AND truth table A B Q ------ 0 0 0 0 1 0 1 0 0 1 1 1 NAND truth table A B Q ------ 0 0 1 0 1 1 1 0 1 1 1 0 Take one NAND for the inputs A & B and another NAND to negate the output of the first to build an AND gate. NAND NAND A-|\ .-|\ | )o--| | )o--- Q B-|/ .-|/

Hi, After some time of thinking and researching this is what i came up with!please comment if i did it right!

Hello Alexander, you do not end up with a correct solution by just painting inversion bubbles in between. One hint already was given: Try to make up a truth table for the inputs (this is in total 6 inputs and quite complex). The next hint was given to use a conventional de morgan notation and to convert this. The third is: You'll have a look onto the schematic. It consist of two blocks: the uppermost left block that consists of an and-or function with the lower "and" function subsituted. And a second and_or function on the right. There are two approaches to solve that: a) You look up http://en.wikibooks.org/wiki/Digital_Circuits/NAND_Logic. You'll find out that a "and" ia a "nand" with following inversion. You'll find out that an "or" is a "nand" with previous inversion. You sketch them together and you'll find out that there are two inversions sequentially which can be eliminated. This will lead you to b) b) You'll find in your literature that a "and-or" combination can substuted by an "nand-nand" conbination without any modification. Now you have to convert the both blocks (uppermost left and right) separately and you will come up with the solution. Any questions? Feel free. rgds