Boolean Algebra and proofs

Prove using Prove algebraically : 1) x'′⊕ y = x⊕y' = (x⊕y)'

2) x⊕1 = x'

3) x⊕x' = 1

4) (A+B)(A'C'+C)(B'+AC') = A'B (*A*+*B*)(*A*′.*C*′+*C*)(*B*′+*A**C*)′=*A*′*B*

I know x⊕y = xy'+x'y But how do i deal with X'⊕Y? What does that become?

And for 4) I did:

(A+B)[(X+Y)'+C] [B.(AC)']

(A+B)[(A+B)'+C] [(B.(A'+B')]

then what do I do from here?

http://i.imgur.com/CPlRB91.png[/FONT][/COLOR]

Been having trouble solving these questions of my homework. Thank You!

Re: Boolean Algebra and proofs

I tried to do the monster one, i have no experience with boolean functions or circuit design but here is what i came up with

if and represent 2, 2-digit binary numbers, then

addition can be done as follows with the result being a 3 bit binary number

(OR

this adds 2 digit binary numbers. I used 3 half adders to get this. I dont know if it could be done more simply

and multiplication can be done by just using the AND operator on every pair you get from this cross product and then adding a 2 bit binary to a one bit binary (addition using the above method by using 0 for the leading digit of one 2 bit number).

For ex

11

x 11

-----------

11 ----(save this digit without any operation and add 11 + 01 using the addition method above)

11x (+)

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

1001

Re: Boolean Algebra and proofs

As for

1)

using the OR,AND,NOT def of XOR is the way to go

for example

^ ^ (take the negation inside) ^ ^ v =

2)

v ^ ^ = 1 ^ = 1 ^ 1 = 1

3)

v ^ ^ ^ ^ ^ v = 1 ^ ( ) =

Re: Boolean Algebra and proofs

thank you, you are a beautiful human being.

How would I solve this question?

imgur: the simple image sharer

I tried and got to:

(A+B)[(A+C)' +C) [(B.A'+C')]

This pic is also part of the "monster question" that I forgot to attach.

Absolutely no idea how to take this, I shouldn't have taken an intro online course.. ><

imgur: the simple image sharer

Re: Boolean Algebra and proofs

Alrighty.

So for that problem

I rewrote is as Now i can rewrite the last term in parenthesis as

So using commutativity and distributavity, i will distribute the

This evaluates to since and I will then distribute this term against the middle parenthetical term. to get again note alot of terms cancel so this equals

Re: Boolean Algebra and proofs

Re: Boolean Algebra and proofs

Re: Boolean Algebra and proofs

Thanks, I'll def look into that.

Are you able to help me with my last two questions? They got me stumped after I write the booleans

imgur: the simple image sharer

imgur: the simple image sharer

For Question 1, I got:

[(A+B).(C+D)] + [(C+D)' . (A+B)] + C

C+(A+B+C+D) + (A+B+D) + (C+D)' + (C+D)+B

A+B+C+(C+D)'

I"m not sure how to simplify this further.

Question 2 is really long and I would appreciate help. I got:

A' + (D'+C') + A + (A' + C'+ D') + B' .D ' + (D+B)' = F

I'm not even sure if this is correct, and these two questions are due as homework tonight.

Any help would be much appreciated, thanks!

Re: Boolean Algebra and proofs

Maybe I'm looking at it wrong but for 1 I got (A+B)(C+D)+~(C+D)(A+B)+C which simplify to A+B+C

Re: Boolean Algebra and proofs

could you please list the steps you did to get to a+b+c? I believe your answer is right, only our first expression is the same :(

I tried distributing the (C+D) ' (A +B) and ended up with BABD +C as the final answer ><

Re: Boolean Algebra and proofs

(a+b)(c+d)+~(c+d)(a+b)+c = (a+b)((c+d)+~(c+d))+c = (a+b)(1)+c= a+b+c

Re: Boolean Algebra and proofs

As for the last one.

The expression is ((~A + ~(C v D))A)((~ ) + (~BC)) = A~B~C~D