Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N, X1, X2]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

2ND(cons(X, XS)) -> HEAD(activate(XS))
2ND(cons(X, XS)) -> ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

Dependency Pairs:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

The following dependency pairs can be strictly oriented:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(n_{_take}(X1, X2)) -> TAKE(X1, X2)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

cons > ACTIVATE
n_{_take} > TAKE

resulting in one new DP problem.
Used Argument Filtering System:

TAKE(x₁, x₂) -> TAKE(x₁, x₂)
ACTIVATE(x₁) -> ACTIVATE(x₁)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
n_{_take}(x₁, x₂) -> n_{_take}(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))

The following usable rules w.r.t. to the AFS can be oriented:

activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

SEL > activate > from > n_{_from}
SEL > activate > from > cons
SEL > activate > from > s
{n_{_take}, take} > nil
{n_{_take}, take} > activate > from > n_{_from}
{n_{_take}, take} > activate > from > cons
{n_{_take}, take} > activate > from > s

resulting in one new DP problem.
Used Argument Filtering System:

SEL(x₁, x₂) -> SEL(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
activate(x₁) -> activate(x₁)
n_{_from}(x₁) -> n_{_from}(x₁)
from(x₁) -> from(x₁)
n_{_take}(x₁, x₂) -> n_{_take}(x₁, x₂)
take(x₁, x₂) -> take(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, n_{_take}(N, activate(XS)))
take(X1, X2) -> n_{_take}(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(n_{_from}(X)) -> from(X)
activate(n_{_take}(X1, X2)) -> take(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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