Termination Proof using AProVE

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)

Termination of R to be shown.

     ↳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.

     ↳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)

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:

     ↳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(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)
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)
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)

     ↳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(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)
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)
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)

     ↳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(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)
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)
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)

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