# Brackets in distributive law?

The (second) distributive law in boolean algebra is defined as

$A + (B C) = (A + B) (A + C)$

But wouldn't it be correct to define it that way:

$(A + (B C)\, ) = (A + B) (A + C)$

Because if you transform

$(\sim D + \sim C) (\sim D + B + A) (C + \sim B + \sim A)$

to

$\sim D + (\sim C (B + A) \, ) (C + \sim B + \sim A)$

it would be incorrect since * has higher precedence than +, so

$(\sim D + (\sim C (B + A)\, )\, ) (C + \sim B + \sim A)$

would be correct, or am I totally mistaken?

• I don't understand your question. The second expression is identical to the first, except that it has a redundant pair of parentheses. – David Richerby Feb 18 '15 at 0:40
• From my view the first one means not D or the rest so if D is low the output is high. The second one binds the D to the rest. – Benjoyo Feb 18 '15 at 0:47
• I believe he means the very first two laws that you wrote. There is literally no difference because A+(BC)=(A+(BC)) because no matter the expression E=(E). This is a rule that goes far far beyond just boolean algebra and holds in every any algebra and logic I can think of – Jake Feb 18 '15 at 3:41

The real problem is you're making a mistake with the precedence rules being used. The sentence: $$(\lnot D + \lnot C)(\lnot D + B + A)(C + \lnot B + \lnot A)$$ is ambiguous until we add rules of precedence, or explicitly parenthesise it. Because logical $\mathsf{AND}$ is associative (and commutative), we normally ignore the fact that we have to decide, one way or the other, which of the two $\mathsf{AND}$s applies first.
The normal rules we use (when nothing else gets in the way, go left to right) implicitly parenthesises the sentence as: $$[(\lnot D + \lnot C)(\lnot D + B + A)](C + \lnot B + \lnot A)$$
If we parenthesise in this explicit way, and apply the distributive law, then you get what you expect: $$[\lnot D + \lnot C(B + A)](C + \lnot B + \lnot A)$$
If we changed our rules of precedence to right-to-left and thought about it as $$(\lnot D + \lnot C)[(\lnot D + B + A)(C + \lnot B + \lnot A)]$$ then we can't apply the distributive law at all, we have to apply the associative property first, then we again get what you expect.