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:
• Dependency Pair:

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

• Dependency Pair:

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

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

• Dependency Pair:

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

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

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

The following remains to be proven:
• Dependency Pair:

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

• Dependency Pair:

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

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

• Dependency Pair:

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

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

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

The following remains to be proven:
• Dependency Pair:

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

• Dependency Pair:

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

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

• Dependency Pair:

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

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

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