[kmill]

Rather than deducing it, you can also look at the truth table, see it is 0 exactly where B is not true and A is true, and get not(A and not B). Then, we can conclude there is a proof such as yours from the completeness theorem, I believe.

[superobserver]

Yes, but it's more fun and edifying to prove it. :) It's so rare I get an opportunity to whip out propositional logic and formal proof methods that I hardly balk at the opportunity. (hears groans from non-nerds)

[Breakthrough]

If you use DeMorgan's law on your result, you can simplify it further to ~A | B