Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Instantiation Transformation

       →DP Problem 2

         ↳Remaining

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pair:

FROM(X) -> FROM(s(X))

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)

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

FROM(X) -> FROM(s(X))

one new Dependency Pair is created:

FROM(s(X'')) -> FROM(s(s(X'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(s(X'')))

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pairs:
2NDSNEG(s(N), cons(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) -> 2NDSNEG(N, Z)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(s(X'')))

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pairs:
2NDSNEG(s(N), cons(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) -> 2NDSNEG(N, Z)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(s(X'')))

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pairs:
2NDSNEG(s(N), cons(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) -> 2NDSNEG(N, Z)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)

     ↳DPs

       →DP Problem 1

         ↳Inst

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
FROM(s(X'')) -> FROM(s(s(X'')))

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pairs:
2NDSNEG(s(N), cons(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) -> 2NDSNEG(N, Z)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)
Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)

Rules:

from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, 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)

Termination of R could not be shown.
Duration:
0:00 minutes