- #1

dagg3r

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**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 :)