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
Polynomial Ordering
→DP Problem 2
Nar

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

The following dependency pair can be strictly oriented:

ACTIVATE(ncons(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__from(x1)) =  x1 POL(n__s(x1)) =  x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 3
Polynomial Ordering
→DP Problem 2
Nar

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

ACTIVATE(ns(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__from(x1)) =  x1 POL(n__s(x1)) =  1 + x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 3
Polo
...
→DP Problem 4
Polynomial Ordering
→DP Problem 2
Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

ACTIVATE(nfrom(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__from(x1)) =  1 + x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 3
Polo
...
→DP Problem 5
Dependency Graph
→DP Problem 2
Nar

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Narrowing Transformation

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

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
→DP Problem 6
Rewriting Transformation

Dependency Pairs:

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

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
Polo
→DP Problem 2
Nar
→DP Problem 6
Rw
...
→DP Problem 7
Narrowing Transformation

Dependency Pairs:

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

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
→DP Problem 6
Rw
...
→DP Problem 8
Rewriting Transformation

Dependency Pairs:

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'')))
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(nnil) -> LENGTH(nil)
LENGTH(ncons(X, nnil)) -> LENGTH1(nnil)

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

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

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

LENGTH1(nnil) -> LENGTH(nnil)

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Nar
→DP Problem 6
Rw
...
→DP Problem 9
Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

LENGTH(ncons(X, nnil)) -> LENGTH1(nnil)
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'')))
LENGTH(ncons(X, nfrom(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(X'') -> LENGTH(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

Innermost Termination of R could not be shown.
Duration:
0:03 minutes