Termination Proof using AProVE

Term Rewriting System R:
[X, Z, N, Y]
from(X) -> cons(X, n_{_from}(s(X)))
from(X) -> n_{_from}(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons(X, Z)) -> ACTIVATE(Z)
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSNEG(s(N), cons(X, Z)) -> ACTIVATE(Z)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons2(X, 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)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, 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, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pairs can be strictly oriented:

2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 0

POL(from(x₁)) = 0

POL(activate(x₁)) = 0

POL(cons(x₁, x₂)) = 0

POL(2NDSNEG(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

POL(cons2(x₁, x₂)) = 0

POL(2NDSPOS(x₁, x₂)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

Dependency Pairs:

2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, 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, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →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, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 5

             ↳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, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →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, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →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, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(n_{_from}(X)) -> from(X)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(n__from(x₁))	= 0
POL(from(x₁))	= 0
POL(activate(x₁))	= 0
POL(cons(x₁, x₂))	= 0
POL(2NDSNEG(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁
POL(cons2(x₁, x₂))	= 0
POL(2NDSPOS(x₁, x₂))	= x₁