# Search the OR of negation between boolean algebra

I have this formula $$(a\cdot b)+(\neg a\cdot \neg b)$$ At first I thought this kind of $a+\neg a = 1$ so the answer is 1, but then I realized $(\neg a\cdot \neg b) \neq \neg (a\cdot b)$.

I try to do De Morgan for each $(a\cdot b)$ and $(\neg a\cdot \neg b)$ so it will be $$(\neg a+\neg b) + (a + b)$$

am I doing it wrong?

(I'm sorry for my bad English)

Best Regards

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## 2 Answers

The last part is incorrect, certainly. If you use De Morgan's Law on $a\cdot b$, you will get $\neg(\neg a + \neg b)$; and if you use De Morgan's law on $\neg a\cdot \neg b$, you will get $\neg(a+b)$, rather than $(\neg a+\neg b)$ and $(a+b)$.

So you would write that what you have is equivalent to $$\neg(\neg a+\neg b) + \neg(a+ b)$$ If you try using De Morgan's Law again, you will get $$\neg\bigl( (\neg a + \neg b)\cdot (a+b)\bigr).$$

In fact, I don't think you can simplify what you have. What you have is an "if and only if": it is true if both $a$ and $b$ are true, or if both $a$ and $b$ are false. It is neither a tautology, nor a contradiction.

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this result is equal to $\bar a\bar b +ab$ disjunctive form...maybe that is a little bit simpler – pedja Oct 5 '11 at 17:18
thanks a lot, the clarification of unable to do of simplification is what I expected for :) – giripp Oct 5 '11 at 17:27
@pedja: That's what he started with: $(a\cdot b) + (\neg a\cdot \neg b)$. – Arturo Magidin Oct 5 '11 at 18:11
@Arturo,I was focused on your answer so I overlooked that fact :) – pedja Oct 5 '11 at 18:20

If you make Karnaugh map (see picture bellow) you will see that your expression is minimal disjunctive form and can't be of more simpler form.

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