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

[dfx]

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?

 

[e(ho0n3]

Sure it does. Suppose a + b = 1. The dual statement is a . b = 0 which is true. (Can you tell why?).

 

[Dick]

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).

 

[e(ho0n3]

Right. Sorry about that.

 

[pooface]

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.