NO

Problem 1: 

(VAR v_NonEmpty:S L:S X:S Y:S)
(RULES
eq(0,0) -> ttrue
eq(s(X:S),s(Y:S)) -> eq(X:S,Y:S)
eq(X:S,Y:S) -> ffalse
inf(X:S) -> cons(X:S,inf(s(X:S)))
length(cons(X:S,L:S)) -> s(length(L:S))
length(nil) -> 0
take(0,X:S) -> nil
take(s(X:S),cons(Y:S,L:S)) -> cons(Y:S,take(X:S,L:S))
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 EQ(s(X:S),s(Y:S)) -> EQ(X:S,Y:S)
 INF(X:S) -> INF(s(X:S))
 LENGTH(cons(X:S,L:S)) -> LENGTH(L:S)
 TAKE(s(X:S),cons(Y:S,L:S)) -> TAKE(X:S,L:S)
-> Rules:
 eq(0,0) -> ttrue
 eq(s(X:S),s(Y:S)) -> eq(X:S,Y:S)
 eq(X:S,Y:S) -> ffalse
 inf(X:S) -> cons(X:S,inf(s(X:S)))
 length(cons(X:S,L:S)) -> s(length(L:S))
 length(nil) -> 0
 take(0,X:S) -> nil
 take(s(X:S),cons(Y:S,L:S)) -> cons(Y:S,take(X:S,L:S))

Problem 1: 

Infinite Processor:
-> Pairs:
 EQ(s(X:S),s(Y:S)) -> EQ(X:S,Y:S)
 INF(X:S) -> INF(s(X:S))
 LENGTH(cons(X:S,L:S)) -> LENGTH(L:S)
 TAKE(s(X:S),cons(Y:S,L:S)) -> TAKE(X:S,L:S)
-> Rules:
 eq(0,0) -> ttrue
 eq(s(X:S),s(Y:S)) -> eq(X:S,Y:S)
 eq(X:S,Y:S) -> ffalse
 inf(X:S) -> cons(X:S,inf(s(X:S)))
 length(cons(X:S,L:S)) -> s(length(L:S))
 length(nil) -> 0
 take(0,X:S) -> nil
 take(s(X:S),cons(Y:S,L:S)) -> cons(Y:S,take(X:S,L:S))
-> Pairs in cycle:
 INF(X:S) -> INF(s(X:S))

The problem is infinite.