Hello everyone hope you're all having a good day,

Here's is the statement I am trying to prove:

Prove that P(Ω \ A) = 1 − P(A), for any Ω, P and A, indicating where Kolmogorov’s axioms are needed.

From what I have thought of so far, P(Ω \ A) = P(Ω) - P(A), by using Axiom II which says P(Ω) = 1, we can deduce that

P(Ω \ A) = P(Ω) - P(A) = 1 - P(A)

My issue is that, I am not sure if the step " P(Ω \ A) = P(Ω) - P(A)" is a correct approach to the question, since it is assuming that it is true without proving it, and no other theorem explicitly states it.

Any help would be appreciated.

Thank you