# Correct Turing machine representation for Rice Theorem proof

Consider the language L1. From Rice Theorem I know L1 is not decidable (i.e. undecidable).

L1 = { R(M) | R(M) is a TM and 1011 ∈ L(M)}

For example if I want to represent by diagram a TM $$M_1$$ that accepts the string 1011 and a TM $$M_2$$ that doesn't accept the string 1011 (e.g., $$M_2$$ accepts only the empty string), following the Rice Theorem not-trivial property, I need to use a (1) acceptance by final states or (2) acceptance by halting or (3) I can use both because I know (by theorem) they are equivalent?