Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2]
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, n_{_from}(n_{_s}(X)))
LENGTH(n_{_cons}(X, Y)) -> S(length1(activate(Y)))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))
LENGTH(n_{_cons}(X, Y)) -> ACTIVATE(Y)
LENGTH1(X) -> LENGTH(activate(X))
LENGTH1(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_nil}) -> NIL
ACTIVATE(n_{_cons}(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(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.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

Dependency Pairs:

ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(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

D_P: empty set
Oriented Rules: none

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

trivial

with Argument Filtering System:

n_{_cons}(x₁, x₂) -> n_{_cons}(x₁, x₂)
n_{_from}(x₁) -> n_{_from}(x₁)
n_{_s}(x₁) -> n_{_s}(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(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.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 4

             ↳Narrowing Transformation

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_nil}) -> nil
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(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(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH1(n_{_nil}) -> LENGTH(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_nil}) -> nil
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

five new Dependency Pairs are created:

LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_nil}) -> nil
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)

one new Dependency Pair is created:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 4

             ↳Nar

...

               →DP Problem 7

                 ↳Negative Polynomial Order

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_nil}) -> nil
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))

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

cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)

Used ordering:
Polynomial Order with Interpretation:

POL( LENGTH(x₁) ) = max{0, x₁ - 1}

POL( n_{_cons}(x₁, x₂) ) = x₂ + 1

POL( n_{_nil} ) = 1

POL( LENGTH1(x₁) ) = max{0, x₁ - 1}

POL( n_{_s}(x₁) ) = 0

POL( s(x₁) ) = 0

POL( cons(x₁, x₂) ) = x₂ + 1

POL( n_{_from}(x₁) ) = 0

POL( from(x₁) ) = 1

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 4

             ↳Nar

...

               →DP Problem 8

                 ↳Negative Polynomial Order

Dependency Pairs:

LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_nil}) -> nil
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)

Strategy:

innermost

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

LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))

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

cons(X1, X2) -> n_{_cons}(X1, X2)
s(X) -> n_{_s}(X)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)

Used ordering:
Polynomial Order with Interpretation:

POL( LENGTH(x₁) ) = max{0, x₁ - 1}

POL( n_{_cons}(x₁, x₂) ) = x₂ + 1

POL( n_{_s}(x₁) ) = 1

POL( LENGTH1(x₁) ) = max{0, x₁ - 1}

POL( s(x₁) ) = 1

POL( cons(x₁, x₂) ) = x₂ + 1

POL( n_{_from}(x₁) ) = 0

POL( from(x₁) ) = 1

This results in one new DP problem.

     ↳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(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(X'') -> LENGTH(X'')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))

Rules:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_nil}) -> nil
s(X) -> n_{_s}(X)
nil -> n_{_nil}
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(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