Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2, Z]
and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
add(0, X) -> activate(X)
add(s(X), Y) -> s(n_{_add}(activate(X), activate(Y)))
add(X1, X2) -> n_{_add}(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), n_{_first}(activate(X), activate(Z)))
first(X1, X2) -> n_{_first}(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_{_add}(X1, X2)) -> add(activate(X1), X2)
activate(n_{_first}(X1, X2)) -> first(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(X)
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

AND(true, X) -> ACTIVATE(X)
IF(true, X, Y) -> ACTIVATE(X)
IF(false, X, Y) -> ACTIVATE(Y)
ADD(0, X) -> ACTIVATE(X)
ADD(s(X), Y) -> S(n_{_add}(activate(X), activate(Y)))
ADD(s(X), Y) -> ACTIVATE(X)
ADD(s(X), Y) -> ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FROM(X) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), X2)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(activate(X1), activate(X2))
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_s}(X)) -> S(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

ADD(s(X), Y) -> ACTIVATE(Y)
ADD(s(X), Y) -> ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FROM(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(activate(X1), activate(X2))
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(0, X) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), X2)

Rules:

and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
add(0, X) -> activate(X)
add(s(X), Y) -> s(n_{_add}(activate(X), activate(Y)))
add(X1, X2) -> n_{_add}(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), n_{_first}(activate(X), activate(Z)))
first(X1, X2) -> n_{_first}(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_{_add}(X1, X2)) -> add(activate(X1), X2)
activate(n_{_first}(X1, X2)) -> first(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(X)
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

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

ACTIVATE(n_{_add}(X1, X2)) -> ADD(activate(X1), X2)

five new Dependency Pairs are created:

ACTIVATE(n_{_add}(n_{_add}(X1'', X2''), X2)) -> ADD(add(activate(X1''), X2''), X2)
ACTIVATE(n_{_add}(n_{_first}(X1'', X2''), X2)) -> ADD(first(activate(X1''), activate(X2'')), X2)
ACTIVATE(n_{_add}(n_{_from}(X'), X2)) -> ADD(from(X'), X2)
ACTIVATE(n_{_add}(n_{_s}(X'), X2)) -> ADD(s(X'), X2)
ACTIVATE(n_{_add}(X1', X2)) -> ADD(X1', X2)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(n_{_add}(X1', X2)) -> ADD(X1', X2)
ACTIVATE(n_{_add}(n_{_s}(X'), X2)) -> ADD(s(X'), X2)
ACTIVATE(n_{_add}(n_{_from}(X'), X2)) -> ADD(from(X'), X2)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(n_{_first}(X1'', X2''), X2)) -> ADD(first(activate(X1''), activate(X2'')), X2)
ADD(0, X) -> ACTIVATE(X)
ACTIVATE(n_{_add}(n_{_add}(X1'', X2''), X2)) -> ADD(add(activate(X1''), X2''), X2)
FROM(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
ACTIVATE(n_{_first}(X1, X2)) -> FIRST(activate(X1), activate(X2))
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(Y)

Rules:

and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
add(0, X) -> activate(X)
add(s(X), Y) -> s(n_{_add}(activate(X), activate(Y)))
add(X1, X2) -> n_{_add}(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), n_{_first}(activate(X), activate(Z)))
first(X1, X2) -> n_{_first}(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_{_add}(X1, X2)) -> add(activate(X1), X2)
activate(n_{_first}(X1, X2)) -> first(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(X)
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

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

ACTIVATE(n_{_first}(X1, X2)) -> FIRST(activate(X1), activate(X2))

10 new Dependency Pairs are created:

ACTIVATE(n_{_first}(n_{_add}(X1'', X2''), X2)) -> FIRST(add(activate(X1''), X2''), activate(X2))
ACTIVATE(n_{_first}(n_{_first}(X1'', X2''), X2)) -> FIRST(first(activate(X1''), activate(X2'')), activate(X2))
ACTIVATE(n_{_first}(n_{_from}(X'), X2)) -> FIRST(from(X'), activate(X2))
ACTIVATE(n_{_first}(n_{_s}(X'), X2)) -> FIRST(s(X'), activate(X2))
ACTIVATE(n_{_first}(X1', X2)) -> FIRST(X1', activate(X2))
ACTIVATE(n_{_first}(X1, n_{_add}(X1'', X2''))) -> FIRST(activate(X1), add(activate(X1''), X2''))
ACTIVATE(n_{_first}(X1, n_{_first}(X1'', X2''))) -> FIRST(activate(X1), first(activate(X1''), activate(X2'')))
ACTIVATE(n_{_first}(X1, n_{_from}(X'))) -> FIRST(activate(X1), from(X'))
ACTIVATE(n_{_first}(X1, n_{_s}(X'))) -> FIRST(activate(X1), s(X'))
ACTIVATE(n_{_first}(X1, X2')) -> FIRST(activate(X1), X2')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

ACTIVATE(n_{_first}(X1, X2')) -> FIRST(activate(X1), X2')
ACTIVATE(n_{_first}(X1, n_{_s}(X'))) -> FIRST(activate(X1), s(X'))
ACTIVATE(n_{_first}(X1, n_{_from}(X'))) -> FIRST(activate(X1), from(X'))
ACTIVATE(n_{_first}(X1, n_{_first}(X1'', X2''))) -> FIRST(activate(X1), first(activate(X1''), activate(X2'')))
ACTIVATE(n_{_first}(X1, n_{_add}(X1'', X2''))) -> FIRST(activate(X1), add(activate(X1''), X2''))
ACTIVATE(n_{_first}(X1', X2)) -> FIRST(X1', activate(X2))
ACTIVATE(n_{_first}(n_{_s}(X'), X2)) -> FIRST(s(X'), activate(X2))
ACTIVATE(n_{_first}(n_{_from}(X'), X2)) -> FIRST(from(X'), activate(X2))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(n_{_first}(n_{_first}(X1'', X2''), X2)) -> FIRST(first(activate(X1''), activate(X2'')), activate(X2))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Y)
ACTIVATE(n_{_first}(n_{_add}(X1'', X2''), X2)) -> FIRST(add(activate(X1''), X2''), activate(X2))
ACTIVATE(n_{_add}(X1', X2)) -> ADD(X1', X2)
ACTIVATE(n_{_add}(n_{_s}(X'), X2)) -> ADD(s(X'), X2)
ADD(s(X), Y) -> ACTIVATE(Y)
ACTIVATE(n_{_add}(n_{_from}(X'), X2)) -> ADD(from(X'), X2)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(n_{_add}(n_{_first}(X1'', X2''), X2)) -> ADD(first(activate(X1''), activate(X2'')), X2)
ADD(0, X) -> ACTIVATE(X)
ACTIVATE(n_{_add}(n_{_add}(X1'', X2''), X2)) -> ADD(add(activate(X1''), X2''), X2)
FROM(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_first}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_add}(X1, X2)) -> ACTIVATE(X1)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)

Rules:

and(true, X) -> activate(X)
and(false, Y) -> false
if(true, X, Y) -> activate(X)
if(false, X, Y) -> activate(Y)
add(0, X) -> activate(X)
add(s(X), Y) -> s(n_{_add}(activate(X), activate(Y)))
add(X1, X2) -> n_{_add}(X1, X2)
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(activate(Y), n_{_first}(activate(X), activate(Z)))
first(X1, X2) -> n_{_first}(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_{_add}(X1, X2)) -> add(activate(X1), X2)
activate(n_{_first}(X1, X2)) -> first(activate(X1), activate(X2))
activate(n_{_from}(X)) -> from(X)
activate(n_{_s}(X)) -> s(X)
activate(X) -> X

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