Term Rewriting System R:
[X, Y, Z]
first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
FROM(X) -> FROM(s(X))

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





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

FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
one new Dependency Pair is created:

FIRST(s(s(X'')), cons(Y, Z')) -> FIRST(s(X''), Z')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

FIRST(s(s(X'')), cons(Y, Z')) -> FIRST(s(X''), Z')


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





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

FIRST(s(s(X'')), cons(Y, Z')) -> FIRST(s(X''), Z')
one new Dependency Pair is created:

FIRST(s(s(s(X''''))), cons(Y, Z'')) -> FIRST(s(s(X'''')), Z'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

FIRST(s(s(s(X''''))), cons(Y, Z'')) -> FIRST(s(s(X'''')), Z'')


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





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

FIRST(s(s(s(X''''))), cons(Y, Z'')) -> FIRST(s(s(X'''')), Z'')
one new Dependency Pair is created:

FIRST(s(s(s(s(X'''''')))), cons(Y, Z''')) -> FIRST(s(s(s(X''''''))), Z''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

FIRST(s(s(s(s(X'''''')))), cons(Y, Z''')) -> FIRST(s(s(s(X''''''))), Z''')


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





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

FIRST(s(s(s(s(X'''''')))), cons(Y, Z''')) -> FIRST(s(s(s(X''''''))), Z''')
one new Dependency Pair is created:

FIRST(s(s(s(s(s(X''''''''))))), cons(Y, Z'''')) -> FIRST(s(s(s(s(X'''''''')))), Z'''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

FIRST(s(s(s(s(s(X''''''''))))), cons(Y, Z'''')) -> FIRST(s(s(s(s(X'''''''')))), Z'''')


Rules:


first(0, X) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))





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

FIRST(s(s(s(s(s(X''''''''))))), cons(Y, Z'''')) -> FIRST(s(s(s(s(X'''''''')))), Z'''')
one new Dependency Pair is created:

FIRST(s(s(s(s(s(s(X'''''''''')))))), cons(Y, Z''''')) -> FIRST(s(s(s(s(s(X''''''''''))))), Z''''')

The transformation is resulting in one new DP problem: 



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




The following remains to be proven:


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




The following remains to be proven:

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