Term Rewriting System R:
[X, Y, L, X1, X2]
eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(ns(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(X))
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(ns(X), ns(Y)) -> EQ(activate(X), activate(Y))
EQ(ns(X), ns(Y)) -> ACTIVATE(X)
EQ(ns(X), ns(Y)) -> ACTIVATE(Y)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)
LENGTH(nil) -> 0'
LENGTH(cons(X, L)) -> S(nlength(activate(L)))
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(ninf(X)) -> INF(activate(X))
ACTIVATE(ninf(X)) -> ACTIVATE(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(activate(X1), activate(X2))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nlength(X)) -> LENGTH(activate(X))
ACTIVATE(nlength(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
ACTIVATE(nlength(X)) -> ACTIVATE(X)
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(activate(X))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
ACTIVATE(ntake(X1, X2)) -> TAKE(activate(X1), activate(X2))
ACTIVATE(ninf(X)) -> ACTIVATE(X)


Rules:


eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(ns(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(X))
activate(X) -> X





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

ACTIVATE(ntake(X1, X2)) -> TAKE(activate(X1), activate(X2))
12 new Dependency Pairs are created:

ACTIVATE(ntake(n0, X2)) -> TAKE(0, activate(X2))
ACTIVATE(ntake(ns(X'), X2)) -> TAKE(s(X'), activate(X2))
ACTIVATE(ntake(ninf(X'), X2)) -> TAKE(inf(activate(X')), activate(X2))
ACTIVATE(ntake(ntake(X1'', X2''), X2)) -> TAKE(take(activate(X1''), activate(X2'')), activate(X2))
ACTIVATE(ntake(nlength(X'), X2)) -> TAKE(length(activate(X')), activate(X2))
ACTIVATE(ntake(X1', X2)) -> TAKE(X1', activate(X2))
ACTIVATE(ntake(X1, n0)) -> TAKE(activate(X1), 0)
ACTIVATE(ntake(X1, ns(X'))) -> TAKE(activate(X1), s(X'))
ACTIVATE(ntake(X1, ninf(X'))) -> TAKE(activate(X1), inf(activate(X')))
ACTIVATE(ntake(X1, ntake(X1'', X2''))) -> TAKE(activate(X1), take(activate(X1''), activate(X2'')))
ACTIVATE(ntake(X1, nlength(X'))) -> TAKE(activate(X1), length(activate(X')))
ACTIVATE(ntake(X1, X2')) -> TAKE(activate(X1), X2')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 3
             ↳Narrowing Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

ACTIVATE(ntake(X1, X2')) -> TAKE(activate(X1), X2')
ACTIVATE(ntake(X1, nlength(X'))) -> TAKE(activate(X1), length(activate(X')))
ACTIVATE(ntake(X1, ntake(X1'', X2''))) -> TAKE(activate(X1), take(activate(X1''), activate(X2'')))
ACTIVATE(ntake(X1, ninf(X'))) -> TAKE(activate(X1), inf(activate(X')))
ACTIVATE(ntake(X1, ns(X'))) -> TAKE(activate(X1), s(X'))
ACTIVATE(ntake(X1, n0)) -> TAKE(activate(X1), 0)
ACTIVATE(ntake(X1', X2)) -> TAKE(X1', activate(X2))
ACTIVATE(ntake(nlength(X'), X2)) -> TAKE(length(activate(X')), activate(X2))
ACTIVATE(ntake(ntake(X1'', X2''), X2)) -> TAKE(take(activate(X1''), activate(X2'')), activate(X2))
ACTIVATE(ntake(ninf(X'), X2)) -> TAKE(inf(activate(X')), activate(X2))
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
ACTIVATE(ntake(ns(X'), X2)) -> TAKE(s(X'), activate(X2))
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
ACTIVATE(ntake(n0, X2)) -> TAKE(0, activate(X2))
ACTIVATE(nlength(X)) -> ACTIVATE(X)
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(activate(X))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ninf(X)) -> ACTIVATE(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)


Rules:


eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(ns(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(X))
activate(X) -> X





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

ACTIVATE(nlength(X)) -> LENGTH(activate(X))
six new Dependency Pairs are created:

ACTIVATE(nlength(n0)) -> LENGTH(0)
ACTIVATE(nlength(ns(X''))) -> LENGTH(s(X''))
ACTIVATE(nlength(ninf(X''))) -> LENGTH(inf(activate(X'')))
ACTIVATE(nlength(ntake(X1', X2'))) -> LENGTH(take(activate(X1'), activate(X2')))
ACTIVATE(nlength(nlength(X''))) -> LENGTH(length(activate(X'')))
ACTIVATE(nlength(X'')) -> LENGTH(X'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:

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