Is P(A . B') = P(A) . P(B')?

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Is this identity $ P(A. B^c) = P(A).P(B^c)$ correct?

My attempt to prove this is as follows:

$$ RHS = P(A). P(B^c) \\ = P(A) . (1 - P(B)) \\ =P(A) - P(A).P(B) \\ =P(A) - P(A \cap B)\\ =P(A \cap B^c) \\ =LHS $$

Also, can this be extended as such: $ P(A^c \cap B^c) = P(A^c) . P(B^c)$ ?

My attempt to prove:

$$ P((A \cup B)^c) = P(A^c). P(B^c) \\ = P((A^c \cap B) \cup (A^c \cap B^c)) \cup P((B^c \cap A) \cup (B^c \cap A^c)) \\ =P(A^c \cap B^c) \\ $$

Are these identities correct? I'd also appreciate any help on the proofs.

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Only valid for independent events. So itself the identity is wrong without explaination of independency of evenrs.