Difference between revisions of "De Morgan's Law"

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(Example 3)
(Example 7)
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===Example 7===
 
===Example 7===
<math> \overline{(\overline{A}+A.(A+B))} + \overline{(B . C)} </math>
+
<math>  
 +
\overline{(\overline{A}+A.(A+B))} + \overline{(B . C)}
 +
abc
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</math>

Revision as of 13:16, 9 May 2018

DeMorgan's laws are the laws of how a NOT gate affects AND and OR statements. They can be easily remembered by "break the line, change the sign". The following image is how to prove De Morgan's Law...

Capture3.jpg

The Process

Step 1 - Reverse the sign

Step 2 - Negate each term

Step 3 - Negate the whole expression

How to apply

Example 1

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

Example 2

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

Example 3

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

Simplifying by using De Morgan's Law:

1. Swap the sign:

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

2. Negate each expression:

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

3. Negate the whole Expression:

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

Using Redundancy law this expression can be simplified to:

[math] A [/math]

This is because if A is 1, the output will always be 1, regardless of the value of B.

Example 4

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

Example 5

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

Example 6

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

Example 7

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