Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N, Y, YS, X1, X2]
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n_{_zWquot}(activate(XS), activate(YS)))
zWquot(X1, X2) -> n_{_zWquot}(X1, X2)
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_zWquot}(X1, X2)) -> zWquot(activate(X1), activate(X2))
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
MINUS(s(X), s(Y)) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> S(quot(minus(X, Y), s(Y)))
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(activate(X1), activate(X2))
ACTIVATE(n_{_zWquot}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ACTIVATE(X2)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

ACTIVATE(n_{_zWquot}(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ACTIVATE(X1)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(activate(X1), activate(X2))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n_{_zWquot}(activate(XS), activate(YS)))
zWquot(X1, X2) -> n_{_zWquot}(X1, X2)
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_zWquot}(X1, X2)) -> zWquot(activate(X1), activate(X2))
activate(X) -> X

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(activate(X1), activate(X2))

eight new Dependency Pairs are created:

ACTIVATE(n_{_zWquot}(n_{_from}(X'), X2)) -> ZWQUOT(from(activate(X')), activate(X2))
ACTIVATE(n_{_zWquot}(n_{_s}(X'), X2)) -> ZWQUOT(s(activate(X')), activate(X2))
ACTIVATE(n_{_zWquot}(n_{_zWquot}(X1'', X2''), X2)) -> ZWQUOT(zWquot(activate(X1''), activate(X2'')), activate(X2))
ACTIVATE(n_{_zWquot}(X1', X2)) -> ZWQUOT(X1', activate(X2))
ACTIVATE(n_{_zWquot}(X1, n_{_from}(X'))) -> ZWQUOT(activate(X1), from(activate(X')))
ACTIVATE(n_{_zWquot}(X1, n_{_s}(X'))) -> ZWQUOT(activate(X1), s(activate(X')))
ACTIVATE(n_{_zWquot}(X1, n_{_zWquot}(X1'', X2''))) -> ZWQUOT(activate(X1), zWquot(activate(X1''), activate(X2'')))
ACTIVATE(n_{_zWquot}(X1, X2')) -> ZWQUOT(activate(X1), X2')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

ACTIVATE(n_{_zWquot}(X1, X2')) -> ZWQUOT(activate(X1), X2')
ACTIVATE(n_{_zWquot}(X1, n_{_zWquot}(X1'', X2''))) -> ZWQUOT(activate(X1), zWquot(activate(X1''), activate(X2'')))
ACTIVATE(n_{_zWquot}(X1, n_{_s}(X'))) -> ZWQUOT(activate(X1), s(activate(X')))
ACTIVATE(n_{_zWquot}(X1, n_{_from}(X'))) -> ZWQUOT(activate(X1), from(activate(X')))
ACTIVATE(n_{_zWquot}(X1', X2)) -> ZWQUOT(X1', activate(X2))
ACTIVATE(n_{_zWquot}(n_{_zWquot}(X1'', X2''), X2)) -> ZWQUOT(zWquot(activate(X1''), activate(X2'')), activate(X2))
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ACTIVATE(n_{_zWquot}(n_{_s}(X'), X2)) -> ZWQUOT(s(activate(X')), activate(X2))
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(n_{_zWquot}(n_{_from}(X'), X2)) -> ZWQUOT(from(activate(X')), activate(X2))
ACTIVATE(n_{_zWquot}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ACTIVATE(X2)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n_{_zWquot}(activate(XS), activate(YS)))
zWquot(X1, X2) -> n_{_zWquot}(X1, X2)
s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_zWquot}(X1, X2)) -> zWquot(activate(X1), activate(X2))
activate(X) -> X

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:06 minutes