Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [boolean-algebra]

Boolean algebras are structures which behave similar to a power set with complement, intersection and union. Use this tag for questions about Boolean algebras as structures, or about functions defined from/to Boolean algebras. For Boolean logic use the tag propositional-calculus.

Boolean algebras are structures which behave similar to a power set with complement, intersection and union. Use this tag for questions about Boolean algebras as structures, or about functions defined from/to Boolean algebras.

A Boolean algebra uses Boolean variables, typically denoted by capital letters, e.g. $A,B$, which can only take the values $0$ or $1$. Operators are $\land$ (conjunction), $\lor$ (disjunction) and $\lnot$ (negation).

For Boolean logic use the tag .

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Help with boolean algebra simplification.

I had the following boolean expression: $$(C\lor D)\land({\sim} B\lor D)\land({\sim} A\lor {\sim} C\lor {\sim} D)$$ I know this can be simplified to: $$(C\land{\sim}B\land{\sim} D)\lor(D\land{\sim} A)\lor(D\land{\sim} C)$$ I have gotten this…
TheAmazingChuck
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Boolean simplication

I'm trying to simplify the following booleans: $$Y=[\overline{ \overline{(A+B)} \quad \overline{(C+D)}}]$$ My solution is: $$Y=[\overline{ \overline{(A+B)} \quad\overline{(C+D)}}]$$ $$ = [\overline{ \overline{(A+B)}} \quad \overline{…
Quixotic
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Proof that $a + 1 = 1$ (Boolean algebra)

The proof works like this: $$a + 1 = (a + 1) * 1 = (a + 1) * (a + \lnot a) = a + (1 * \lnot a) = a + \lnot a = 1$$ It only uses the axioms of Boolean algebra. I understand every step besides the third one. He uses the distributive law here but I…
Julian
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Simplify ABC+AB'C+A'BC+A'B'C+A'B'C'

ABC+AB'C+A'BC+A'B'C+A'B'C', AC + A'BC + A'B'C + A'B'C', C(A+A'B+A'B') + A'B'C', C(A + A'(B+B') + A'B'C', C ( A + A') + A'B'C', C + A'B'C', but true answer is C+A'B. Help me, what I missed?
MurelloS
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How can I express xNORy solely with NAND operations?

I've tried every which way I can think of to manipulate the algebra using the various laws I was given, but I cannot figure out a way to get $\overline{x+y}$ to convert to only NAND operations using $\overline{x\cdot x}$, $\overline{x\cdot y}$, or…
Mirrana
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Is this absorption law if not by what rule is it simplified?

In https://www.youtube.com/watch?v=0as464WmfCo it is mentioned that the following is simplified via the "absorption" law: $$A + \widehat{A}B = A + B.$$ However https://www.youtube.com/watch?v=tWU1mGCT_34 at around 4:40 it states this is not an…
awgtek
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Simplifying a function via K-Map

So I have been trying to figure this out for about an hour but I am obviously not understanding what I have been reading and watching so I thought I'd ask my own question, I would appreciate a very layman explanation of the following. I am trying to…
AFC
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Equivalence Relation of $(x\land y)\lor ( \lnot x \land \lnot y)=1$

Define $ R \subseteq T xT $ as follows: $ (x,y) \in R \iff (x\land y)\lor ( \lnot x\land \lnot y)=1$. Show, using the laws of Boolean Algebra, that R is an equivalence relation. Hint: if $ A = B = 1 \;then\; A \land B \land C = A \land B \implies…
DSt_FTW
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Finding the Complement of a Boolean Expression

The expression I was given was $$F=(a+c)(a+b')(a+b+c')$$ My first thought was to use DeMorgans so I expanded the two left parenthesis to the following: $$(a+ac+b'c)(a+b+c')$$ Afterwards I applied DeMorgans to get the following: $$…
AFC
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Simplifying the following expression algebraically

This is for a class on logic gates and boolean expressions, the expression I was given is the following: $$x'y'z'+w'x'yz'+wx'yz'$$ I have done one other question like this so far and I struggled with it, so far I have only been able to do the…
AFC
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Boolean Algebra Distributive Property over More Than Two Literals?

I am currently studying Boolean Algebra. While solving my practice questions, I came to wonder if I am properly using distributive law. So, distributive law states that a. $·$ is distributive over $+$ : $x ( y + z ) = xy + xz$ b. $+$ is distributive…
WhaSukGO
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DNF to CNF conversion

So i tried to convert this (ABCD+AB¬C¬D+¬A¬CD+¬AC¬D+¬B¬CD+¬BC¬D) into CNF form but im stuck at (¬A+¬B+¬C+D).(¬A+¬B+C+¬D).(A+¬B+C+D).(A+¬B+¬C+¬D).(A+C+D).(A+¬C+¬D).(B+C+D).(B+¬C+¬D) and the right answer should be…
KusKus
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Simplifying the expression using Boolean Algebra

Simplifying the expression using Boolean Algebra into sum-of-products (SOP) expressions . refers to AND + refers to OR (y' + x) ∙ (z + z') ∙ (y' + x') + (z + x'∙z) ∙ (x + y) This is what I have so far. (y' + x) ∙ (z + z') ∙ (y' + x') + (z + x'∙z) ∙…
Lawrence Wong
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What is wrong with this truth table?

I started by making a truth table for A and B, with $\begin{array}{c:c|c}A&B&A \wedge B\\\hline T&T& T\\ T&F& F\\ F&T& F\\ F&F& F\end{array}$ To see what would happen, I made a table for ~A and ~B: $\begin{array}{c:c|c}\neg A&\neg B&\neg A…
GaneGoe
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Boolean Algebra Missing Step

$A$$\bar{B}$ + $\bar{B}$$C$ + $\bar{A}$$C$ is the equation I am stuck on. It says it can be simplified down to $A$$\bar{B}$ + $\bar{A}$$C$. I can't find any rules that can match this. The steps(and the name of the steps eg. $1$+$A$ = $1$) to solve…
John Smith
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