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

         ↳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

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: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

         ↳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

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

         ↳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

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: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = x₁ + x₂

POL(0) = 0

POL(s(x₁)) = 1 + x₁

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

         ↳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

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

         ↳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

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)

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = x₁ + x₂

POL(n__zWquot(x₁, x₂)) = 1 + x₁ + x₂

POL(ZWQUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(ACTIVATE(x₁)) = x₁

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

         ↳Nar

Dependency Pair:

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

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

         ↳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

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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳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

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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 9

                 ↳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

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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 10

                 ↳Argument Filtering and Ordering

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

The following dependency pairs 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'))))
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'''))))

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)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from) = 0

POL(from) = 0

POL(activate(x₁)) = x₁

POL(SEL(x₁, x₂)) = 1 + x₁ + x₂

POL(n__zWquot(x₁, x₂)) = x₁ + x₂

POL(zWquot(x₁, x₂)) = x₁ + x₂

POL(nil) = 0

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Nar

...

               →DP Problem 11

                 ↳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

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

POL(QUOT(x₁, x₂))	= x₁ + x₂
POL(0)	= 0
POL(s(x₁))	= 1 + x₁

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

POL(n__from)	= 0
POL(from)	= 0
POL(activate(x₁))	= x₁
POL(SEL(x₁, x₂))	= 1 + x₁ + x₂
POL(n__zWquot(x₁, x₂))	= x₁ + x₂
POL(zWquot(x₁, x₂))	= x₁ + x₂
POL(nil)	= 0
POL(s(x₁))	= 1 + x₁