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First of all, sorry for my bad english. I'm not very comfortable writting about Boolean algebra in english but I'll try my best.

So my teacher came up with this problem. Express the following function using only AND + NOT gates, then, NAND only.

The function: $$ f = {a'}{b'}+{a}{b}+{b'}{c} $$

After converting f to its cannonical form I got the truth table so I can check if I made some mistakes when doing the circuit.

But my question is, do I have to apply the same process to express it as AND+NOT and NAND? They both use the same logic.

My solutions were:

NAND: $$ f'' = [({a'}{b'})'*({a}{b})'*({b'}{c})']' $$

AND + NOT: $$ f' = ({a'}{b'})'*({a}{b})'*({b'}{c})' $$

And here are the circuits:

NAND only

AND+NOT

Albert Lorenzo
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    I don't understand your notations : 1) what is the meaning of "*" in both cases ? 2) Why don't you use a special notation for nand 3) What do you mean by $f''$ (not(not(f))=f !) – Jean Marie May 24 '21 at 22:46
  • Sorry for these mistakes. * equals AND. Following my teacher notes f’’ means the involutive rule. Thanks – Albert Lorenzo May 24 '21 at 23:41
  • if "*" means "AND", what is the meaning of the product in expression a'b' ? is it a' OR b' ? Probably not... – Jean Marie May 24 '21 at 23:45
  • Its a AND b too, I just wanted to make clear the difference between + and * for the terms of the function – Albert Lorenzo May 25 '21 at 00:30

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If I understand the root of your question, you are wondering why AND + NOT and NAND are different in circuits but look more similar in boolean algebra.

You already mentioned that they are logically equivalent. The difference is in how you encode it. Logic gates have an operation for NAND.

You can emphasise the difference if you introduce a function or symbol that performs NAND in boolean algebra.

$$f = (a'\uparrow b')\uparrow(a\uparrow b)\uparrow(b'\uparrow c)$$

Which still means the same as

$$f = [(a'b')'\land(ab)'\land(b'c)']'$$

So, yes, whatever method you use to convert to AND + NOT can be used to convert to NAND.

NeRoboto
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