Term Rewriting System R:
[X, XS, N, Y, YS]
from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FROM(X) -> FROM(s(X))
SEL(s(N), cons(X, XS)) -> SEL(N, 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)) -> ZWQUOT(XS, YS)

Furthermore, R contains five SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Instantiation Transformation`
`       →DP Problem 2`
`         ↳Remaining`
`       →DP Problem 3`
`         ↳Remaining`
`       →DP Problem 4`
`         ↳Remaining`
`       →DP Problem 5`
`         ↳Remaining`

Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

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:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳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)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳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)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳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)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳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)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Inst`
`       →DP Problem 2`
`         ↳Remaining Obligation(s)`
`       →DP Problem 3`
`         ↳Remaining Obligation(s)`
`       →DP Problem 4`
`         ↳Remaining Obligation(s)`
`       →DP Problem 5`
`         ↳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)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

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

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

• Dependency Pair:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

Rules:

from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, 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), zWquot(XS, YS))

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