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

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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
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

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

The following remains to be proven:

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
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

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