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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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

The following dependency pairs can be strictly oriented:

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))

The following usable rules using the Ce-refinement can be oriented:

activate(n_{_from}(X)) -> from(X)
activate(n_{_cons}(X1, X2)) -> cons(X1, X2)
activate(X) -> X
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
cons(X1, X2) -> n_{_cons}(X1, X2)

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

{2NDSPOS, 2NDSNEG} > activate > from > n_{_from}
{2NDSPOS, 2NDSNEG} > activate > from > {n_{_cons}, cons}
{2NDSPOS, 2NDSNEG} > activate > from > s

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

2NDSPOS(x₁, x₂) -> 2NDSPOS(x₁, x₂)
2NDSNEG(x₁, x₂) -> 2NDSNEG(x₁, x₂)
s(x₁) -> s(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)
activate(x₁) -> activate(x₁)
n_{_cons}(x₁, x₂) -> n_{_cons}(x₁, x₂)
n_{_from}(x₁) -> n_{_from}(x₁)
from(x₁) -> from(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

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 using the Ce-refinement 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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and 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 using the Ce-refinement 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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳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