Termination Proof using AProVE

Term Rewriting System R:
[X, XS, N, Y, YS, X1, X2]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), n_{_zWquot}(activate(XS), activate(YS)))
zWquot(X1, X2) -> n_{_zWquot}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_zWquot}(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
MINUS(s(X), s(Y)) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_zWquot}(X1, X2)) -> ZWQUOT(X1, X2)

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

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

The following dependency pair can be strictly oriented:

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

The following usable rules w.r.t. to the AFS can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

QUOT > 0
minus > 0
s > 0

resulting in one new DP problem.
Used Argument Filtering System:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

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

The following dependency pairs can be strictly oriented:

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)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

cons > ACTIVATE > ZWQUOT

resulting in one new DP problem.
Used Argument Filtering System:

ZWQUOT(x₁, x₂) -> ZWQUOT(x₁, x₂)
ACTIVATE(x₁) -> ACTIVATE(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
n_{_zWquot}(x₁, x₂) -> n_{_zWquot}(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

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

The following dependency pair can be strictly oriented:

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

The following usable rules w.r.t. to the AFS can be oriented:

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)
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

SEL > {zWquot, activate, from} > nil > 0
SEL > {zWquot, activate, from} > n_{_from} > 0
SEL > {zWquot, activate, from} > cons > 0
SEL > {zWquot, activate, from} > s > 0
quot > 0
minus > 0
n_{_zWquot} > 0

resulting in one new DP problem.
Used Argument Filtering System:

SEL(x₁, x₂) -> SEL(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
activate(x₁) -> activate(x₁)
n_{_from}(x₁) -> n_{_from}(x₁)
from(x₁) -> from(x₁)
n_{_zWquot}(x₁, x₂) -> x₁
zWquot(x₁, x₂) -> zWquot(x₁)
quot(x₁, x₂) -> x₁
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 8

             ↳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

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes