Term Rewriting System R:
[X, XS, N, Y, YS, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.

`   R`
`     ↳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(nfrom(X)) -> FROM(X)
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

MINUS(s(X), s(Y)) -> MINUS(X, Y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

MINUS(s(s(X'')), s(s(Y''))) -> MINUS(s(X''), s(Y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(MINUS(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
two new Dependency Pairs are created:

QUOT(s(X''), s(0)) -> QUOT(0, s(0))
QUOT(s(s(X'')), s(s(Y''))) -> QUOT(minus(X'', Y''), s(s(Y'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Narrowing Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

QUOT(s(s(X'')), s(s(Y''))) -> QUOT(minus(X'', Y''), s(s(Y'')))
two new Dependency Pairs are created:

QUOT(s(s(X''')), s(s(0))) -> QUOT(0, s(s(0)))
QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(minus(X', Y'), s(s(s(Y'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 8`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

QUOT(s(s(s(X'))), s(s(s(Y')))) -> QUOT(minus(X', Y'), s(s(s(Y'))))
no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
one new Dependency Pair is created:

ZWQUOT(cons(X, nzWquot(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(nzWquot(X1'', X2''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ZWQUOT(cons(X, nzWquot(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(nzWquot(X1'', X2''))
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
one new Dependency Pair is created:

ZWQUOT(cons(X, XS), cons(Y, nzWquot(X1'', X2''))) -> ACTIVATE(nzWquot(X1'', X2''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 11`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ZWQUOT(cons(X, XS), cons(Y, nzWquot(X1'', X2''))) -> ACTIVATE(nzWquot(X1'', X2''))
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)
ZWQUOT(cons(X, nzWquot(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(nzWquot(X1'', X2''))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)
two new Dependency Pairs are created:

ACTIVATE(nzWquot(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))
ACTIVATE(nzWquot(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ACTIVATE(nzWquot(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))
ZWQUOT(cons(X, nzWquot(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(nzWquot(X1'', X2''))
ACTIVATE(nzWquot(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(X1'', X2''))) -> ACTIVATE(nzWquot(X1'', X2''))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ZWQUOT(cons(X, nzWquot(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(nzWquot(X1'', X2''))
two new Dependency Pairs are created:

ZWQUOT(cons(X, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ZWQUOT(cons(X, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 13`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ZWQUOT(cons(X, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ZWQUOT(cons(X, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(X1'', X2''))) -> ACTIVATE(nzWquot(X1'', X2''))
ACTIVATE(nzWquot(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ZWQUOT(cons(X, XS), cons(Y, nzWquot(X1'', X2''))) -> ACTIVATE(nzWquot(X1'', X2''))
two new Dependency Pairs are created:

ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 14`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))
ZWQUOT(cons(X, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))
ZWQUOT(cons(X, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(nzWquot(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', nzWquot(X1'''', X2'''')), cons(Y'', YS''))
four new Dependency Pairs are created:

ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 15`
`                 ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))
ZWQUOT(cons(X, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))
ZWQUOT(cons(X, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

ACTIVATE(nzWquot(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', nzWquot(X1'''', X2'''')))
four new Dependency Pairs are created:

ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', nzWquot(X1''''', X2''''')))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', nzWquot(X1''''', X2''''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', nzWquot(X1''''', X2''''')))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', nzWquot(X1''''', X2''''')))
ACTIVATE(nzWquot(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ACTIVATE(nzWquot(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 16`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ACTIVATE(nzWquot(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))
ACTIVATE(nzWquot(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', nzWquot(X1''''', X2''''')))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', nzWquot(X1''''', X2''''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', nzWquot(X1''''', X2''''')))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', nzWquot(X1''''', X2''''')))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))
ZWQUOT(cons(X, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))
ZWQUOT(cons(X, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pairs can be strictly oriented:

ACTIVATE(nzWquot(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))
ACTIVATE(nzWquot(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', nzWquot(X1''''', X2''''')))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', nzWquot(X1''''', X2''''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', nzWquot(X1''''', X2''''')))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', nzWquot(X1''''', X2''''')))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))), cons(Y''', YS'''))
ACTIVATE(nzWquot(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', nzWquot(cons(X'''''', nzWquot(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))
ACTIVATE(nzWquot(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', nzWquot(X1''''', X2''''')), cons(Y''', nzWquot(cons(X'''''', XS''''''), cons(Y'''''', nzWquot(X1'''''''', X2'''''''')))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(n__zWquot(x1, x2)) =  1 + x1 + x2 POL(cons(x1, x2)) =  x2 POL(ZWQUOT(x1, x2)) =  x1 + x2 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 10`
`             ↳FwdInst`
`             ...`
`               →DP Problem 17`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pairs:

ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ZWQUOT(cons(X, XS), cons(Y, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ZWQUOT(cons(X, nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', XS''''), cons(Y'''', nzWquot(X1'''''', X2''''''))))
ZWQUOT(cons(X, nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(nzWquot(cons(X'''', nzWquot(X1'''''', X2'''''')), cons(Y'''', YS'''')))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Narrowing Transformation`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
three new Dependency Pairs are created:

SEL(s(N), cons(X, nfrom(X''))) -> SEL(N, from(X''))
SEL(s(N), cons(X, nzWquot(X1', X2'))) -> SEL(N, zWquot(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Narrowing Transformation`

Dependency Pairs:

SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, nzWquot(X1', X2'))) -> SEL(N, zWquot(X1', X2'))
SEL(s(N), cons(X, nfrom(X''))) -> SEL(N, from(X''))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

SEL(s(N), cons(X, nfrom(X''))) -> SEL(N, from(X''))
two new Dependency Pairs are created:

SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))
SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, nfrom(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Narrowing Transformation`

Dependency Pairs:

SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))
SEL(s(N), cons(X, nzWquot(X1', X2'))) -> SEL(N, zWquot(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

SEL(s(N), cons(X, nzWquot(X1', X2'))) -> SEL(N, zWquot(X1', X2'))
four new Dependency Pairs are created:

SEL(s(N), cons(X, nzWquot(X1'', nil))) -> SEL(N, nil)
SEL(s(N), cons(X, nzWquot(nil, X2''))) -> SEL(N, nil)
SEL(s(N), cons(X, nzWquot(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), nzWquot(activate(XS'), activate(YS'))))
SEL(s(N), cons(X, nzWquot(X1'', X2''))) -> SEL(N, nzWquot(X1'', X2''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Nar`
`             ...`
`               →DP Problem 20`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

SEL(s(N), cons(X, nzWquot(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), nzWquot(activate(XS'), activate(YS'))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

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

SEL(s(N), cons(X, XS')) -> SEL(N, XS')
three new Dependency Pairs are created:

SEL(s(s(N'')), cons(X, cons(X'', XS'''))) -> SEL(s(N''), cons(X'', XS'''))
SEL(s(s(N'')), cons(X, cons(X'', nfrom(X''''')))) -> SEL(s(N''), cons(X'', nfrom(X''''')))
SEL(s(s(N'')), cons(X, cons(X'', nzWquot(cons(X'''', XS'''), cons(Y''', YS'''))))) -> SEL(s(N''), cons(X'', nzWquot(cons(X'''', XS'''), cons(Y''', YS'''))))

The transformation is resulting in three new DP problems:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Nar`
`             ...`
`               →DP Problem 21`
`                 ↳Polynomial Ordering`

Dependency Pair:

SEL(s(N), cons(X, nzWquot(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), nzWquot(activate(XS'), activate(YS'))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, nzWquot(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), nzWquot(activate(XS'), activate(YS'))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(n__from(x1)) =  0 POL(from(x1)) =  0 POL(activate(x1)) =  0 POL(0) =  0 POL(n__zWquot(x1, x2)) =  0 POL(zWquot(x1, x2)) =  0 POL(cons(x1, x2)) =  0 POL(SEL(x1, x2)) =  x1 POL(minus(x1, x2)) =  0 POL(nil) =  0 POL(quot(x1, x2)) =  0 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Nar`
`             ...`
`               →DP Problem 24`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Nar`
`             ...`
`               →DP Problem 22`
`                 ↳Polynomial Ordering`

Dependency Pair:

SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, nfrom(X'''))) -> SEL(N, cons(X''', nfrom(s(X'''))))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(n__from(x1)) =  0 POL(SEL(x1, x2)) =  x1 POL(cons(x1, x2)) =  0 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 18`
`             ↳Nar`
`             ...`
`               →DP Problem 23`
`                 ↳Polynomial Ordering`

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(SEL(x1, x2)) =  x1 POL(cons(x1, x2)) =  0 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

Innermost Termination of R successfully shown.
Duration:
0:13 minutes