Laws of Boolean Algebra

Boolean Algebra can be thought of as a subset or variant of ordinary Elementary Algebra. The numbers and variables in Boolean Algebra are just 0 and 1. That is, any variable, any expression, can eventually by evaluated to either 0 or 1. Example,

A
AB
A+C
A(D+EF)+BC

Note that since we are dealing with the set {0,1} here, so additions and multiplications are not the same as ordinary algebraic additions and multiplications. The addition sign denotes logical OR, and the multiplication sign denotes logical AND. There is an additional operator - the inverter, which denotes logical complement. The 'bar' over the variable is denoted by the single quotation mark:

not(A) = A'

There are differences between Boolean Algebra and Elementary Algebra in terms of the operations and laws, but a large number of laws are common to both. Example,

Commutative:

AB = BA
A+B = B+A

Distributive:

A(B+C) = AB + AC

Some common "laws" of Boolean Algebra

Most of these are not actually laws, but can be readily derived from the fundamentals. Students often misunderstood/forget how they are derived.

AND

A.0 = 0
A.1 = A
A.A = A
A.A' = 0

OR

A+0 = A
A+1 = 1
A+A = A
A+A' = 1

These are not laws/rules but basics derived from the definitions of AND and OR gate. Students who do not understand these should refer back to fundamental of AND and OR gates, their respective truth tables etc. When one input is connected to signal A, and the other input connected to 0/1/A/A', what is the output?

For example, A.A' = 0 can be seen from the AND gate's truth table:

A	B	Out=A.B
0	0	0
0	1	0
1	0	0
1	1	1

In this case, B = A', and it means we need only look at rows 2 and 3, where the two inputs are complement of each other. In both cases, the output is 0. Hence A.A' = 0.


Absorption

A+(A')B = A + B

This "law" can be proven using the following "law":

A+AB = A(1+B) = A

Hence, we replace A by the complicated version (A+AB) and factorize out B. The term (A + A')=1. Hence we have

A+(A')B = A + AB + (A')B = A + (A + A')B = A + B

Similarly,

A' + AB = A' + B

DeMorgan's Theorem

(A+B)' = A' . B'
(AB)' = A' + B'