Understanding Boolean Duality: The Power of Bar and Breaking the Bar

  • Thread starter Thread starter dfx
  • Start date Start date
  • Tags Tags
    Dual Law
AI Thread Summary
Boolean duality states that every valid Boolean expression has a dual, achieved by swapping operators and constants: "+" with ".", "." with "+", and "0" with "1". The discussion clarifies that while one might think these replacements can be applied universally, they require careful consideration of negation. For example, the dual of the expression "a + b = 1" is correctly represented as "not a . not b = 0" rather than simply switching operators. The method of "breaking the bar" helps in simplifying Boolean equations by applying De Morgan's laws. Understanding these principles is essential for tackling more complex Boolean expressions, such as those analyzed using Karnaugh maps.
dfx
Messages
59
Reaction score
1

Homework Statement



An extract from my notes reads that "every boolean law has a dual: any valid statement is also valid with:"

. replaced with +
+ replaced with .
0 replaced with 1 and vice versa.

Homework Equations



None

The Attempt at a Solution



I have no clue what this means. Surely it doesn't mean that you can switch around the + and . in any boolean statement and it will still hold true? Can anyone explain, and what are the implications of it?
 
Physics news on Phys.org
Sure it does. Suppose a + b = 1. The dual statement is a . b = 0 which is true. (Can you tell why?).
 
e(ho0n3 said:
Sure it does. Suppose a + b = 1. The dual statement is a . b = 0 which is true. (Can you tell why?).

It's not true at all. The dual statement to a+b=1 is (not a).(not b)=0. Replacing 0 by 1and vice versa implies you should replace a by (not a).
 
Last edited:
Right. Sorry about that.
 
Easiest way to analyze this is like this.

a or b = 1

not (a or b) =0

now you break the bar(not) by changing the 'or' to an 'and' and you are left with:

not a 'and' not b = 0

This strategy of bar and breaking the bar is key in simplifying boolean equations. For much more complex eqns there is always the k-map.
 

Similar threads

Basic Boolean Algebra Proof: Validity of the Cancellation Law?
Replies
3
Views
6K
Use source transformation, find Vo
Replies
19
Views
8K
Heating Effect of Electric Current-Transmit at High Voltage
Replies
5
Views
2K
Why Does a Metal Bar Move in a Magnetic Field?
Replies
8
Views
2K
Engineering RL Circuit Power Computation: What Am I Missing in My Solution?
Replies
1
Views
2K
Why Does My C Program Segmentation Fault Without a Filename?
Replies
5
Views
3K
I Understanding Physical Processes: EM, Gravitational, Mechanical & Entropic
Replies
1
Views
1K
B Showing that formulae are invariant for all values in the domain
Replies
2
Views
2K
Using Ampere's Law for these two different integration paths
Replies
1
Views
1K
Power consumption costs of a fridge with bad lights
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top