De Morgan's Law

Does x!=0 simplify to x>0?

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

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

Consider this rules :

1. (second De Morgan's laws )

2. 

By 1., it implies

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

By 2., it implies

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

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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+")");
    }
}

DeMorgans Law states the following:

!(A & B) = !A | !B    (I)
!(A | B) = !A & !B    (II)

In your case (II) applies: !(x!=0 || y!=0) => !(x!=0) && !(y!=0) => (x==0) && (y==0)

PS: Your question: "Does x!=0 simplify to x>0?" can be answered with "no" unless x can not take negative values (for example if the type of x is unsigned).