Term Rewriting System R:
[X, XS, N, Y, YS]
from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FROM(X) -> FROM(s(X))
SEL(s(N), cons(X, XS)) -> SEL(N, XS)
MINUS(s(X), s(Y)) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

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)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))


Strategy:

innermost




As we are in the innermost case, we can delete all 10 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