What does it mean that "the constants true and false can represented with only nand"?

Question

Does this statement mean that you can represent true and false without using AND or OR?

I figured out that $ \lnot A\,\text{nand}\, A = \text{true}$

But how can I represent $ \text{false} $ without using AND or OR... or does the question not imply this restriction?

score 4 · Accepted Answer · edited May 06 '14 at 03:58

4

$(A \mid A)$ is the same as $\neg A$, so to get "false", just do $\top \mid \top \iff ( ( A \mid A ) \mid A ) \mid ( ( A \mid A ) \mid A )$

edited May 06 '14 at 03:58

Michael Hardy

answered May 06 '14 at 02:42

A.S

1

What is the | symbol? – May 06 '14 at 02:44
Was an accident. The $|$ symbol is our not-and symbol. – A.S May 06 '14 at 02:44
@JoelReyesNoche MSE doesn't like duplicates, even of deleted answers. I might hit the rep cap today, so it might not matter anyway, but thanks. – A.S May 06 '14 at 02:46

score 1 · Answer 2 · answered Jun 01 '21 at 13:55

¬A can be represented as (A nand A).

So, basically swap ¬A with (A nand A) ==> (A nand A) nand A) which will always be true.

Now, to represent false in only nand ==> you simply take the previous answer and add at the beginning not.

¬( (A nand A) nand A)), since we know nand can only be false if both cases is True then we use the

[ (A nand A) nand A)] nand [ (A nand A) nand A)] = false

score 0 · Answer 3 · edited May 06 '14 at 03:36

0

How about this:

$(\lnot A\operatorname{nand}A)\operatorname{nand}(\lnot A\operatorname{nand}A)$

As each expression in the brackets will evaluate to true, the overall expression would be the negation after the and which would always be false.

edited May 06 '14 at 03:36

Graham Kemp

answered May 06 '14 at 02:47

JB King

3 Answers3