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

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)) -> 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

         ↳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, 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

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:

     ↳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, 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

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:

     ↳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, 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

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 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.

     ↳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, 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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳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, 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

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:

     ↳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, 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

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:

     ↳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, 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

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.

     ↳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(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

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, n_{_zWquot}(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳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, n_{_zWquot}(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(X1'', X2''))
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(X1, X2)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)

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

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

The transformation is resulting in one new DP problem:

     ↳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, n_{_zWquot}(X1'', X2''))) -> ACTIVATE(n_{_zWquot}(X1'', X2''))
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(X1, X2)
ZWQUOT(cons(X, n_{_zWquot}(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(n_{_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

Strategy:

innermost

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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳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(n_{_zWquot}(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))
ZWQUOT(cons(X, n_{_zWquot}(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(X1'', X2''))
ACTIVATE(n_{_zWquot}(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(X1'', X2''))) -> ACTIVATE(n_{_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

Strategy:

innermost

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

ZWQUOT(cons(X, n_{_zWquot}(X1'', X2'')), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(X1'', X2''))

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳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, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(X1'', X2''))) -> ACTIVATE(n_{_zWquot}(X1'', X2''))
ACTIVATE(n_{_zWquot}(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', n_{_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

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳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, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_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

Strategy:

innermost

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

ACTIVATE(n_{_zWquot}(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))) -> ZWQUOT(cons(X'', n_{_zWquot}(X1'''', X2'''')), cons(Y'', YS''))

four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳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(n_{_zWquot}(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))), cons(Y''', YS'''))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_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

Strategy:

innermost

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

ACTIVATE(n_{_zWquot}(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))) -> ZWQUOT(cons(X'', XS''), cons(Y'', n_{_zWquot}(X1'''', X2'''')))

four new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳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(n_{_zWquot}(cons(X''', XS'''), cons(Y''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))))
ACTIVATE(n_{_zWquot}(cons(X''', XS'''), cons(Y''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', XS'''), cons(Y''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))), cons(Y''', n_{_zWquot}(X1''''', X2''''')))) -> ZWQUOT(cons(X''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))), cons(Y''', n_{_zWquot}(X1''''', X2''''')))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', n_{_zWquot}(X1''''', X2''''')))) -> ZWQUOT(cons(X''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', n_{_zWquot}(X1''''', X2''''')))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))) -> ZWQUOT(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))), cons(Y''', YS'''))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))) -> ZWQUOT(cons(X''', n_{_zWquot}(cons(X'''''', n_{_zWquot}(X1'''''''', X2'''''''')), cons(Y'''''', YS''''''))), cons(Y''', YS'''))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ACTIVATE(n_{_zWquot}(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_zWquot}(X1'''''''', X2'''''''')))))) -> ZWQUOT(cons(X''', n_{_zWquot}(X1''''', X2''''')), cons(Y''', n_{_zWquot}(cons(X'''''', XS''''''), cons(Y'''''', n_{_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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost 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.

     ↳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, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ZWQUOT(cons(X, XS), cons(Y, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2'''''')))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', XS''''), cons(Y'''', n_{_zWquot}(X1'''''', X2''''''))))
ZWQUOT(cons(X, n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS''''))), cons(Y, YS)) -> ACTIVATE(n_{_zWquot}(cons(X'''', n_{_zWquot}(X1'''''', X2'''''')), cons(Y'''', YS'''')))

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳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, 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

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, n_{_from}(X''))) -> SEL(N, from(X''))
SEL(s(N), cons(X, n_{_zWquot}(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:

     ↳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, n_{_zWquot}(X1', X2'))) -> SEL(N, zWquot(X1', X2'))
SEL(s(N), cons(X, n_{_from}(X''))) -> SEL(N, from(X''))

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

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, n_{_from}(X''))) -> SEL(N, from(X''))

two new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))
SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳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, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))
SEL(s(N), cons(X, n_{_zWquot}(X1', X2'))) -> SEL(N, zWquot(X1', X2'))
SEL(s(N), cons(X, XS')) -> SEL(N, 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

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, n_{_zWquot}(X1', X2'))) -> SEL(N, zWquot(X1', X2'))

four new Dependency Pairs are created:

SEL(s(N), cons(X, n_{_zWquot}(X1'', nil))) -> SEL(N, nil)
SEL(s(N), cons(X, n_{_zWquot}(nil, X2''))) -> SEL(N, nil)
SEL(s(N), cons(X, n_{_zWquot}(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), n_{_zWquot}(activate(XS'), activate(YS'))))
SEL(s(N), cons(X, n_{_zWquot}(X1'', X2''))) -> SEL(N, n_{_zWquot}(X1'', X2''))

The transformation is resulting in one new DP problem:

     ↳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, n_{_zWquot}(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), n_{_zWquot}(activate(XS'), activate(YS'))))
SEL(s(N), cons(X, XS')) -> SEL(N, XS')
SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))

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

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'', n_{_from}(X''''')))) -> SEL(s(N''), cons(X'', n_{_from}(X''''')))
SEL(s(s(N'')), cons(X, cons(X'', n_{_zWquot}(cons(X'''', XS'''), cons(Y''', YS'''))))) -> SEL(s(N''), cons(X'', n_{_zWquot}(cons(X'''', XS'''), cons(Y''', YS'''))))

The transformation is resulting in three new DP problems:

     ↳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, n_{_zWquot}(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), n_{_zWquot}(activate(XS'), activate(YS'))))

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, n_{_zWquot}(cons(X'', XS'), cons(Y', YS')))) -> SEL(N, cons(quot(X'', Y'), n_{_zWquot}(activate(XS'), activate(YS'))))

Additionally, the following usable rules for innermost can be oriented:

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X
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)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 0

POL(from(x₁)) = 0

POL(activate(x₁)) = x₁

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(quot(x₁, x₂)) = x₁

POL(nil) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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, 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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳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, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, n_{_from}(X'''))) -> SEL(N, cons(X''', n_{_from}(s(X'''))))

There are no usable rules for innermost 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.

     ↳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, 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

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 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:
0:14 minutes

POL(MINUS(x₁, x₂))	= 1 + x₁
POL(s(x₁))	= 1 + x₁

POL(n__zWquot(x₁, x₂))	= 1 + x₁ + x₂
POL(cons(x₁, x₂))	= x₂
POL(ZWQUOT(x₁, x₂))	= x₁ + x₂
POL(ACTIVATE(x₁))	= x₁

POL(n__from(x₁))	= 0
POL(from(x₁))	= 0
POL(activate(x₁))	= x₁
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(quot(x₁, x₂))	= x₁
POL(nil)	= 0
POL(s(x₁))	= 1 + x₁