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(s(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(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(nlength(X)) -> length(X)
activate(X) -> X

Innermost 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)
INF(X) -> S(X)
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(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)
ACTIVATE(nlength(X)) -> LENGTH(X)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Nar


Dependency Pairs:

LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(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(s(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(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(nlength(X)) -> length(X)
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(X)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
ACTIVATE(x1) -> x1
nlength(x1) -> nlength(x1)
LENGTH(x1) -> x1
cons(x1, x2) -> cons(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

EQ(ns(X), ns(Y)) -> EQ(activate(X), activate(Y))


Rules:


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


Strategy:

innermost




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

EQ(ns(X), ns(Y)) -> EQ(activate(X), activate(Y))
12 new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:44 minutes