Basic Ideas and Concepts

What is a truth table and what can it do
Mapping a function on a truth table and determining all possible outputs
Applying principles of Boolean Algebra to minimize given function
Learn how to apply minterms and maxterms expansion to a truth table

Truth Tables

A truth table is basically a representation of all the possible input combinations and the functional output of each of those combinations. It will tell you on a case-by-case basis, what will the functional output be in every input instance.

We need a way to represent the 3 basic logic operations in algebraic formulas. (AND, OR, NOT)
Therefore we adopt the following standards for those representations:
A AND B = $A\cdot B=AB\!$
A OR B = $A+B\!$
NOT A (Inverted A) = ${\overline {A}}\!$

For a simple example, a truth table of the AND gate is given below.

**a and b**
a	b	f = a b
0	0	0
0	1	0
1	0	0
1	1	1

In general the number of rows in a truth table will be $2^{n},\!$ , where n is the number of variables. So in a 3 variable function where all the inputs are products, the following would be the truth table.

**a and b and c**
a	b	c	f = a b c
0	0	0	0
0	0	1	0
0	1	0	0
0	1	1	0
1	0	0	0
1	0	1	0
1	1	0	0
1	1	1	1

Boolean Algebra

We introduce rules of Boolean Algebra to help us deal with simplifying more complex functions. The rules that should be learned are as follows.

**Boolean Algebra Theorems and Axioms**
Single Variable	Multi-Variable
$0\cdot 0=0\!$	$x\cdot y=y\cdot x\!$
$1+1=1\!$	$x+y=y+x\!$
$1\cdot 1=1\!$	$x\cdot (y\cdot z)=(x\cdot y)\cdot z\!$
$0+0=0\!$	$x+(y+z)=(x+y)+z\!$
$0\cdot 1=1\cdot 0=0\!$	$x\cdot (y+z)=x\cdot y+x\cdot z\!$
$1+0=0+1=1\!$	$x+(y\cdot z)=(x+y)\cdot (x+z)\!$
If $x=0\!$ then ${\overline {x}}=1\!$	$x+x\cdot y=x\!$
If $x=1\!$ then ${\overline {x}}=0\!$	$x\cdot (x+y)=x\!$
$x\cdot 0=0\!$	$x\cdot y+x\cdot {\overline {y}}=x\!$
$x+1=1\!$	$(x+y)\cdot (x+{\overline {y}})=x\!$
$x\cdot 1=x\!$	${\overline {x\cdot y}}={\overline {x}}+{\overline {y}}\!$
$x+0=x\!$	${\overline {x+y}}={\overline {x}}\cdot {\overline {y}}\!$
$x\cdot x=x\!$	$x+{\overline {x}}\cdot y=x+y\!$
$x+x=x\!$	$x\cdot ({\overline {x}}+y)=x\cdot y\!$
$x\cdot {\overline {x}}=0\!$	$x\cdot y+y\cdot z+{\overline {x}}\cdot z=x\cdot y+{\overline {x}}\cdot z\!$
$x+{\overline {x}}=1\!$	$(x+y)\cdot (y+z)\cdot ({\overline {x}}+z)=(x+y)\cdot ({\overline {x}}+z)\!$
${\overline {\overline {x}}}=x\!$