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

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Instantiation Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining


Dependency Pair:

FROM(X) -> FROM(s(X))


Rules:


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


Strategy:

innermost




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

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

FROM(s(X'')) -> FROM(s(s(X'')))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

FROM(s(X'')) -> FROM(s(s(X'')))


Rules:


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


Strategy:

innermost




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

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

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Inst
           →DP Problem 4
Inst
             ...
               →DP Problem 5
Instantiation Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining


Dependency Pair:

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))


Rules:


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


Strategy:

innermost




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

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

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))


Rules:


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


Strategy:

innermost




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

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

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))

The transformation is resulting in one new DP problem:



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


Dependency Pair:

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))


Rules:


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


Strategy:

innermost




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

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

FROM(s(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(s(s(X'''''''''')))))))

The transformation is resulting in one new DP problem:



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




The following remains to be proven:


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




The following remains to be proven:


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




The following remains to be proven:

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