with only two variables it's easy enough to just calculate all the permutations

(numbers at top of columns denote order, so column 4 is result of first, 2 result of 2nd)

     1  4  3 1  2  1
A B  !A || !(!A && B)
0 0  1  1  1 1  0  0
0 1  1  1  0 1  1  1
1 0  0  1  1 0  0  0
1 1  0  1  1 0  0  1

      1 2  1
A B  !A || !B
0 0  1  1  1
0 1  1  1  0
1 0  0  1  1
1 1  0  0  0

actually, if you know these things, that last is, using de morgan's law, equiv to !(a && b)

They're almost equivalent, but fails when both A and B = true.

Hello!

If you got crazy stuff going on, it might help you to use a flag to
monitor the checks you are doing.
So you can do it one step at the time...

1. boolean wowIReallyShouldDoThisNow = false;
2. 
   
3. // do the checks
4. if (!A) {
5.   wowIReallyShouldDoThisNow = true;
6.   println ("ok, it's !A now.");
   
7. }
8. if ( !(!A && B)  ) {
9.   wowIReallyShouldDoThisNow = true;
10.   println ("ok, this is !(!A && B) here...   A is " + A + " and B is " + B + ".");
11. }
12. // checks are all done
13. 
    
14. // finally evaluate result
    
15. if (wowIReallyShouldDoThisNow) {
16.     // finally do it
17. }

Please note that in this way wowIReallyShouldDoThisNow is false at startup and can only become
true - not false again.

Since you do it step by step, the steps are connected by a logical OR.

So in hard parts this might be easier to program and to read, especially if you have to do 3 or 4 tests...

Greetings, Chrisir    



P.S.
Also, when you need e.g. a for-loop instead of (!A) you can use a function that returns boolean (when you e.g. need a for-loop to check each member of an array and return true if all members have a certain value):

1. If (allEnemyCowsAreFinallyRed()) {
2.   wowIReallyShouldDoThisNow = true;
3.   println ("All enemy cows are red.");
4. }

With a function allEnemyCowsAreFinallyRed() containing a for-loop over the array of cows.

They are not equivalent expressions, as GoToLoop pointed out, they differ when A and B are both true.

Using De Morgan's laws (and maybe a few other laws of boolean algebra), the first expression can be simplified as follows: