Can you Simplify These Boolean Expressions Using the Laws of Boolean Algebra?

[Paymemoney]

1. Homework Statement and attempted solutions
Simply each of the following Boolean expressions:
i)
http://img69.imageshack.us/img69/4438/equa1.jpg

ii)
http://img69.imageshack.us/img69/9907/equa2.jpg

iii)Expand the following SOP form to proper miniterms (don't forget to delete identical terms):
http://img69.imageshack.us/img69/1740/equa3.jpg

Homework Equations

Laws of Boolean Algebra:

A + A = A A . A = A NOT NOT A = A
A + 0 = A A . 0 = 0
A + 1 = 1 A . 1 = A
A + NOT A = 1 A . NOT A = 0

(A + B) + C = A + (B + C) (A . B) . C = A . (B . C)
(A + B) = (B + A) (A . B) = (B . A)

A . (B + C) = (A . B) + (A . C) A + (B . C) = (A + B) . (A + C)

NOT (A . B) = NOT A + NOT B
NOT (A + B) = NOT A . NOT B

X . Y + X . NOTY

P.S

 

[magwas]

in 1) you made the first step incorrectly
NOT (A + B) = NOT A . NOT B

same in ii)

 

[Paymemoney]

so for question i) would the first line be NOT A . NOT A . B then i went A . NOT A(B) = B because A . NOT A = 0
i don't think this is correct.

 

[magwas]

Try to be very punctual about what you do, do one step at a time, and it helps to note the relevant equality while you are not sure.
You can check each step by drawing the Karnaugh table for it. I do the first step for you to see what I mean.

\overline{\overline{A} + A*\overline{B}}
Its table:
<br /> \halign{\hfil # &amp; # &amp; # &amp; #\hfil \cr<br /> &amp; &amp; A &amp; \cr<br /> &amp; &amp; 0 &amp; 1 \cr<br /> B &amp; 0&amp; 0 &amp; 0 \cr<br /> &amp; 1 &amp; 0 &amp; 1 \cr }

Now using \overline{x+y} = \overline{x} * \overline{y} we got:
\overline{\overline{A}} * (\overline{A*\overline{B}})
Now if you fill the table for it, you will see whether the transformation is done right.

Of course you can deliver a normal form, and a simplified expression directly from the table. However I recommend to master the algebraic solution also, because there are cases where the algebraic way is much easier.

 

[Paymemoney]

Well by using the kmap or truth table i found out that

0 * 0 equals (A * \overline{A}B)
which equals to 0 so answer is (A* B)

Now how would you find this without using a kmap or truth table?

 

[pheeesics]

I believe you violated DeMorgan's Theorem in step one, like magwas said.

I see in i) you complemented the overall output, starting with the + as the outermost operation. Yet you complement A AND NOT B individually, instead of as a whole. Remember that (A . NOTB) is an input of itself. Once you change the operation inside of (A . NOT B) to (A + NOT B), then you can complement the inputs A and B individually.

Additionally, remember that when you complement output and inputs, you go from OR to AND, yet I see you never changed the OR.

Perhaps my explanation is confusing, so here is a picture.
http://img519.imageshack.us/img519/5554/boolean.jpg

Hopefully someone will back me up, as I am learning this as well.

 

[magwas]

Well by using the kmap or truth table i found out that

0 * 0 equals (A * \overline{A}B)
which equals to 0 so answer is (A* B)

Now how would you find this without using a kmap or truth table?

Using algebra. Bool algebra is very similar to usual algebra. Only we have one additional operator above + and *. Remember that NOT() is a function. For arbitrary f, you cannot do f(a + b) = f(a) + f(b), or any other meaningful thing like that.
With f being NOT, we have the luxury of having the DeMorgan rules:
\neg(a+b)=\neg(a)*\neg(b)
and
\neg(a*b) = \neg(a) + \neg(b).

So all you have to do is do simple algebra, and adhere to the rules.

 

[Paymemoney]

yeh i got the answer thanks for help.