Boolean algebra, unsure if I am doing it correct.

Question

So I am in my second year of university, and am being shown examples of logic gates. I know that I have the output correct, but am trying to get it simplified to decrease the amount of gates needed.

Here is the expression

$$\overline{\overline{A.\overline{AB}}.\overline{B.\overline{AB}}}$$ $$\big((A\land (A\land B)^\prime)^\prime \land (B\land (A\land B)^\prime)^\prime\big)^\prime$$

I have a solution for it but am sure it's not correct; it's a lot of flipping on the operators. If anyone can post a solution with workings that would be great.

(probably easier if you do it on pen and paper then post a picture)

Answer 1

I'll use $\land$ to denote conjunction (sometimes you'll see it expressed as $AB$ or $A\cdot B$).

I use $\lor$ to mean disjunction, (sometimes represented where $A\lor B$ means the same thing as $A+B$).

For negation, for example $\overline{A}$ = "not A", I'll apply $^\prime : A^\prime$

You'll need to make use of DeMorgan's Laws,, and lot's of it. $$(A\land B)^\prime = A^\prime \lor B^\prime$$ $$(A\lor B)^\prime = A^\prime \land B^\prime$$

For example I'll use it in my first step:

$$\big(((A\land (A\land B)^\prime)^\prime \land (B\land (A\land B)^\prime)^\prime\big)^\prime\equiv (A \land (A\land B)^\prime)\lor (B \land (A\land B)^\prime)\tag 1$$

Next, you can use DeMorgans to show that $(A\land B)^\prime = A^\prime \lor B^\prime$

Knowing this, we have $$(A\land (A^\prime \lor B^\prime))\lor (B \land (A^\prime \lor B^\prime))\tag 2$$

In the end, you will also need to know how distribution works: $$p \lor (q \land r) = (p\lor q) \land (p\lor r)$$ $$p \land (q\lor r) = (p\land q)\lor (p\land r)$$

If you apply all the above correctly, you should arrive at $$(A\land B^\prime) \lor (B \land A^\prime) = A\,.\,\overline{B}\, + \,\overline{A}\,. B$$

Answer 2

Use De-Morgan's laws:

$\overline {A\cdot B}=\overline B+\overline A $

$\overline {A+B}=\overline B\cdot \overline A $

Answer 3

$X=\lnot(\lnot(A\land\lnot(A\land B))\land\lnot(B\land\lnot(A\land B)))$
$C=A\land B$
$X=\lnot(\lnot(A\land\lnot C)\land\lnot(B\land\lnot C))$
$D=\lnot C$
$X=\lnot(\lnot(A\land D)\land\lnot(B\land D))$
$E=A\land D$
$X=\lnot(\lnot E\land\lnot(B\land D))$
$F=\lnot E$
$X=\lnot(F\land\lnot(B\land D))$
$G=B\land D$
$X=\lnot(F\land\lnot G)$
$H=\lnot G$
$X=\lnot(F\land H)$
$I=F\land H$
$X=\lnot I$

$\mathtt{AB⋅CDEFGHI⋅X}$
$\mathtt{00⋅0101011⋅0}$
$\mathtt{01⋅0101100⋅1}$
$\mathtt{10⋅0110010⋅1}$
$\mathtt{11⋅1001011⋅0}$

$X=((A\land\lnot B)\lor(\lnot A\land B))$
$\text{or }X=\lnot(A\equiv B)$