# Boolean expression to NOR-gates

I'm having some trouble understanding how I can convert a boolean expression to a NOR-gate only expression. What I'm working with looks like this:

$$T = BD + \overline{A}B\overline{C} + \overline{A}CD$$

I know you're supposed to use deMorgan's theorem, but I'm not sure how to use it. Can I just select parts of the expression I want to use the theorem on, or does this change the result of the expression?

It would also be nice to see a step-by-step solution for the expression above.

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.
The most straightforward way would be just to replace each operation with it's implementation with NOR gates: $$NOT(A)=\bar{A} = \overline{(A+A)} = NOR(A,A)$$ $$OR(A,B) = A+B=\overline{\overline{A+B}}=NOT(NOR(A,B))$$ $$AND(A,B) = AB = \overline{\overline{AB}}=\overline{\bar{A}+\bar{B}}=NOR(NOT(A), NOT(B))$$ From here you can just substitute the operations.