Boolean Arithmetic Simplification

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The discussion revolves around proving the equivalence of two Boolean expressions, specifically (\sim x)\vee z and a more complex expression involving multiple terms. Initial attempts included applying DeMorgan's rule and the absorption rule, but the simplification process was challenging. Ultimately, the solution was achieved by first eliminating the Y terms using DeMorgan's and the distributive rule, which simplified the problem significantly. The final steps fell into place easily after this initial simplification. This approach highlights the effectiveness of methodically applying Boolean algebra rules for simplification.
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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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