# How To Simplify Boolean Expressions

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According to WolframAlpha, $$(A_3cdot (A_2+A_1+A_0))+(overlinecdot A_2 cdotoverlinecdot overline)$$ can be simplified to $$=A_3cdot A_0+A_3 cdot should be A_1+A_2cdotoverlinecdotoverline$$. but how? Here is my attempt:

## How To Simplify Boolean Expressions

In particular, I don’t know how to get from $$A_2cdot(A_3+overlinecdotoverline)$$ to $$A_2cdotoverlinecdotoverline$$ did I miss something? Any help is appreciated!

## I Am Solving A Boolean Expression And Getting More Than One Simplified Expression, On Checking, Both The Expressions Are Correct. Please Explain! Is Simplification Of A Boolean Expression, Not Unique?

The fact that this subexpression appears in the context of the larger expression as a whole means that you can simplify it.

Actually, how do you go from $A_3+A_3′A_1′A_0′$ to $A_3+A_1′A_0′$ itself in the previous step… Wow, is it because $A_3′A_1′A_0′$ is equal to $to A_1’A_0’$? No, not at all! Rather the fact that you also have $A_3$ means that you can reduce the expression $A_3’A_1’A_0’$ to $A_1’A_0’$

Something similar is going on here: to simplify $A_2(A_3+A_1’A_0′)$ to $A_2A_1’A_0’$ we need the fact that we also have $A_3A_1$ and $A_3A_0$ .

begin cdot A_1tag*\ =&A_2cdot A_3+A_2cdotoverlinecdotoverline+A_3cdot (A_0+A_1)tag*\ =&A_2cdot A_3+A_2cdotoverline +A_3 cdot (A_0+A_1)tag*\ =&A_2cdotoverline+A_3cdot (A_0+A_1)tag*\ =&A_2cdotoverlinecdotoverline + A_3cdot A_0+ A_3cdot A_1tag*\end

#### Solved 17) What Is The Primary Motivation For Using Boolean

So far your efforts are correct, try applying the distributive law for the first part, can you see why we don’t need A2⋅A3 in this statement?

begin cdot A_1tag*\ =&A_2cdot A_3cdot(1)+A_2cdotoverlinecdotoverline+A_3cdot A_0+A_3cdot A_1tag*\ =&A_2cdot A_3 cdot (A_1+overline)+A_2cdotoverlinecdotoverline+A_3cdot A_0+A_3cdot A_1tag*\ =&A_2cdot A_3cdot A_1+A_2cdot A_3cdotoverline+ A_2 cdotoverlinecdotoverline+A_3cdot A_0+A_3cdot A_1tag*\ =&A_2cdot A_3cdot A_1+A_3cdot A_1+A_2cdot A_3cdotoverline+A_2cdot over cdotoverline+A_3cdot A_0tag*\ =&A_3cdot A_1+A_2cdot A_3cdotoverline+A_2cdotoverlinecdotoverline+A_3cdot A_0tag*\ =&A_3 cdot A_1+A_2cdot A_3cdotoverline(1)+A_2cdotoverlinecdotoverline+A_3cdot A_0tag*\ =&A_3cdot A_1+A_2cdot A_3cdot overline( A_0+overline)+A_2cdotoverlinecdotoverline+A_3cdot A_0tag*\ =&A_3cdot A_1+A_2cdot A_3cdotoverlinecdot A_0+A_2cdot A_3cdot over cdotoverlinetag*\&+A_2cdotoverlinecdotoverline+A_3cdot A_0\ =&A_3cdot A_1+A_2cdotoverlinecdot A_3cdot A_0+A_3cdot A_0tag*\& +A_2cdot A_3cdotoverlinecdotoverline+A_2cdotoverlinecdotoverline\ =&A_3cdot A_1+A_3cdot A_0tag*+A_2cdot overlinecdotoverline\ end

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## Solved Problems From Boolean Algebra U Assignment

Background: I’m teaching myself computer architecture because I don’t have a technical degree and I want to learn it, even though my economic situation will never allow me to have a formal education. I’m doing all the problems in a book, but I encountered a problem where the problem seems too complicated for me. The Karnaugh map will not work here because it has 5 inputs instead of 4. Also, a truth table looks a bit confusing as 5 bits means 32 entries.

Q: Can someone show me how to solve this problem using the theorems of Boolean algebra? And logically explain how it works? I have spent many hours trying to solve this problem. I’m sure once I decide how to simplify the algebra I’ll be able to design the circuit.

// Well this is a boolean logic that can be simplified so that its elements (a, b, c) are repeated over multiple words, I’m not sure but let’s try to summarize what this boolean function is Only necessary for work.

// You should know that c+c’ gives 1 (check with truth table 1+(1)’ = 1 or 0 + (0)’=0+1=1. // Also note, cc’ (C* C ‘) returns 0. (Check all possible values ​​0*1 or 1*0 from truth table. Always return zero // Let’s continue =b(1)+a’b’c’=b+ a’ b’ c’ = b’. (abc)’

### If A, B And C Are The Elements Of Boolean Algebra, Simplify The Expression (a’ + B’) (a + C’) + B'(b + C) . Draw The Simplified Circuit

Y = b’a +b’b+b’c = b’a +0+b’c=b’a+b’c=b'(a+c)=b+(a+c)’=b+( a’*c’) The final answer is y =b+(a’*c’) which seems to be generated by the tool y = ( !a * !c) + b Now you can handle (b)! great!

(I solved (c) by hand but it’s a lot of writing (ie a lot of reading). You can get the correct answer directly using the tool, but I think that’s how you need it)

// You can see that d+e is showing too much, let’s try to factor it.

4- The term in which (ab+ac+1+……) // +1 will give 1 in this bracket, whatever terms are in brackets. So, multiply this parenthesis by 1 and change it.

#### Solved] This Is For Discrete Structure Simplify Each Boolean Expression ,…

Look, you learn how to simplify this by practicing many examples, I leave it to you. You can plug this function into the tool above to get quick answers if you want.

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