Boolean Algebra Simplification

[conorordan]

Homework Statement

$R=(A \cdot \overline{B})+(\overline{A} \cdot B)$​

Using Boolean algebra, including the complement relation $D\cdot \overline{D}=0$, convert R to a form that uses one NAND, one AND and one OR gate (a total of three gates).

Homework Equations

de Morgan's theorem...

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

The Attempt at a Solution

By some trial and error I arrived at this;

$(\overline{A}+\overline{B})\cdot(A+B)$

And then using the second of de Morgan's equations on the first OR I got to this;

$\overline{A\cdot B}\cdot(A+B)$

Which works as a solution, it has one NAND, one AND and one OR, however, I can't just skip to that bit, it's the working out in between that I'm missing, I'm not sure how to fill in the gaps.

Thanks for any help you can offer!

 

[mfb]

You can show that the last statement is equivalent to the original one if you work backwards - it is easier in that direction, but the other direction works as well.