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I have no idea how to do boolean algebra,

First question is

x'y + x(x + y')

I need to first draw a circuit diagram(logic gate) and then simplify it and draw a simplified logic gate.

As of now I only have it down to

x'y + xx + xy'

Am I right in simplifying it down to

x'y + xy' = 0 as x+x' = 0 and y+y' = 0 ?? and then having + xx which is equivalent to x?

Ending up with the answer of 0 + 0 + x which = x?

MichaelMariani
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  • $\bar{x}y + x\bar{y}$ describes an XOR gate. Either $x = 1 \land y=0$ or $x =0 \land y=1$ (exclusive or, either $x$ or $y$ must be $\text{True}$, but not both at the same time. short: $x \oplus y$). And you're right, by the tautology, $xx = x$ for boolean $x$. So what you have should be $\bar{x}y + x + x\bar{y}$, and you can simplify that further. (Hint: pull out an $x$ on the right) – Maximilian Gerhardt May 12 '16 at 14:21
  • How do i simplify that further? thats where I'm stuck at the moment.

    Am i right in assuming that i can turn x+xy' into just xy or am i still lost

     – MichaelMariani May 12 '16 at 14:23
  • Actually, Would i be right by using Absorption where x + (x * y') = x ? – MichaelMariani May 12 '16 at 14:30
  • Maximilian am I correct on this next Question? – MichaelMariani May 12 '16 at 14:58
  • Next time put seperate questions in seperate threads, the answer is now in my comment to my first answer. – Maximilian Gerhardt May 12 '16 at 15:09

1 Answers1

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In the comments I already simplified it to

$$\bar{x}y + x + x\bar{y}$$

(distribute $x$ into the parenthesis, then by tautology $xx = x$)

Pull out an $x$ on the right side.

$$\bar{x}y + x + x\bar{y}$$ $$\bar{x}y + x(1 + \bar{y})$$ $$\bar{x}y + x(1)$$ $$\bar{x}y + x$$

We can analyse this expression by the means of the truth table, i.e., we ask ourselves when this expression becomes true. This means: If $x = \text{True}$, then the whole expression becomes true. If $x =\text{False}$, then $\bar{x} = 1$ and the expression only becomes $y$. Then $y = 1$ makes the expression true.

$\Longrightarrow$ If $x = 1$ the expression is true

$\Longrightarrow$ If $y = 1$ the expression is true

$\Longrightarrow$ The expression cannot be made true through other means

$\Longrightarrow$ The whole expression equals $$x + y$$

Maximilian Gerhardt
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  • Ohh okay so you can simplify x + xy into x(1 + y), i wasn't sure if that was possible. Which law is that? Thank you so much for the help! How do you end up at x + y – MichaelMariani May 12 '16 at 14:33
  • It's the OR distributive law backwards, then I used $1 + x = 1$ (OR identity law / Annulment law). You can check out http://www.electronics-tutorials.ws/boolean/bool_6.html for a set of rules in boolean logic. And the end, I use the truth table to decude that it's $x + y$ (since that expression has the same truth table as $\bar{x}y+x$. I'm sure that it can be shown by using boolean laws only too, though. – Maximilian Gerhardt May 12 '16 at 14:36
  • Ohh right, yea i get it now. Thanks for the link I'll get into it. Im probably going to have trouble with a couple more questions down the line, as this is just the second one out of about 6 (progressively getting harder) am I able to get in contact with you in any way? or are you able to look out for my Q's/Should I comment them here? – MichaelMariani May 12 '16 at 14:39
  • No personal help service - revise your laws carefully, and if you have more questions, ask them properly here on math exchange :). – Maximilian Gerhardt May 12 '16 at 14:39
  • Thats all good :) Will do. – MichaelMariani May 12 '16 at 14:41
  • Careful, it's true that $\overline{a + b} = \bar{a}\cdot \bar{b}$, but for the must first group the expressions. We set $a = x$ and $b = xy$, we then have $\overline{x + xy} = \bar{x}\cdot \overline{xy}$. We can then again use demorgan's law and say $\overline{xy} = \bar{x} + \bar{y}$. So we have $\overline{x + xy} = \bar{x}(\bar{x} + \bar{y}) = \bar{x}\bar{x} + \bar{x}\bar{y} = \bar{x} +\bar{x}\bar{y}$. Then we multiply by $y$ and get $y\bar{x} + y\bar{x}\bar{y}$ Which is $y\bar{x}$ since $y \cdot \bar{y} = 0$. So the whole thing is $\bar{x} \cdot y$ – Maximilian Gerhardt May 12 '16 at 15:04