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I'd like to express $AB$ just in terms of sum, I have already tried this: $(A′B+AB′)′\cdot(A+B)=AB$, but, of course, $A'B$, for example, is still multiplying.

And also, I'd like to do the reverse process, expressing $ab + a'(b+c)$ just in terms of multiplication, not a single sum operator, but I'm not getting it, trying any one of the Boole's theorems...

Which would be the correct way to express both of them just with the indicated operators?

Thanks so much everyone in advance!

postmortes
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edusola93
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  • @edusola93 Why does your title say "without sum or multiplication"? What precisely are the operations that you are allowed? – Erick Wong May 21 '17 at 06:32
  • @ErickWong On second reading, my assumption does not seem justified, because the OP has $(AB'+A'B)'$, which is not two-level. I'm going to delete my other comments, which at this point are just confusing. – Fabio Somenzi May 21 '17 at 06:38
  • @FabioSomenzi Correspondingly I'll delete my first two comments as well :). – Erick Wong May 21 '17 at 06:43
  • thanks for commenting guys @ErickWong, I'm sorry if somethings are not understood correctly, I'm not perfect at English, sorry. To summarize it, I just have to express a·b in + (sum) operator and ab + a'(b+c) just with · (multiply) operator. – edusola93 May 21 '17 at 06:46
  • @FabioSomenzi (I tag you here, 'cause I couldn't in the last comment) – edusola93 May 21 '17 at 06:49
  • As Erik suggested, you can write $AB$ as $(A'+B')'$. You can similarly write $ab+a'(b+c)$ as $(ab')'\cdot(b'c')'$. – Fabio Somenzi May 21 '17 at 07:13

1 Answers1

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Answered in the question comments section, by @ErickWong and @FabioSomenzi.

AB     as    (A′+B′)′.
ab+a′(b+c)    as    (ab′)′⋅(b′c′)′
edusola93
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  • This is written very confusingly. Why the redundant ABAB? The resulting expression contains a multiplication. Anyway, the point is that any "and" can be converted into an "or" and vice versa, using deMorgan's laws. Applying this recursively allows you to eliminate all "and"s from any given expression (resp. all "or"s). – Erick Wong May 22 '17 at 20:12
  • Damn, there was some kind of error when sending the answer @ErickWong, somehow the message has been written x2 times, at least what was inside the "code" tags, gonna edit it right now, don't know what happened, sorry! – edusola93 May 23 '17 at 16:19
  • The first line is good but the second line still has some problems. I think it should be $((ab)'(a'(b'c')')')'$. Alternatively you get a simpler expression by first simplifying to $b+a'c$ using other rules of logic. – Erick Wong May 24 '17 at 01:49