How does this Boolean simplification work?

Question

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.

Answer 1

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

Answer 2

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

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

From the comments

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.

Answer 3

I don't think it needs to go so far.

The first equation A+AB is true, if either A is true, or, A and B are both true.

So if B is true, A needs to be true too for the equation to be true. But the equation is true also when only A is true regardless of if B is true or not.