NOT ((NOT A AND NOT B) OR (A AND NOT B)) simplification using de morgans law

Question

I had my AS mock exam today, and this question came up. I've checked it on calculators and it says it simplifies to B, which is what I got in the exam, but I'm not entirely sure how I got there.

My understanding is that if you want to change the sign (in this case, the or), then you negate the terms on both sides, and then negate the overbar. You then simplify and solve.

I've sat here trying to solve it for ages, and I keep getting different results. I've written out the truth table and it matches, but I don't see how you simplify it NOT((NOT A AND NOT B) OR (A AND NOT B)) From what I can see, I negate the terms in the two brackets, then negate the not at the beginning, and change the sign in the middle. So what I have is (A AND B) AND (NOT A AND B)

A.NOTA.B.B

A.NOTA=0 right?

then B.B=B so its B(0)=0, which isn't the answer.

I'm not sure how to apply the rule.

Answer 1

Since $\neg(p\lor q)$ becomes $(\neg p)\land(\neg q)$, de Morgan rewrites your truth-function as $(\neg(\neg A\land\neg B))\land\neg(A\land\neg B)$. But de Morgan lets us rewrite your conjuncts, so we get the further simplification $(A\lor B)\land(\neg A\lor B)$, which of course reduces by the law of the excluded middle to $B$. (You may want to try, as an exercise, simplifying your original expression this way first to see what happens).

Answer 2

$$\begin{align} \overline{(\overline{A}\cdot\overline{B})+(A\cdot\overline{B})} &=\overline{\overline{A}\cdot\overline{B}}\cdot\overline{A\cdot\overline{B}}\tag{De Morgan's}\\ &=(A+B)\cdot(\overline{A}+B)\tag{De Morgan's}\\ &=A\cdot\overline{A}+A\cdot B+B\cdot\overline{A}+B\cdot B\tag{Expanding}\\ &=0+A\cdot B+B\cdot\overline{A}+B\\ &=B+A\cdot B+B\cdot\overline{A}\tag{Commutative}\\ &=B\cdot(1+A+\overline{A})\tag{Factorise}\\ &=B\cdot1\\ &=B\\ \end{align}$$