# Why is A(A+B) = A [Absorption Law]?

Could you proof it to me that A(A+B) = A?

AA + BA [AA = A]

A + AB

Then what?

• Use a truth table. Commented Jan 17, 2017 at 12:00

I presume you are looking for a way to prove the identity using a calculus. So far you have used distributivity an idempotency.

Recall that A = A1 so you get A1+AB and you can use distributivity again, this time in the other direction. Then two obvious steps.

Here is a proof: $$A \stackrel{(1)}= A \cdot 1 \stackrel{(2)}= A \cdot (1+B) \stackrel{(3)}= A \cdot 1 + A \cdot B \stackrel{(4)}= A + A \cdot B \stackrel{(5)}= A \cdot A + A \cdot B \stackrel{(6)}= A \cdot (A+B).$$

Axioms used:

(1),(4) multiplicative identity

(2) absorption

(3),(6) distributivity

(5) idempotence

I always liked the $\min, \max$ definitions of $\cdot$ and $+$, since some courses in boolean algebra just give those laws and ask you to accept them. [Boolean Algebra: Basic Operations]

\begin{align*} x \land y &= x \cdot y = \min(x,y)\\ x \lor y &= x + y = \max(x,y) \end{align*} where $0 \leq x,y \leq 1$.

So $A(A+B)$ becomes: $$\min(A, \max(A,B))$$

• Case 1: If $A \leq B$ then

\begin{align*} \min(A, \color{red}{\max(A,B)}) &= \min(A, \color{red}{B})\\ &= \min(A, B) = A\\ \end{align*}

• Case 2: If $A \geq B$ follows the same way as the first one.
• A(A+B)         as The AND operation distributes
•                      over the OR operation
•                   for all x, y, z ∈ B
•                    x • (y + z) = (x • y) + (x • z) so
• A(A+B) = AA+AB      Idempotent x • x = x
• = A+AB
• = A(1+B)      as 1+B=1 so
• = A(1)           as A.1=A so
• = A
• I don't see any significant difference between this answer and Raiyan's existing answer. Commented Feb 22, 2019 at 8:14
• the question was "Could you proof it to me that A(A+B) = A?" not "Could you proof it to me that A=A(A+B) ?" is it right? Commented Feb 23, 2019 at 5:07