NO

Problem 1: 

(VAR v_NonEmpty:S X:S Y:S Z:S)
(RULES
first(0,Z:S) -> nil
first(s(X:S),cons(Y:S,Z:S)) -> cons(Y:S,first(X:S,Z:S))
from(X:S) -> cons(X:S,from(s(X:S)))
sel(0,cons(X:S,Z:S)) -> X:S
sel(s(X:S),cons(Y:S,Z:S)) -> sel(X:S,Z:S)
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 FIRST(s(X:S),cons(Y:S,Z:S)) -> FIRST(X:S,Z:S)
 FROM(X:S) -> FROM(s(X:S))
 SEL(s(X:S),cons(Y:S,Z:S)) -> SEL(X:S,Z:S)
-> Rules:
 first(0,Z:S) -> nil
 first(s(X:S),cons(Y:S,Z:S)) -> cons(Y:S,first(X:S,Z:S))
 from(X:S) -> cons(X:S,from(s(X:S)))
 sel(0,cons(X:S,Z:S)) -> X:S
 sel(s(X:S),cons(Y:S,Z:S)) -> sel(X:S,Z:S)

Problem 1: 

Infinite Processor:
-> Pairs:
 FIRST(s(X:S),cons(Y:S,Z:S)) -> FIRST(X:S,Z:S)
 FROM(X:S) -> FROM(s(X:S))
 SEL(s(X:S),cons(Y:S,Z:S)) -> SEL(X:S,Z:S)
-> Rules:
 first(0,Z:S) -> nil
 first(s(X:S),cons(Y:S,Z:S)) -> cons(Y:S,first(X:S,Z:S))
 from(X:S) -> cons(X:S,from(s(X:S)))
 sel(0,cons(X:S,Z:S)) -> X:S
 sel(s(X:S),cons(Y:S,Z:S)) -> sel(X:S,Z:S)
-> Pairs in cycle:
 FROM(X:S) -> FROM(s(X:S))

The problem is infinite.