Termination Proof using AProVE

Term Rewriting System R:
[X, Z, N, Y, X1, X2]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, n_{_from}(s(X)))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z))) -> ACTIVATE(Y)
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z))) -> ACTIVATE(Z)
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z))) -> ACTIVATE(Y)
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z))) -> ACTIVATE(Z)
PI(X) -> 2NDSPOS(X, from(0))
PI(X) -> FROM(0)
PLUS(s(X), Y) -> PLUS(X, Y)
TIMES(s(X), Y) -> PLUS(Y, times(X, Y))
TIMES(s(X), Y) -> TIMES(X, Y)
SQUARE(X) -> TIMES(X, X)
ACTIVATE(n_{_from}(X)) -> FROM(X)
ACTIVATE(n_{_cons}(X1, X2)) -> CONS(X1, X2)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSNEG(N, activate(Z))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

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

2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSNEG(N, activate(Z))

three new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSNEG(N, from(X''))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSNEG(N, cons(X1', X2'))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSNEG(N, Z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 4

             ↳Narrowing Transformation

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSNEG(N, cons(X1', X2'))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSNEG(N, from(X''))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSPOS(N, activate(Z))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

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

2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z))) -> 2NDSPOS(N, activate(Z))

three new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSPOS(N, from(X''))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSPOS(N, cons(X1', X2'))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSPOS(N, Z')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSNEG(N, cons(X1', X2'))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSPOS(N, cons(X1', X2'))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSNEG(N, from(X''))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSPOS(N, from(X''))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSNEG(N, Z')

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

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

2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSNEG(N, from(X''))

two new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSNEG(N, cons(X''', n_{_from}(s(X'''))))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSNEG(N, n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Narrowing Transformation

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSNEG(N, cons(X''', n_{_from}(s(X'''))))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSPOS(N, cons(X1', X2'))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSPOS(N, from(X''))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSNEG(N, cons(X1', X2'))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSPOS(N, Z')

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

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

2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X'')))) -> 2NDSPOS(N, from(X''))

two new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSPOS(N, cons(X''', n_{_from}(s(X'''))))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSPOS(N, n_{_from}(X'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSPOS(N, cons(X''', n_{_from}(s(X'''))))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSNEG(N, cons(X1', X2'))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSPOS(N, cons(X1', X2'))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSNEG(N, cons(X''', n_{_from}(s(X'''))))

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

The following dependency pairs can be strictly oriented:

2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSPOS(N, cons(X''', n_{_from}(s(X'''))))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, n_{_cons}(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSNEG(N, cons(X1', X2'))
2NDSNEG(s(N), cons(X, n_{_cons}(Y, n_{_cons}(X1', X2')))) -> 2NDSPOS(N, cons(X1', X2'))
2NDSPOS(s(N), cons(X, n_{_cons}(Y, n_{_from}(X''')))) -> 2NDSNEG(N, cons(X''', n_{_from}(s(X'''))))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__cons(x₁, x₂)) = 0

POL(n__from(x₁)) = 0

POL(cons(x₁, x₂)) = 0

POL(2NDSNEG(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(2NDSPOS(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

Dependency Pair:

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

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(PLUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

Dependency Pair:

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

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(TIMES(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Nar

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 10

             ↳Dependency Graph

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, n_{_cons}(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(n__cons(x₁, x₂))	= 0
POL(n__from(x₁))	= 0
POL(cons(x₁, x₂))	= 0
POL(2NDSNEG(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁
POL(2NDSPOS(x₁, x₂))	= x₁