Converting boolean function (dnf) to only 2-input NANDs

Question

I'm currently preparing for an exam and was doing some exercices regarding boolean logic. Commonly asked question is to build a function from a truth table and simplify it with a karnaugh map.

After that we mostly have to convert the dnf to a circuit with NAND or NOR logic only, usually not the problem, but im struggling to convert this dnf to a circuit with NAND-Gates that only use 2 inputs:

(¬B ∧ D) ∨ (A ∧ ¬B ∧ C) ∨ (¬A ∧ ¬C ∧ ¬D) ∨ (¬A ∧ B ∧ ¬C)

The inputs are available in a double-rail-system, so normal and inverted.

I tried using the rules for boolean logic to extract a variable, so i'd only have 2 variables/inputs per gate left, but was getting confused due to having initial 3 inputs/variable for the AND gates.

Appreciate any help on this!

Answer 1

Don't worry much about three variable cases, you can always use two NANDs:

https://www.quora.com/How-can-this-be-simplified-using-only-NAND-gate-A+B-C-D+E

Writing it like this might help you:

(¬B ∧ D) ∨ (A ∧ (¬B ∧ C)) ∨ (¬A ∧ (¬C ∧ ¬D)) ∨ (¬A ∧ (B ∧ ¬C))