Verified answer

Just for my benefit, I'm going to change notation into Java syntax:

A and B = A && B

A or B = A || B

not A = !A

1 = TRUE

0 = FALSE

Good, now, I'm going to inject some parentheses into it to clarify the order in which we're going to approach it. That means our expression is now:

((!A && B) || (A && !B)) || (A && !D) || ((A && D) && !(TRUE || B && D))

Notice this is a string of three OR statements. Thus for it to be TRUE, only one of the three needs to be TRUE.

I'm going to start with the last of the three:

(A && D) && !(TRUE || B && D)

And the second half of that:

TRUE || B && D

This is always TRUE, regardless of B&&D, since it's an OR.

But this statement is negated by a NOT. NOT TRUE is FALSE. So now we have:

(A && D) && FALSE

This now always FALSE, regardless of A&&D because just one of the statements is FALSE always.

Now we have:

((!A && B) || (A && !B)) || (A && !D) || FALSE

We can now eliminate that last part since it's always FALSE:

((!A && B) || (A && !B)) || (A && !D)

Now let's look at the first part:

(!A && B) || (A && !B)

This is TRUE as long as A and B are opposite. If they're the same, it's FALSE. Now, if this were coding, I would say:

A != B

But I don't think Boolean algebra tends to make use of that. Otherwise, I think just leave it as-is, but I'm going to do a bit more of the problem using A!=B.

The middle can't be simplified.

(A != B) || (A && !D)

So, this is true when:

A, B, !D

A, !B, D

A, !B, !D

!A, B, D

!A, B, !D

!A, !B, D

And false when:

A, B, D

!A, !B, !D