Electronic – How to write this boolean expression using only NOR gates

There is a trick for converting sum-of-products to nand gates and a trick for converting product-of-sums to nor gates. (In diagram below).

But you want nor gates and you are starting in sum-of-products form. So how do you convert sum-of-products to product-of-sums? (Also this one.)

Well, in your case the product-of-sums form is trivially derived just by factoring out the \$\overline{B}\$ to get \$\overline{B}(\overline{A}+C)\$. More generally though, the plug-n-chug way you convert sum-of-products to product-of-sums is

1. double negate your formula. \$\overline{\overline{\overline{A} \, \overline{B} + \overline{B} \, C}} \$
2. Use Demorgan to push the first not through the sum. \$\overline{(\overline{\overline{A}\,\overline{B}})\,(\overline{\overline{B}\,C})} \$
3. And one more Demorgan to turn the products into sums. \$ \overline{(A+B)(B+\overline{C})} \$
4. Now multiply the product-of-sums to get a negated sum-of-products. \$ \overline{AB+A\overline{C}+B+B\overline{C}} = \overline{A\overline{C} + B}\$
5. Almost there! A negated sum of products is a product of sums by two more Demorgans: \$ (\overline{A\,\overline{C}})(\overline{B}) = (\overline{A}+C)(\overline{B}).\$

Now you've got a product of sums.

Finally, here's how you convert product-of-sums to nor gates. It's sometimes called "pushing bubbles".

^{simulate this circuit – Schematic created using CircuitLab}

So I get what you get (and what @jippie got). (\$\overline{A}\$ NOR C) NOR B. \$\overline{\overline{\overline{A+B}+C}+B}\$ is logicially equivalent, of course, but requires 3 2-input nor gates instead of 2 2-input nor gates and a "1-input nor gate" (not gate).

So Complement Law says \$\overline{\overline{X}} = X\$

We start with AND - OR. $$BD + \overline{A}B\overline{C} + \overline{A}CD$$ Double Complement. $$\overline{\overline{BD + \overline{A}B\overline{C} + \overline{A}CD}}$$ Use DeMorgan's Theorem to remove lower complement. $$\overline{\overline{BD} • \overline{\overline{A}B\overline{C}} • \overline{\overline{A}CD}}$$ AND - OR has become NAND - NAND. Use DeMorgan's on terms. $$\overline{(\overline{B} + \overline{D}) • (A + \overline {B} + C) • (A + \overline{C} + \overline{D})}$$ NAND - NAND has become OR - NAND. Use DeMorgan's Theorem to remove complement. $$\overline{(\overline{B} + \overline{D})} + \overline{(A + \overline {B} + C)} + \overline{(A + \overline{C} + \overline{D})}$$ OR - NAND has become NOR - OR. Double Complement again. $$\overline{\overline{\overline{(\overline{B} + \overline{D})} + \overline{(A + \overline {B} + C)} + \overline{(A + \overline{C} + \overline{D})}}}$$ NOR - OR become NOR - NOR. With an extra NOR connected as a NOT gate.