Term Rewriting System R:
[X, Y, L]
eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(s(X), s(Y)) -> EQ(X, Y)
INF(X) -> INF(s(X))
TAKE(s(X), cons(Y, L)) -> TAKE(X, L)
LENGTH(cons(X, L)) -> LENGTH(L)

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

EQ(s(X), s(Y)) -> EQ(X, Y)


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





The following dependency pair can be strictly oriented:

EQ(s(X), s(Y)) -> EQ(X, Y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(EQ(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

INF(X) -> INF(s(X))


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





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

INF(X) -> INF(s(X))
one new Dependency Pair is created:

INF(s(X'')) -> INF(s(s(X'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

INF(s(X'')) -> INF(s(s(X'')))


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





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

INF(s(X'')) -> INF(s(s(X'')))
one new Dependency Pair is created:

INF(s(s(X''''))) -> INF(s(s(s(X''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 7
                 ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

INF(s(s(X''''))) -> INF(s(s(s(X''''))))


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





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

INF(s(s(X''''))) -> INF(s(s(s(X''''))))
one new Dependency Pair is created:

INF(s(s(s(X'''''')))) -> INF(s(s(s(s(X'''''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 8
                 ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

INF(s(s(s(X'''''')))) -> INF(s(s(s(s(X'''''')))))


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





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

INF(s(s(s(X'''''')))) -> INF(s(s(s(s(X'''''')))))
one new Dependency Pair is created:

INF(s(s(s(s(X''''''''))))) -> INF(s(s(s(s(s(X''''''''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 6
             ↳Inst
             ...
               →DP Problem 9
                 ↳Instantiation Transformation
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

INF(s(s(s(s(X''''''''))))) -> INF(s(s(s(s(s(X''''''''))))))


Rules:


eq(0, 0) -> true
eq(s(X), s(Y)) -> eq(X, Y)
eq(X, Y) -> false
inf(X) -> cons(X, inf(s(X)))
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(Y, take(X, L))
length(nil) -> 0
length(cons(X, L)) -> s(length(L))





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

INF(s(s(s(s(X''''''''))))) -> INF(s(s(s(s(s(X''''''''))))))
one new Dependency Pair is created:

INF(s(s(s(s(s(X'''''''''')))))) -> INF(s(s(s(s(s(s(X'''''''''')))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
       →DP Problem 3
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:

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