The maths book give the example question:
Simplify $(A \cap B \cap C) \cup (A \cap B \cap C') \cup (A \cap B\,' \cap C)$.
($C'$ is the complement of $C$.)
The first few steps they give I understand, they simplify it to:
$$(A \cap B \cap C) \cup (A \cap C \cap B\,') \cup (A \cap B \cap C')$$
using commutative laws.
However the next step I don't understand (I know it uses distributive laws, but that's it). It goes from the above to:
$$((A \cap C) \cap (B \cup B\,')) \cup (A \cap B \cap C')$$
I understand they only touched the first 2 brackets, but I can't figure out what they've done to get there :/
The 2 Distributive laws they've given me are:
$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
I know I'm missing something really simple, but I just can't figure it out:/ any help will be much appreciated.
Best Answer
We can apply commutativity again to get:
$A \cap B \cap C = A \cap C \cap B$ (technically we're applying associativity and commutativity, but I hope this "short-cut" is clear).
Let's replace $A \cap C$ with the letter $X$. Then your "first two brackets" (as you call them) are:
$(X \cap B) \cup (X \cap B')$, so using the second of your distributive laws, we have:
$(X \cap B) \cup (X \cap B') = X\cap (B \cup B')$
Replacing $X$ with $A \cap C$, this becomes:
$(A \cap C) \cap (B \cup B')$, which is the part you say you didn't understand.