DHARM
N-COM\APPENDIX.PM5BOOLEAN ALGEBRA 357
xyzF 1 F 2 F 3 F 4
0001000
0010100
0100010
0110001
1000011
1010101
1100111
1110101
Fig. A.5. Truth table for F 1 , F 2 , F 3 , and F 4.
We can represent any Boolean function by using truth table. An n variable Boolean
function must contain 2n rows in the truth table starting from 0 to 2n – 1. It is also possible that
two Boolean expressions can be instigating from same Boolean function because, by applying
theorems and axioms of Boolean algebra, we can simplify the expression that return to some
common function. Alternatively, two Boolean functions are said to be similar to each other if
the truth values of both functions are same for all possible interpretations.
While, evaluating the Boolean function the precedence of the parenthesis is the highest,
next precedence is of the complement, then the operator AND and finally the operator OR.
Alternatively, the expression written inside the parenthesis will be evaluated first, follows the
complement, then follows the ‘. ’ and finally follows the ‘+’ operations.
A.5 Simplification of Boolean Functions..............................................................................
The objective of the simplification of a Boolean function is to minimize the number of literals.
Since, each term of Boolean function is implemented with a logic gate, and then each literal in
the function designates an input to a gate. To, minimize the number of literals and the number
of terms certainly trim down the circuit and equipments. Although it is not always possible to
minimize both factors simultaneously.
We can simplify the Boolean function by using most common postulates and basic theo-
rems of Boolean algebra through trail & chance. Of course, there is no specific procedure of
deduction that yields the minimized expression. In the following example we can see that, by
way of trail & chance we can simplify the Boolean function.
Example A.1 Simplify the following Boolean functions to a minimum number of literals.
- f = x y +x y′
= x (y + y′) P4
= x. 1 P2
= xP 1 - f = y (w z′ + w z) + x y
= y. w (z′ + z) + x y P2
= y. w. 1 + x y P1
= y w + x y P1
= y w + y x P3
= y (w + x) P4