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)

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, 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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)


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)


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 5
             ↳Non Termination


Dependency Pair:

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


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

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

R = none

s = FROM(X)
evaluates to t =FROM(s(X))

Thus, s starts an infinite chain as s matches t.

Innermost Non-Termination of R could be shown.
Duration:
0:01 minutes