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I could really use some help with basic boolean algebra manipulation

rocadelpunk

rocadelpunk

Diamond Member
This is our first homework on the stuff and I dunno if I'm really missing something (probably) or I'm on right track..

problem 1

so i guess first thing was using demorgan's, then expanded it.

I dunno if the idenity x bar times x bar follows the same as x*x=x, but that's what I assumed and then since you have the two identical terms you can get rid of one of them, so final answer would be Abar*Bbar?

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problem 2

so, i guess I'm just confused about how to simplify the (AB + Abar)
I tried to use the identity X + Xbar= (x + xbar)(x+y)= x+y...maybe it's b/c i woke up too early/too dumb 😛/too hungry

but i'm just stuck :/

-----------------

problem 3

I hope what I did is correct...other than further factoring and doign A(C+D)+D on inside...i don't know what else I can do

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problem 4

On this one I'm not quite sure what to do with the right side (it goes back to problem two where I'm not sure how to handle the xy+xbarz








Any help would be greatly appreciated. Thanks.


edit; i'm gonna go to my t.a.'s office hours for a few and then come back in case someone has a question for me.
 
There are multiple ways to do all of these problems. What method are you expeced to use?

You can simplify these formulats using truth tables. Take AB+!A (problem 2) for example. Note I'm using ! in place of bar.

A|B|AB|!A|AB+!A
---------------
T|T|T |F |T
T|F|F |F |F
F|T|F |T |T
F|F|F |T |T

That table looks better using a fixed width font.

You can see from the table that the answer is only false for the case A!B. So your answer is !(A!B).

Using demorgans law you can simplify it to:
!(A!B) = !(A)+!(!B) = !A+B


Alternatively you could use the distributive law (X+YZ = (X+Y)(X+Z))
AB+!A
=!A+AB //Communitive law (X+Y = Y+X)
=(!A+A)(!A+B) //Distributive law (X+YZ = (X+Y)(X+Z))
=(T)(!A+B) //I forget the name of this law, but trust me, it's a law. 😉 (X+!X = T)
=!A+B //Identity law (T*X = X)


I hope that gives you a better idea how to do the other problems, because I'm certainly not going to do all of your homework for you. If you are still stuck I'll try to help you out later. I graded for this class in college.

Edit: Also, your reasoning for problem 1 is correct. It is called the Idempotent law. X+X = X.

Edit2: Just looked closer at you solution for problem 3 and you havbe a huge mistake going from step 2 to step 3. You use the distributive law (X(Y+Z) = XY+XZ) incorrectly. In your case X=B, Y=C, Z=D(A+!C), so step 3 should read B(C+D(A+!C)) instead of B((C+D)(A+!C)). Big difference. If you work it out correctly you should get B(C+D) as the correct answer.
 
yah i didn't expect people to do the problems for me, just wanted some hints/strategy.

The truth table is a good idea, t.a. said it would help check my answers.

I've been just using distributive laws/basic idenities so far.

And the only reason i posted multiple problems were they are related so I figured that someone could notice a general theme or something.


anyway for the second one, i got after the C(AB+!A) ...C(!A+A)(!A+B)...C(!A+B)

third one i have better idea how to do, but i just realized i missed a or sign so need to go back.

yah i need more work with the distributive law/other laws. kinda hard to remember at first (was trying to not look back at pages)
Thanks for the tip kyteland.



edit: thanks yah i noticed that hah, can't do this stuff rushing, i worked out third on bus ride home and got same answer. thanks!
 
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