Converting logic gates to a purely NAND implementation, how come this is valid?

Question

So Im doing some exercises with answers provided on conversion to a purely NAND gate implementation.

enter image description here

I understand how the 2 AND gates and the OR gate connecting them were converted, by double negation but how come the OR gate connecting b and c can be converted to a NAND gate? (how can you just put a negation on b and c without double negating it?) i dont understand how it could still be equal to the previous implementation.

kjhughes · Accepted Answer · 2014-03-22T13:21:55.713

1

You are correct. The shown transformation of the (b + c) OR gate is a mistake.

The proper transformation of (b + c) would be as follows:

(b + c)    Given
(b + c)''  Apply double negative.
(b'c')'    Apply De Morgan's Law.

Which is not the same as the shown (b' + c').

The intermediate diagram would most constructively show canceling inversion bubbles added to the output of the b+c OR gate and the input of the a(b+c) AND gate (as was done in the later stages). This corresponds to the application of the double negative in the above derivation. Then, the final diagram should be as shown plus b inverted and c inverted.

If inversions of b and c are not otherwise available, they can be implemented via NANDs wired in a NOT configuration.

edited Mar 22 '14 at 13:21

answered Mar 21 '14 at 20:07

kjhughes

106,133
27
181
240

so the correct answer putting a inverter (NAND implementation of inverter) on B and C and making the OR gate into a NAND gate? – jantristanmilan Mar 22 '14 at 07:25
Right. I updated the answer to say specifially how the diagrams should be adjusted. – kjhughes Mar 22 '14 at 13:19

Converting logic gates to a purely NAND implementation, how come this is valid?

1 Answers1