Need Help with Boolean Algebra Equations - Proof and Explanation

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Homework Help Overview

The discussion revolves around Boolean algebra equations, specifically focusing on proofs and explanations related to various identities and operations. The original poster presents several equations and proofs, seeking verification of their correctness.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to prove several Boolean identities, while some participants question the clarity and format of the proofs presented. Others suggest that the original poster may need to reconsider their understanding of disproof and the structure of their equations.

Discussion Status

Participants are actively engaging with the original poster's work, providing feedback on the clarity of the proofs and the correctness of the reasoning. There is a recognition of the need for clearer formatting and justification in the proofs, but no consensus has been reached on the correctness of the original poster's conclusions.

Contextual Notes

Some participants note that the original poster's formatting follows their teacher's style, which may differ from typical conventions. There is also mention of potential confusion arising from the use of symbols and notation.

chobo2
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I would like you guys to check over my Boolean Algebra equations to see if I did them right.

My teacher uses the same symbols as the computer science books

Legend:
[PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL]=[/URL] XOR

XOR = a[PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL]b[/URL] = a * b' + a' * B

+ = OR
* = And
' = Not

a) X[PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL]1[/URL] = X

proof:

X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] + 1 = Def
X * 1' + X' * 1 = P5a
X * 0 + X' 1 = T9a, T10a
0 + X' = T9b, L6b

Conclusion X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 1 ≠ X

b) X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 1 = X’

proof:

X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 1 = Def

X* 1' + X' * 1 = P5a
X * 0 + X' * 1 = T9a,T10a
0 + X’ = L6b, T9b
X' =

Conclusion X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 1 = X’

c)
Proof:

X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 0 = Def

X * 0' + X' * 0 = P5b
X * 1 + X' * 0 = T10a
X + X' * 0 = T9a
X + 0 = T9b
X=

Conclusion X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 0 = X

d)
X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] ) = X'

Proof:

X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 0 = Def
X * 0' + X' * 0 = P5b
X * 1 + X' * 0 = T10a
X + X' * 0 = T9a
X + 0 = T9b
X =

Conclusion X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] 0 ≠ X'

e)
X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] X' = 0

Proof:

X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] X' = Def

X * X'' + X' * X' = T13a
X * X + X' * X' = T11a
X + X’ * X' = T12a
X + 0 = T9b
X =

Conclusion X [PLAIN][PLAIN][PLAIN][PLAIN][PLAIN][PLAIN]http://img218.imageshack.us/img218/2948/xnor5qv.jpg [/URL][/URL][/URL][/URL][/URL] X' ≠ 0

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

Postulates:

P1a X = 1 if X ≠ 0 ----- P1b X = 0 if X ≠ 1
P2a 0*0 = 0 ----- P2b 0+0 = 0
P3a 1*1 = 1 ---- P3b 1+1 = 1
P4a 1*0 = 0 ----- P4b 1+0 = 1
P5a 1’ = 0 ----- P5b 0’ = 1

Algebraic Laws

Commutative laws

L6a x*y = y*x ----- L6b x+y = y+x

Associative laws

L7a x*(y*z) = (x*y)*z ----- L7b x+(y+z) = (x+y)+z

Distributive laws

L8a x*(y+z) = x*y+x*z ----- L8b x+y*z = (x+y)*(x+z)

Thermos

T9a x*0 = 0 ------ T9b x+0 = 0
T10a x*1 = x ------ T10b x+1 = 1
T11a x*x = x ------ T11b x+x = x
T12a x*x’ = 0 ------ T12b x+x’ = 1
T13a x’’ = x ------ T13b x = x’’

Absorption Theorems

T14a x+x*y = x -------- T14b x*(x+y)=x
T14c X*(x’ + y) = x*y ------- T14d x+x’*y= x+y

De Morgan’s Theorems

T15a (x*y*z)’ = x’ + y’ + z’
T15b (x+y+z)’ = x’ * y’ * z’
 
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This was amazingly unreadable.

Carl
 
Well, I understand that your link is supposed to be the symbol [itex]\oplus[/itex]. (Click on the image to see what you can type to generate that image) So, for example, the thing at the top is supposed to say [itex]a \oplus b = a \cdot b' + a' \cdot b[/itex]. (Or are you using [itex]\bar{a}[/itex]?)


Oh, I think I get what you're trying to ask here -- each "x) _____" is the problem you were given, and what's below is your proof or disproof, and you're looking for us to check over your work.

I will say one thing: you have the idea of a disproof wrong. To disprove something, what you should be doing is to come up with a particular example for which the statement is incorrect. E.G. because [itex]0 \oplus 1 = 1[/itex], that is sufficient for disproving [itex]x \oplus 1 = x[/itex], because this counterexample shows that it fails when x = 0.

Now, if you can prove a statement is always wrong, that is enough to disprove that statement, but many times it is impossible to do that. For example, [itex]x + 0 = 1[/itex] is not a tautology, but it is impossible to prove that it is always wrong.


Now, the format of your proofs are very confusing. :frown: For (a), for example, it looks like you're trying to say:

[tex]x \oplus 1 = x \cdot 1' + x' \cdot 1 = x \cdot 0 + x' \cdot 1 = 0 + x'[/tex]

And giving a justification for each equality. Aside from the typos you made, the way you've actually written that is extremely confusing. Normally, when you spread an equation out over multiple lines, you put the equals signs on the left. E.G.

x [itex]\oplus[/itex] 1 = x 1' + x' 1
= x 0 + x' 1
= 0 + x'
= x'

(Normally you would align the equals signs, but that's difficult in this text format)

And when you're giving justifications for steps, you shouldn't make them look like part of the equation. In a text format like this forum, I might have written it something like:

x [itex]\oplus[/itex] 1 = x 1' + x' 1 (by definition)
= x 0 + x' 1 (by P5a)
...

I suppose you can write it in a two-column format when you turn in your homework, though. In text, though, you really need some sort way to visually separate the justifications from the equations.


Anyways, the steps of your work generally look right, though I haven't tried matching up the steps with the justification.
 
Yes I did not know that this forum did have that symbol since another fourm I posted it did not have the symbol and it was just a copy and paste from it.

As far as you saying I don't understand how to disprove stuff or whatever. It maybe true but from the examples given from my teacher that is they way she writes the conclusions up so I am just following the way she does it.

And the format I use is the same as my teacher.

So I rather keep it the way she does since she seems to like it like that.
 

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