Can Boolean Algebra Simplify Complex Logical Expressions?

[dagg3r]

Boolean Algebra Hard!

hey all, got stuck on some boolean algebra just wondering if you all can check my working out thanks :)
basically the ` represent bar's and in the example no. 1 the p is barred, r is barred,and the overall function is barred hope you get the gist of things thanks

1. [(p` + qr)(pq+r`)]`

my working out is, using de morgan's rule
= [(p` + qr)]` + [(pq + r`)]
= p``*(qr)` + r``*(pq)`
=p*(qr)` + r*(pq)` \\ De morgan's rule again
= P( q` + r` ) + r(P` + q`) \\ expanded out
= pq` + r`p + rp` + q`r
\\ i use the rule that r`p + rp` = `
thus = q`(R + p)

is that right hopefully i did it correctly :)

2. (z + (x*y`)) + yx + (x*(y` + z))
thats the function my working out is i expanded it out.
thus
= (z +xy`) + yx + xy` + zx
\\ then i left it as it is as use many of the boolean rules and got
xy` +Z + YX
X(Y` + y)+ z
=X + z
\\i used the karnaugh maps and got x + z to be the simpliest function as well but was wondering ifanyone can check this out for me thanks.


3. [x` + (y`*z`)][yz` + x`][y`+z`]
= [x` + z`y`][x` + yz`][y` + z`]
\\then i used the rule that P(P+Q)=P so that means taking x` as a common factor takinga look at the first 2 functions out ofthe 3
= x`(yz` + x`)
=x` \\ now we have (y` + z`) left as a function thus
= x`(y`+z`)
\\ using de morgan's rule
=(x+yz)`

is that the simpliestform and how would i draw this as a simplified switchinig circuit because i believe the whole function barred you can't draw it?

thats all hope this isn't a load of gibberish :)

 

[AKG]

\\ i use the rule that r`p + rp` = `

I'm not sure of this rule. Suppose r = 0, p = 1, then r'p = 1*1 = 1, so 1 + rp' = 1 which is not equal to 0 or '.

Number 2 is correct. You will just use the fact that x + xA where A can be anything will just be x, and you'll find you have an x swallowing up everything, there's no need for Karnaugh maps. 3 is right. You can always bar a whole circuit. How you bar it depends on what gates you have available to you, but the following two ways work:

Use an AND and an OR gate to make x+yz, then use an inverter gate to bar it. Otherwise, use a NAND gate and NAND x+yz with itself. If you can bar a single thing like x, y, or z, then you can bar any larger expression. And I believe this way will be better then doing x'(y' + z') because you'd need to bar 3 things.

 

[tacman]

Hi there,

I can definitely check your working out for these boolean algebra problems. Let's take a look at each one:

1. [(p` + qr)(pq+r`)]`
Your working out looks correct to me! You have correctly applied De Morgan's rule to simplify the expression. Nice job.

2. (z + (x*y`)) + yx + (x*(y` + z))
Similarly, your working out here also looks correct. You have correctly expanded the expression and applied boolean rules to simplify it. Good job using Karnaugh maps as well to confirm your answer.

3. [x` + (y`*z`)][yz` + x`][y`+z`]
Your working out is correct so far, but you can simplify it a bit further. After applying De Morgan's rule, you have (x+yz)` as the simplified expression. This can be further simplified to just x` using the rule that P(P+Q)=P. So the final simplified expression would be x`. As for drawing it as a switching circuit, you are correct that you cannot draw a barred function. So in this case, you would just draw a circuit with an input for x and outputs for y and z.

Overall, great job on solving these boolean algebra problems! Keep practicing and you'll become a pro in no time.