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Let $S$ a boolean algebra; $a,b,c \in S$ prove that $(a'+b)'+(a'+b')'=a$

Then:

$(a'+b)'+(a'+b')'= (a')'+b'+(a')'+(b')'=a+b'+a+b = (a+a)+(b+b')=a+1=1$

Maybe i'm wrong but I think that the problem is bad written, since $a+1=1$, it never can be equal to $a$.

There is a way to solve this exercise?

Thanks!

EDIT


As @amWhy says, when I'm applying the De Morgan's law, then $+$ turns to $\cdot$ and $\cdot$ turns to $+$

So (...)

$(ab)'=a'+b'$

$(a+b)'=a'b'$

then:

$(a'+b)'+(a'+b')'=(a')'\cdot b'+(a')'\cdot(b')'=ab'+ab=a(b+b')=a \cdot 1 = a$

Then, $(a'+b)'+(a'+b')'=a$

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    $\begingroup$ You've got the laws wrong. $(a+b)^\prime=a^\prime.b^\prime$. $\endgroup$
    – Sudarsan
    Sep 25 '13 at 0:48
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When you apply DeMorgan's, the operation changes from + to multiplication.

$$(a'+b)'+(a'+b')'= (a')'b'+(a')'(b')' = ab' + ab = a(b' + b) = a\cdot 1 = a$$

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  • $\begingroup$ Even ack'ed you in the answer! :-) +1 $\endgroup$
    – Amzoti
    Sep 25 '13 at 0:56
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For an alternate proof, try factoring instead of expanding: $$ (a' + b)' + (a' + b')' = ((a' + b)(a' + b'))' = (a' + bb')' = (a' + 0)' = (a')' = a $$

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