Difference between revisions of "Boolean Algebra"

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(Example 14)
(Solving Boolean Equations)
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===Example 12===
 
===Example 12===
  
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===Example 13===
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===Example 14===
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===Example 15===
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===Example 16===
  
===Example 13===
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===Example 17===
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===Example 18===
 
'''Simplify:'''
 
'''Simplify:'''
 
<math> X.(\overline{X}+Y) </math>
 
<math> X.(\overline{X}+Y) </math>
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Fully simplified the equation
 
Fully simplified the equation
  
===Example 14===
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===Example 19===
 
<math> (\overline{A} + A) . (\overline{A} + C) </math>
 
<math> (\overline{A} + A) . (\overline{A} + C) </math>
  
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<math>= (\overline{A} + C)</math>
 
<math>= (\overline{A} + C)</math>
  
===Example 15===
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===Example 20===
 
<math>(X + Y) . (X + \overline{Y})</math>
 
<math>(X + Y) . (X + \overline{Y})</math>
 
<br>Distributive:  
 
<br>Distributive:  
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<math> X </math>
 
<math> X </math>
  
==Example 16==
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===Example 21===

Revision as of 10:21, 11 February 2019

Boolean Algebra Precedence

the order of precedence for boolean algebra is:

  1. Brackets
  2. Not
  3. And
  4. Or

Boolean Identities

Using AND

[math] A.1 = A [/math]

This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.

[math] 0.A = 0 [/math]

Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.

[math] A.A = A[/math]

The output is determined by A alone in this equation. This can be simplified to just "A".

[math] A.\overline{A}=0 [/math]

Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.

Using OR

[math] 0+A = A [/math]

0 or A can be simplified as just A.

[math] 1+A = 1 [/math]

1 or A can be simplified as just 1.

[math] A+A=A[/math]

A or A can be simplified as just A.

[math] \overline{A}+A=1[/math]

NOT A or A can be simplified as just 1.

Boolean Laws

Commutative Law

The Commutative Law is where equations are the same no matter what way around the letters are written. For example

[math] A+B = B+A [/math]

or

[math] A.B = B.A [/math]

Associate Law

If all of the symbols are the same it doesn't matter which order the equation is evaluated.

[math] A+(B+C) = B + (A+C) [/math]

[math] A+(B+C) = B + (A+C) [/math]

[math] A+(B+C) = C + (A+B) [/math]

So:

[math] A.(B.C) = B . (A.C) [/math]

[math] A.(B.C) = B . (A.C) [/math]

[math] A.(B.C) = C . (A.B) [/math]

Distributive Law

The distributive law is these two equations.

[math] A.(B+C) = A.B + A.C [/math]

[math] A+(B.C) = (A+B).(A+C) [/math]

This is essentially factorising or expanding the brackets, but you can also:

[math] A.B + A.C = A.(B+C)[/math]

[math] A+B.A+C = A+(B.C) [/math]

Redundancy Law

Law 1 :

[math] A + \overline{A} B = A + B [/math]

Proof :

[math]= A + \overline{A} B \\ = (A + \overline{A})(A + B) \\ = 1 . (A + B) \\ = A + B [/math]

Law 2:

[math] A.(\overline{A} + B) = A.B[/math]

Proof :

[math]= A.(\overline{A} + B) \\ = A.\overline{A} + A.B \\ = 0 + A.B \\ = A.B [/math]

Law 3:

[math] A.(A + B) = A[/math]

Proof :

[math] A.(A + B) = (0+A) . (A + B)\\ [/math]

so:

[math] A + (0 . B) [/math]

So:

[math] A + 0 = A [/math]

Law 4:

[math] A+(A . B) = A[/math]

Proof :

[math] A+(A . B) = (1 . A) + (A . B)\\ [/math]

so:

[math] A . (1 + B) [/math]

So:


[math] A . 1 = A [/math]

Identity Law

This is also in the identities section:

[math] A.A = A [/math]

[math] A+A = A [/math]

Negation Law

Just like in any other logic negating a negative is a positive so:

[math] \overline{ \overline{A} } = A [/math]

Solving Boolean Equations

Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Simplify: [math] X.(\overline{X}+Y) [/math]

[math] (X.\overline{X})+(X.Y) [/math] using the Distributive Law

[math] 0 + (X.Y) [/math] using the identity [math]A.\overline{A}=0[/math]

[math] X.Y [/math] using the identity [math]A+0=A[/math]

Fully simplified the equation

Example 19

[math] (\overline{A} + A) . (\overline{A} + C) [/math]

OR Identity

[math](\overline{A} + A) = 1[/math]

AND Identity

[math]= 1.(\overline{A} + C)[/math]

Simplify

[math]= (\overline{A} + C)[/math]

Example 20

[math](X + Y) . (X + \overline{Y})[/math]
Distributive:
[math]X . (Y + \overline{Y})[/math]
Identity laws:
[math]Y + \overline{Y} = 1[/math]

[math] X.1 = X[/math]

Alternative

[math] X.X + X.\overline{Y} + Y.X + Y.\overline{Y} [/math] Expanding the brackets

[math] X + X.\overline{Y} + Y.X + 0 [/math] Use of [math] X.X = X [/math] and [math] Y.\overline{Y} = 0 [/math]

[math] X + X(\overline{Y}+Y) [/math] Taking X out of the brackets

[math] X + X(1) [/math] Use of [math] Y + \overline{Y} = 1 [/math]

[math] X(1) [/math]

[math] X [/math]

Example 21