
`       →DP Problem 2`
`         ↳UsableRules`

Dependency Pairs:

ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
2. ACTIVATE(ns(X)) -> ACTIVATE(X)
3. ACTIVATE(nfrom(X)) -> ACTIVATE(X)
and get the following Size-Change Graph(s):
{1, 2, 3} , {1, 2, 3}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3} , {1, 2, 3}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
ncons(x1, x2) -> ncons(x1, x2)
nfrom(x1) -> nfrom(x1)
ns(x1) -> ns(x1)

We obtain no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳Usable Rules (Innermost)`

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Narrowing Transformation`

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))

Rules:

cons(X1, X2) -> ncons(X1, X2)
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(nfrom(X)) -> from(activate(X))
activate(nnil) -> nil
s(X) -> ns(X)
nil -> nnil
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Strategy:

innermost

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

LENGTH1(X) -> LENGTH(activate(X))
five new Dependency Pairs are created:

LENGTH1(ns(X'')) -> LENGTH(s(activate(X'')))
LENGTH1(ncons(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(nfrom(X'')) -> LENGTH(from(activate(X'')))
LENGTH1(nnil) -> LENGTH(nil)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 5`
`                 ↳Narrowing Transformation`

Dependency Pairs:

LENGTH1(nfrom(X'')) -> LENGTH(from(activate(X'')))
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(ncons(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(ns(X'')) -> LENGTH(s(activate(X'')))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))

Rules:

cons(X1, X2) -> ncons(X1, X2)
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(nfrom(X)) -> from(activate(X))
activate(nnil) -> nil
s(X) -> ns(X)
nil -> nnil
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Strategy:

innermost

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

LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))
five new Dependency Pairs are created:

LENGTH(ncons(X, ns(X''))) -> LENGTH1(s(activate(X'')))
LENGTH(ncons(X, ncons(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH(ncons(X, Y')) -> LENGTH1(Y')
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))
LENGTH(ncons(X, nnil)) -> LENGTH1(nil)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 6`
`                 ↳Rewriting Transformation`

Dependency Pairs:

LENGTH(ncons(X, nnil)) -> LENGTH1(nil)
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(ncons(X, Y')) -> LENGTH1(Y')
LENGTH1(ncons(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(ncons(X, ncons(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(ns(X'')) -> LENGTH(s(activate(X'')))
LENGTH(ncons(X, ns(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(nfrom(X'')) -> LENGTH(from(activate(X'')))

Rules:

cons(X1, X2) -> ncons(X1, X2)
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(nfrom(X)) -> from(activate(X))
activate(nnil) -> nil
s(X) -> ns(X)
nil -> nnil
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Strategy:

innermost

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

LENGTH(ncons(X, nnil)) -> LENGTH1(nil)
one new Dependency Pair is created:

LENGTH(ncons(X, nnil)) -> LENGTH1(nnil)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

LENGTH(ncons(X, nnil)) -> LENGTH1(nnil)
LENGTH1(nfrom(X'')) -> LENGTH(from(activate(X'')))
LENGTH(ncons(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH(ncons(X, ncons(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(ncons(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(ncons(X, ns(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(ns(X'')) -> LENGTH(s(activate(X'')))
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))

Rules:

cons(X1, X2) -> ncons(X1, X2)
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(nfrom(X)) -> from(activate(X))
activate(nnil) -> nil
s(X) -> ns(X)
nil -> nnil
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Strategy:

innermost

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

LENGTH(ncons(X, nnil)) -> LENGTH1(nnil)
LENGTH(ncons(X, ncons(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))

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

cons(X1, X2) -> ncons(X1, X2)
s(X) -> ns(X)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Used ordering:
Polynomial Order with Interpretation:

POL( LENGTH(x1) ) = max{0, x1 - 1}

POL( ncons(x1, x2) ) = x2 + 1

POL( nnil ) = 1

POL( LENGTH1(x1) ) = max{0, x1 - 1}

POL( ns(x1) ) = 0

POL( s(x1) ) = 0

POL( cons(x1, x2) ) = x2 + 1

POL( nfrom(x1) ) = 0

POL( from(x1) ) = 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

LENGTH1(nfrom(X'')) -> LENGTH(from(activate(X'')))
LENGTH(ncons(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(ncons(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(ncons(X, ns(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(ns(X'')) -> LENGTH(s(activate(X'')))
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))

Rules:

cons(X1, X2) -> ncons(X1, X2)
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(nfrom(X)) -> from(activate(X))
activate(nnil) -> nil
s(X) -> ns(X)
nil -> nnil
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Strategy:

innermost

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

LENGTH(ncons(X, ns(X''))) -> LENGTH1(s(activate(X'')))

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

cons(X1, X2) -> ncons(X1, X2)
s(X) -> ns(X)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Used ordering:
Polynomial Order with Interpretation:

POL( LENGTH(x1) ) = max{0, x1 - 1}

POL( ncons(x1, x2) ) = x2 + 1

POL( ns(x1) ) = 1

POL( LENGTH1(x1) ) = max{0, x1 - 1}

POL( s(x1) ) = 1

POL( cons(x1, x2) ) = x2 + 1

POL( nfrom(x1) ) = 0

POL( from(x1) ) = 1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

LENGTH1(nfrom(X'')) -> LENGTH(from(activate(X'')))
LENGTH(ncons(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(ncons(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(ns(X'')) -> LENGTH(s(activate(X'')))
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))

Rules:

cons(X1, X2) -> ncons(X1, X2)
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(nfrom(X)) -> from(activate(X))
activate(nnil) -> nil
s(X) -> ns(X)
nil -> nnil
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)

Strategy:

innermost

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