Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N]
from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FROM(X) -> FROM(s(X))
2ND(cons(X, XS)) -> HEAD(XS)
TAKE(s(N), cons(X, XS)) -> TAKE(N, XS)
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

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)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

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)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
Dependency Pair:
TAKE(s(N), cons(X, XS)) -> TAKE(N, XS)

Rules:

from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
Dependency Pair:
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

     ↳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)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
Dependency Pair:
TAKE(s(N), cons(X, XS)) -> TAKE(N, XS)

Rules:

from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
Dependency Pair:
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

     ↳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)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
Dependency Pair:
TAKE(s(N), cons(X, XS)) -> TAKE(N, XS)

Rules:

from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
Dependency Pair:
SEL(s(N), cons(X, XS)) -> SEL(N, XS)

Rules:

from(X) -> cons(X, from(s(X)))
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(XS)
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, take(N, XS))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)

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