Termination Proof using AProVE

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

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

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X
Dependency Pairs:
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(X1, X2)

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(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)
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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X
Dependency Pairs:
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(X1, X2)

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(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)
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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X
Dependency Pairs:
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(X1, X2)

Rules:

from(X) -> cons(X, n_{_from}(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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(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)
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)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

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