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.
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)
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+")");
}
}