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
     ↳Removing Redundant Rules for Innermost Termination



Removing the following rules from R which left hand sides contain non normal subterms

take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))


   R
     ↳RRRI
       →TRS2
         ↳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)
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)
INF(X) -> S(X)
LENGTH(nil) -> 0'
LENGTH(cons(X, L)) -> S(nlength(activate(L)))
LENGTH(cons(X, L)) -> ACTIVATE(L)

Furthermore, R contains two SCCs. 


   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Size-Change Principle
           →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
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
inf(X) -> cons(X, ninf(s(X)))
inf(X) -> ninf(X)
s(X) -> ns(X)
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0





We number the DPs as follows:
  1. LENGTH(cons(X, L)) -> ACTIVATE(L)
  2. ACTIVATE(nlength(X)) -> LENGTH(X)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
{2} , {2}
1>1

which lead(s) to this/these maximal multigraph(s):
{2} , {1}
1>1
{1} , {2}
1>1

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
nlength(x1) -> nlength(x1)
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →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
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
inf(X) -> cons(X, ninf(s(X)))
inf(X) -> ninf(X)
s(X) -> ns(X)
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0





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
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Negative Polynomial Order


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





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

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


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

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


Used ordering:
Polynomial Order with Interpretation:

POL( EQ(x1, x2) ) = x2

POL( ns(x1) ) = x1 + 1

POL( activate(x1) ) = x1 + 1

POL( ninf(x1) ) = 0

POL( inf(x1) ) = 0

POL( ntake(x1, x2) ) = 0

POL( take(x1, x2) ) = 0

POL( s(x1) ) = x1 + 1

POL( nlength(x1) ) = 0

POL( length(x1) ) = 1

POL( n0 ) = 0

POL( 0 ) = 0

POL( cons(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Negative Polynomial Order


Dependency Pairs:

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





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

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


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

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


Used ordering:
Polynomial Order with Interpretation:

POL( EQ(x1, x2) ) = x1

POL( ns(x1) ) = x1 + 1

POL( ninf(x1) ) = 0

POL( inf(x1) ) = 0

POL( s(x1) ) = x1 + 1

POL( ntake(x1, x2) ) = 0

POL( take(x1, x2) ) = 0

POL( activate(x1) ) = x1 + 1

POL( nlength(x1) ) = 0

POL( length(x1) ) = 1

POL( n0 ) = 0

POL( 0 ) = 0

POL( cons(x1, x2) ) = 0


This results in one new DP problem. 


   R
     ↳RRRI
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

EQ(ns(X), ns(nlength(X''))) -> EQ(activate(X), length(X''))
EQ(ns(nlength(X'')), ns(Y)) -> EQ(length(X''), activate(Y))


Rules:


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




The Proof could not be continued due to a Timeout.
Innermost Termination of R could not be shown.
Duration:
1:03 minutes