De Morgan's Law

boolean-logic, demorgans-law, java

Solution

Does x!=0 simplify to x>0?

No that's not true. Because integers are signed.

How to simplify : `!(x!=0 || y !=0)` ?

Consider this rules :

(second De Morgan's laws )

By 1., it implies

`!(x!=0 || y !=0) <=> (!(x!=0)) && (!(y != 0))`

By 2., it implies

`(!(x!=0)) && (!(y != 0)) <=> (x == 0) && (y == 0)`

To test you can write the following loop :

for(int x = -5; x < 5; x++){
     for(int y = -5; y < 5; y++){
         if(!(x!=0 || y !=0))
            System.out.println("True : ("+x+","+y+")");
    }
}

Problem

I am trying to simplify the following using DeMorgan's Law: ! (x!=0 || y !=0) Does x!=0 simplify to x>0? Or am I wrong in the following: ``` !(x>0 || y>0) !(x>0) && !(y>0) ((x<=0) && (y<=0)) ``` Thanks.

Original source