Commutative and associative laws; A+A'B = A+B

Commutative laws

The order of variables in an OR or AND operation does not change the result.

• $A + B = B + A$
• $A \cdot B = B \cdot A$

Example: $\bar A + C = C + \bar A$; with A = 1, B = 0: $1 \cdot 0 = 0 \cdot 1 = 0$. In hardware, swapping the inputs of an AND or OR gate does not change its output.

Associative laws

The grouping of variables in an OR or AND operation does not change the result.

• $A + (B + C) = (A + B) + C$
• $A \cdot (B \cdot C) = (A \cdot B) \cdot C$

Example: with A = 1, B = 0, C = 1: $1 \cdot (0 \cdot 1) = 0$ and $(1 \cdot 0) \cdot 1 = 0$. So a 3-input AND can be built from two 2-input AND gates in any order.

Proof of $A + \bar A B = A + B$

Using the distributive law $X + YZ = (X + Y)(X + Z)$:

$$\begin{aligned} A + \bar A B &= (A + \bar A)(A + B) \\ &= 1 \cdot (A + B) \\ &= A + B \end{aligned}$$

Truth-table check: