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

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

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

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_cons}(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.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_s}(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.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_from}(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.

     ↳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, 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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Narrowing Transformation

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

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Innermost Termination of R could not be shown.
Duration:
0:00 minutes

POL(n__cons(x₁, x₂))	= 1 + x₁
POL(n__from(x₁))	= x₁
POL(n__s(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁