simplification in boolean algebra

Question

Please could someone help me understand how; A + ~AB = A+B Actually, I don't understand how from A + ~AB we arrive at A+B Anyways thanks for helping me in advance!

Answer 1

I will use $\bar A$ for ~A.

$A + \bar A B = \overline{\overline{A + \bar{A} B}} = \overline{\bar A \cdot \overline{\bar A B}} =\overline{\bar A \cdot(\bar {\bar A} + \bar B)} = \overline{\bar A \cdot (A + \bar B)} = \overline{\bar A \cdot A + \bar A \cdot \bar B} = \overline{0+\bar A \cdot \bar B} = \overline{\bar A \cdot \bar B} = \overline{\overline{A+B}} = A+B$.

We have used De-Morgan's Laws: $\overline{X + Y} = \bar X \cdot \bar Y$ and $\overline{XY} = \bar X + \bar Y$. Also, $0 + X = X$ for any boolean variable $X$, and $\overline X \cdot X = 0$.

Answer 2

From

$A \lor \bar{A} = T$

$T \lor B = T$

$B = B \land T = B \land (A \lor \bar{A}) = BA \lor B\bar{A}$

we can rewrite

$A \lor B = A \lor BA \lor B\bar{A} = A(T \lor B) \lor B\bar{A} = A \lor B\bar{A}$

In plain English:
Regardless of $B$, $A \lor \bar{A}B$ is true, if $A$ is true. If $A$ is false, the expression is equal to $B$.

Answer 3

$$A+A'B=(A+A')(A+B)=1(A+B)=A+B$$

The first step is Distribution: $A+BC=(A+B)(A+C)$