# Every Boolean law has a dual?

1. Apr 6, 2007

### dfx

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

None

3. 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?

2. Apr 13, 2007

### e(ho0n3

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

3. Apr 13, 2007

### Dick

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: Apr 13, 2007
4. Apr 14, 2007

### e(ho0n3

5. Apr 14, 2007

### 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.