Need Help with Boolean Algebra Equations - Proof and Explanation

In summary: In any case, the fact that one specific case is a counterexample is enough to disprove a statement. It is not necessary to show that every case is a counterexample.In summary, the conversation is about Boolean Algebra equations and using them to prove or disprove statements. The symbols used in the equations are similar to those used in computer science books. The conversation includes proofs or disproofs of various statements using postulates, algebraic laws, and other theorems. The idea of a disproof is clarified, stating that a single counterexample is enough to disprove a statement, rather than showing that every case is a counterexample.
  • #1
chobo2
2
0
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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  • #2
This was amazingly unreadable.

Carl
 
  • #3
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.
 
  • #4
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.
 

1. What is Boolean algebra and why is it important?

Boolean algebra is a mathematical system that deals with binary variables and logical operations. It is important because it is the foundation of digital electronics, computer programming, and many other fields where logical operations are used.

2. How is Boolean algebra used in digital circuits?

Boolean algebra is used in digital circuits to represent and manipulate logical operations. This allows for the design and analysis of complex digital systems, such as computers, calculators, and other electronic devices.

3. What are the basic operations in Boolean algebra?

The basic operations in Boolean algebra are AND, OR, and NOT. AND represents logical conjunction, OR represents logical disjunction, and NOT represents logical negation.

4. How can Boolean algebra be simplified?

Boolean algebra can be simplified using various algebraic rules and theorems. This involves reducing expressions to their simplest form by combining terms and eliminating redundant variables.

5. What are some real-world applications of Boolean algebra?

Boolean algebra has many real-world applications, including in computer programming, digital circuit design, database querying, and automated reasoning. It is also used in fields like artificial intelligence, robotics, and telecommunications.

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