# Boolean Algebra Hard

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 + rp + rp + qr
\\ i use the rule that rp + 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 + zy][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 :)

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