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I have the following boolean algebra, where union is $+$ and intersection is $\cdot$ :

$(x\cdot y)+((z+y)\cdot \bar{z})+y=y$

Is there a systematic way of doing this, or do you need to puzzle? My train of thought would be to see if I could get rid of $z$ and $\bar{z}$ by putting them together, but am unsure if the operation for this needs to be $+$ or $\cdot$.

The other thing I tried to do is use the commutative and distributive property on $\bar{z}$ and $(z+y)$:

$(\bar{z} \cdot z)+(\bar{z} \cdot y) = \textit{0} + (\bar{z} \cdot y) = (\bar{z} \cdot y)$

but then I'm stuck with $(\bar{z} \cdot y)$, and there seems to be no way to get rid of $\bar{z}$ anymore.

I could also use the distributive property on $y$ and $((z+y)\cdot \bar{z})$ in $((z+y)\cdot \bar{z})+y$:

$(y+(z+y)) \cdot (y + \bar{z})=\\((y+z)+(y+y)) \cdot (y+ \bar{z})=\\((y+z)+y)\cdot(y+ \bar{z})=\\y+((y+z) \cdot(y+\bar{z}))=\\y+y+(z\cdot \bar{z})=\\y+\textit{0}=\\y$

This is nice (though I'm not sure step 3 and 4 are legit), but then I'm stuck with the $x$!

I would appreciate any help in this matter!

user94450
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1 Answers1

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You are going down the right path, you need to use the distributive law but in reverse i.e. factoring. I'm not sure of an explicit algorithm to follow other than just playing with the equation.

$$(x \cdot y) + ((z + y) \cdot \bar{z}) + y = (x \cdot y) + (z \cdot \bar{z} + y \cdot \bar{z}) + y = (x \cdot y) + (y \cdot \bar{z}) + y$$ $$= (x \cdot y) + (\bar{z} \cdot y) + y = (x + \bar{z} + 1) \cdot y = 1 \cdot y = y$$

Ryan Sullivant
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  • Thank you for your answer! I don't know if you read my previous comment, if so forget it! I got it! Thank you very much! – user94450 Sep 12 '13 at 06:23
  • Yeah sure, no problem. The $1$ is from the last $+y$. Make sure to factor on the same side you are multiplying from. What I am really doing is $(x \cdot y) + (\bar{z} \cdot y) + (1 \cdot y) = (x + \bar{z} + 1) \cdot y$. Then $1 + a$ = $1$ for any $a$. Its like one of the properties you used above, we know in boolean algebras that $a + a = a, a \cdot a = a, 1 \cdot a = a, a + 0 = a$, etc. – Ryan Sullivant Sep 12 '13 at 06:26
  • Yes, these are the identities if I'm correct. – user94450 Sep 12 '13 at 06:32