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

AA + BA [AA = A]

A + AB

Then what?

  • 4
    $\begingroup$ Use a truth table. $\endgroup$ – Yuval Filmus Jan 17 '17 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
  • 1
    $\begingroup$ I don't see any significant difference between this answer and Raiyan's existing answer. $\endgroup$ – Apass.Jack Feb 22 at 8:14
  • $\begingroup$ 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? $\endgroup$ – Mohamed Fathy Alhedhed Feb 23 at 5:07

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