# How does this Boolean simplification work?

I've just started learning about Boolean identities and simplification, and I'm already puzzled by one rule. I'm following this tutorial, which presents the following rule:

A + AB = A


The proof is as follows:

A + AB
↓         Factoring A out of both terms
A(1 + B)
↓         Applying identity A + 1 = 1
A(1)
↓         Applying identity 1A = A
A


I understand the second and third steps. B + 1 must be equivalent to 1 because if one input to an OR gate is 1, the output will always be 1. And the third step makes sense too, because whether 1 x A evaluates to 1 or 0 depends on the value of A.

What I don't understand is the first step. It's been many years since I've studied algebra at school, and the intuitive leap that "factoring A out of both terms" leads from A + AB to A(1 + B) is beyond my rusty skills.

I would greatly appreciate an explanation to help my understanding of how you "factor out" a Boolean variable for an equation, and how I would know to do this for any given equation.

• Can you see it's true for integer (not boolean) arithmetic? That might help seeing it's true for boolean arithmetic too.
– user16324
Commented Dec 22, 2020 at 19:09
• I guess so. 2 + (2 * 4) = 2 + 8 = 10, and 2 * (1 + 4) = 2 * 5 = 10. So it makes sense when you check it, I'm just not sure how I would determine that I need to change A + AB to A(1 + B) in the first place.
– Lou
Commented Dec 23, 2020 at 9:06

It's just an identity. It's true. You can do that straight away. A + AB = A(1 + B)

More generally: AC + AB = A(C+B)

In this case, the boolean algebra identity is similar to what you usually do in normal algebra.

I don't think there is any proof for this except you can verify this with a truth table.

And I think there is a simpler and more intuitive way to perform this simplification in one step instead of three. If you consider AB, it can only be true if A is true. If A is false, then AB is also false. So if A is true, then A + AB is obviously true. And if A is false, then A + AB is also false.

So with this understanding, we can directly say A + AB = A

• Well, what I don't get is why removing the second "A" from "AB" leads to there being a 1, and not a 0, or A, or anything else? I'm sure this is involving basic algebra, but I haven't done any such in coming up 10 years now, and I'm pretty rusty.
– Lou
Commented Dec 29, 2020 at 16:46
• @Lou A·1 = A, right? A·0 = 0, so it's not the same. Commented Dec 29, 2020 at 18:18
• @Hearth I'm probably just understanding this in a janky way (and/or tired at the end of a work day), but none of the answers have really helped me to understand the algebra going on here. I'm still confused by what exactly is happening when you change A + AB into A(1 + B). Are you ... dividing by A or something? And getting 1? I would be grateful for an ELI5 here.
– Lou
Commented Dec 29, 2020 at 18:27
• @Lou Yes, dividing by A is a good analogy. Commented Dec 29, 2020 at 18:31
• So, correct me if I'm misunderstanding ... it seems like if you "divide" A + AB by A, then you take the first A away and leave it with nothing. But you can't have nothing in Boolean terms, so you're left with a 1. You also take the second away from the "A and B" clause, but because there's a B left, you're left with B. So if you factor out A, that's why you get 1 and B inside the brackets. Am I anywhere near?
– Lou
Commented Dec 30, 2020 at 17:03

$$\xy+xz = x(y+z) \$$ is known as the distributive posulate (OR-version).

In your case: $$\A \cdot1+AB=A(1+B)=A \$$

So this is what I didn't get - why do you make A into A . 1? How do you "know" that there's a hidden "and 1" in that expression

This is known as posulate 2 of Boolean algebra: $$\x \cdot1=x \$$

Testing this postulate by plugging in $$\x=0 \$$ and $$\x=1\$$ proves its validity. $$0 \cdot1=0 \: \: \: \text{for} \: \:x=0$$ $$1 \cdot1=1 \: \: \: \text{for} \: \:x=1$$

And moreover, why is it 1? Not 0, or A?

If it was 0 then the Boolean equation would be false. $$A \cdot1 \neq A \cdot0$$ Actually, the expression $$\x \cdot0=0 \$$ is known as theorem 2 of Boolean Algebra.

• So this is what I didn't get - why do you make A into A . 1? How do you "know" that there's a hidden "and 1" in that expression, that only reveals itself when you factor out A? And moreover, why is it 1? Not 0, or A?
– Lou
Commented Dec 29, 2020 at 16:45