Boolean Arithmetic Simplification

  • Thread starter BraedenP
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In summary, the problem is to prove that the statement (\sim x)\vee z = \sim(x\vee y)\vee\sim(y\vee\sim z)\vee\sim(x\vee\sim y)\vee\sim(\sim y\vee\sim z) is true. The equations used are DeMorgan's Rule: \sim(p\wedge q) = \sim p\vee\sim q and Absorption Rule: p\vee(p\wedge q) = p. The solution involves converting OR's within brackets to AND's, simplifying using DeMorgan's and the distributive rule, and ultimately getting rid of all the Ys
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BraedenP
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Homework Statement



I am asked to prove that [itex](\sim x)\vee z = \sim(x\vee y)\vee\sim(y\vee\sim z)\vee\sim(x\vee\sim y)\vee\sim(\sim y\vee\sim z)[/itex].

I've tried using all combinations of DeMoran's rule, the distributive rule to get the y terms together, and the absorption rule to get rid of the y (which is required in order to simplify it down in terms of x and z.


Homework Equations



DeMorgan's Rule: [itex]\sim(p\wedge q) = \sim p\vee\sim q[/itex]
Absorption Rule: [itex]p\vee(p\wedge q) = p[/itex]

The Attempt at a Solution



I can post some of the steps I've taken, but none really lead anywhere. Where is a good place to start for a question like this?
 
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  • #2
Convert the OR's contained within brackets to AND's using DeMorgan's rules.
 
  • #3
Thanks -- yeah, I was trying that before, but couldn't get it simplified down enough.

I got the solution now (and it wasn't too hard). I just had to use DeMorgan's and the distributive rule to get rid of all the Ys first, and then everything else just fell into place without much effort.
 

1. What is Boolean Arithmetic Simplification?

Boolean Arithmetic Simplification is a method used to simplify Boolean expressions by reducing the number of terms and operators, while still maintaining the same logical value.

2. Why is Boolean Arithmetic Simplification important?

Boolean Arithmetic Simplification is important because it helps make complex Boolean expressions easier to understand and evaluate. It also helps in reducing the number of logic gates needed in a circuit, thus making it more efficient.

3. What are some common techniques used in Boolean Arithmetic Simplification?

Some common techniques used in Boolean Arithmetic Simplification include the use of Boolean laws and rules, De Morgan's laws, Karnaugh maps, and truth tables.

4. How do I know when to use Boolean Arithmetic Simplification?

You can use Boolean Arithmetic Simplification when you need to simplify complex Boolean expressions, such as those used in logic circuits or programming. It can also be used to find the minimum number of logic gates needed to implement a given Boolean expression.

5. Are there any limitations to Boolean Arithmetic Simplification?

Yes, there are limitations to Boolean Arithmetic Simplification. It may not always be possible to simplify a given Boolean expression, and even when it is possible, the resulting simplified expression may not be the most efficient in terms of logic gates. It is important to carefully consider the trade-offs between simplicity and efficiency when using Boolean Arithmetic Simplification.

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