0️⃣Boolean Algebra

Just like normal algebra, Boolean algebra uses letters to represent values and uses them as expressions. Boolean algebra uses the values TRUE and FALSE (i.e. 1 and 0).

Notation

Expression	Meaning
A, B, C, etc.	Used to represent an unknown value, just like x or y in normal algebra.
$\overline A$	NOT A. An overline represents the NOT operation being applied to what is below the line.
$A \cdot B$	A AND B. The dot represents AND operation.
$A + B$	A OR B. The plus sign represents OR operation.

Order of Precedence

The following list shows the order of precedence (Highest to Lowest):

• Brackets

• NOT

• AND

• OR

Boolean Identities

There are numerous Boolean identities that can help simplify Boolean expressions:

$$A \cdot 0 = 0$$

$$B \cdot 1 = B$$

$$C \cdot C = C$$

$$D + 0 = D$$

$$E + 1 = 1$$

$$F + F = F$$

$$\overline{\overline{C}} = C$$

De Morgan's Law

De Morgan's law can be very useful when simplifying logical expressions. It can be remembered by recalling the phrase:

Break the bar and change the sign

The "bar" refers to the overline representing the NOT sign and the sign refers to changing the dot or plus sign.

Example of using De Morgan's law on $\overline{A + B}$:

• Step 1: Break the bar $\overline{A} + \overline{B}$

• Step 2: Change the sign: $\overline{A} \cdot \overline{B}$

De Morgan's law can also be used in reverse, by changing the sign and building the bar. Following is an example of changing $\overline{C} + \overline{D}$:

• Step 1: Change the sign $\overline{C} \cdot \overline{D}$

• Step 2: Build the bar $\overline{C \cdot D}$

Distributive Rules

Just like mathematics, you can expand and factorise Boolean expressions.

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

Example Questions

Simplify the Boolean expression $A + \overline{B\cdot A}$

1. Use De Morgan's Law to break bar $A + \overline{B} + \overline{A}$

2. Use $A + \overline{A} = 1$ -> $\overline{B} + 1$

3. = 1

Simplify the Boolean expression $C \cdot B + \overline{C} \cdot B$

1. Factorise: $B \cdot (C + \overline{C})$

2. $= B \cdot (1) = B$