Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 6

             ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

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))

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 7

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 8

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

           →DP Problem 6

             ↳Inst

...

               →DP Problem 9

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

       →DP Problem 5

         ↳Remaining

Dependency Pair:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳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(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(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))

     ↳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(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(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))

     ↳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(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(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))

     ↳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(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(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))

     ↳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(s(s(s(s(X'''''''''')))))) -> FROM(s(s(s(s(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