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Boolean Algebra

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Boolean Algebra

Boolean algebra: set of elements, set of operators, and axioms

Axioms:

- Closure
- Associative Law
- Commutative Law
- Identity Element
- Inverse
- Distributive Law

A set B with operators + and •

1)closure + and •

2)identity

x+0 = 0+x = x

x•1 = 1•x = x

3)commutative

x + y = y + x

x•y = y•x

4)distributive

x•(y + z) = (x•y) + (x•z)

x + (y•z) = (x + y)•(x + z)

5)for x Œ B there exist x’ Œ B (complement)

x + x’ = 1 and x•x’ = 0

6) at least two element x,y Œ B such that x ≠ y

- Boolean algebra requires
- elements of the set B
- rules of operation for + and •
- they satisfy the six postulates

- Two-Valued Boolean Algebra
- B = {0,1}
- AND, OR, NOT operations
- check postulates

- Duality
- interchange OR and AND
- interchange 0 and 1
- eg
- x•1 = x
- x + 0 = x

- see table 2-1
- operator precedence
- ()
- NOT
- AND
- OR

- Venn Diagrams

- consider the functions:
F1 = x’yz’

F2 = z + x’y’

F3 = x’yz’ + x’z + xy’z

F4 = x’y + y’z

- show truth table (like table 2-2)
- note: F3 = F4
- obtain F4 by manipulating F3

- literal ==> primed or unprimed variable
- simplify (minimize number of literals)
x’ + xy’

x(x’+y)

xy’z + x’y’z + xz’

xy + x’z + yz

(x + y)(x’ + z)(y + z)

x’ + xy’ = x’1 + xy’

= x’(y + y’) + xy’

= x’y + x’y’ + xy’

= x’y + x’y’ + x’y’ + xy’

= x’(y + y’) + y’(x’ + x)

= x’ + y’

x(x’+y) = xx’ + x y’ = 0 + xy’ = xy’

xy’z + x’y’z + xz’ = y’z(x + x’) + xz’

= y’z + xz’

xy + x’z + yz

= xyz’ + xyz + x’y’z + x’yz + xyz + x’yz

= xyz’ + xyz + x’y’z + x’yz (eliminate duplicates)

= xy(z + z’) + x’z(y + y’)

= xy + x’z

(x + y)(x’ + z)(y + z)

= (x + y)(x’ + z) (dual of previous example)

minterms

- how can we represent a 1 in the truth table?

maxterms

- how can we represent a 0 in the truth table?