Boolean Arithmetic Simplification

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SUMMARY

The discussion centers on the simplification of the Boolean expression (\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). The user initially struggled with applying DeMorgan's Rule and the Absorption Rule effectively. Ultimately, the solution was achieved by first eliminating the Y terms using DeMorgan's and the distributive rule, leading to a straightforward simplification of the expression.

PREREQUISITES
  • Understanding of Boolean algebra concepts
  • Familiarity with DeMorgan's Rule
  • Knowledge of the Absorption Rule
  • Experience with the distributive property in Boolean expressions
NEXT STEPS
  • Study advanced applications of DeMorgan's Theorem in Boolean simplification
  • Explore the use of the Absorption Rule in complex Boolean expressions
  • Learn about the distributive property in Boolean algebra
  • Practice simplifying Boolean expressions using truth tables
USEFUL FOR

This discussion is beneficial for students studying Boolean algebra, computer science majors focusing on logic design, and anyone interested in mastering Boolean simplification techniques.

BraedenP
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Homework Statement



I am asked to prove that (\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).

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: \sim(p\wedge q) = \sim p\vee\sim q
Absorption Rule: p\vee(p\wedge q) = p

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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Convert the OR's contained within brackets to AND's using DeMorgan's rules.
 
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.
 

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