Term Rewriting System R:
[X, Y, Z]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfrom(X)) -> FROM(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(X) -> X





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

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
two new Dependency Pairs are created:

SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

SEL(s(X), cons(Y, Z')) -> SEL(X, Z')
SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(X) -> X





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

SEL(s(X), cons(Y, nfrom(X''))) -> SEL(X, from(X''))
two new Dependency Pairs are created:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, nfrom(X'''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(X) -> X





The following dependency pairs can be strictly oriented:

SEL(s(X), cons(Y, nfrom(X'''))) -> SEL(X, cons(X''', nfrom(s(X'''))))
SEL(s(X), cons(Y, Z')) -> SEL(X, Z')


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(n__from(x1))=  0  
  POL(SEL(x1, x2))=  x1  
  POL(cons(x1, x2))=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes