NO

Problem 1: 

(VAR v_NonEmpty:S N:S X:S XS:S Y:S YS:S)
(RULES
from(X:S) -> cons(X:S,from(s(X:S)))
minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
minus(X:S,0) -> 0
quot(0,s(Y:S)) -> 0
quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S)))
sel(0,cons(X:S,XS:S)) -> X:S
sel(s(N:S),cons(X:S,XS:S)) -> sel(N:S,XS:S)
zWquot(cons(X:S,XS:S),cons(Y:S,YS:S)) -> cons(quot(X:S,Y:S),zWquot(XS:S,YS:S))
zWquot(nil,XS:S) -> nil
zWquot(XS:S,nil) -> nil
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 FROM(X:S) -> FROM(s(X:S))
 MINUS(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
 QUOT(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
 QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S))
 SEL(s(N:S),cons(X:S,XS:S)) -> SEL(N:S,XS:S)
 ZWQUOT(cons(X:S,XS:S),cons(Y:S,YS:S)) -> QUOT(X:S,Y:S)
 ZWQUOT(cons(X:S,XS:S),cons(Y:S,YS:S)) -> ZWQUOT(XS:S,YS:S)
-> Rules:
 from(X:S) -> cons(X:S,from(s(X:S)))
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
 minus(X:S,0) -> 0
 quot(0,s(Y:S)) -> 0
 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S)))
 sel(0,cons(X:S,XS:S)) -> X:S
 sel(s(N:S),cons(X:S,XS:S)) -> sel(N:S,XS:S)
 zWquot(cons(X:S,XS:S),cons(Y:S,YS:S)) -> cons(quot(X:S,Y:S),zWquot(XS:S,YS:S))
 zWquot(nil,XS:S) -> nil
 zWquot(XS:S,nil) -> nil

Problem 1: 

Infinite Processor:
-> Pairs:
 FROM(X:S) -> FROM(s(X:S))
 MINUS(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
 QUOT(s(X:S),s(Y:S)) -> MINUS(X:S,Y:S)
 QUOT(s(X:S),s(Y:S)) -> QUOT(minus(X:S,Y:S),s(Y:S))
 SEL(s(N:S),cons(X:S,XS:S)) -> SEL(N:S,XS:S)
 ZWQUOT(cons(X:S,XS:S),cons(Y:S,YS:S)) -> QUOT(X:S,Y:S)
 ZWQUOT(cons(X:S,XS:S),cons(Y:S,YS:S)) -> ZWQUOT(XS:S,YS:S)
-> Rules:
 from(X:S) -> cons(X:S,from(s(X:S)))
 minus(s(X:S),s(Y:S)) -> minus(X:S,Y:S)
 minus(X:S,0) -> 0
 quot(0,s(Y:S)) -> 0
 quot(s(X:S),s(Y:S)) -> s(quot(minus(X:S,Y:S),s(Y:S)))
 sel(0,cons(X:S,XS:S)) -> X:S
 sel(s(N:S),cons(X:S,XS:S)) -> sel(N:S,XS:S)
 zWquot(cons(X:S,XS:S),cons(Y:S,YS:S)) -> cons(quot(X:S,Y:S),zWquot(XS:S,YS:S))
 zWquot(nil,XS:S) -> nil
 zWquot(XS:S,nil) -> nil
-> Pairs in cycle:
 FROM(X:S) -> FROM(s(X:S))

The problem is infinite.