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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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


   R
     ↳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, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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)


Strategy:

innermost




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: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 5
             ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


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, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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)


Strategy:

innermost




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

FROM(s(X'')) -> FROM(s(s(X'')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 5
             ↳Inst
             ...
               →DP Problem 6
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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)


Strategy:

innermost




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

FROM(s(s(X''''))) -> FROM(s(s(s(X''''))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 5
             ↳Inst
             ...
               →DP Problem 7
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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)


Strategy:

innermost




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

FROM(s(s(s(X'''''')))) -> FROM(s(s(s(s(X'''''')))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 5
             ↳Inst
             ...
               →DP Problem 8
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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)


Strategy:

innermost




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

FROM(s(s(s(s(X''''''''))))) -> FROM(s(s(s(s(s(X''''''''))))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳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:


   R
     ↳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:


   R
     ↳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:


   R
     ↳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:

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