I have a Boolean algebra with some elements $a,b,c$. I have to show this:
$(a ∧ b) ∨ (a′ ∧ c) ∨ (b ∧ c) = (a ∧ b) ∨ (a′ ∧ c)$.
Now I have done other such proofs before but this one I got lost in. I see Boolean algebras have a cancellation law, so my guess was I had to cancel the terms that are on both sides, to get:
$b ∧ c = O_B.$
Since this ended up looking more like an equation than an identity - I could use some help. I would especially like to know why the cancellation method fails. Thanks!
Here's a hint: In a Boolean algebra, $a \vee a = a \vee a \vee a$ for all elements $a$, but that doesn't mean that $O_B = a$ for all $a$. Thus I think you should look back at what kind of cancellation laws are valid.
http://en.wikipedia.org/wiki/Boolean_algebra