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 .

3083 questions
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Non-principal ideal in Boolean ring

Does anyone know a simple example of a Boolean ring with a non-principal ideal? Every finitely generated ideal in a Boolean ring is principal, hence such an ideal cannot be finitely generated...
Evariste
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Simplifying the boolean expression AB' + B'C + ABC' to B'C + AC'

I'm currently learning about boolean expressions. Currently, we're looking at how to simplify them. For this question I need to show the intermediate steps using the laws of boolean algebra. However, I'm not sure where to start with this one…
qtto
  • 21
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Boolean Simplification using Algebra

Having trouble showing the following relation: $$A\cdot B + A'\cdot B' + B\cdot C = A\cdot B + A'\cdot B' + A'\cdot C $$ using Boolean Algebra. Any help is appreciated.
user385631
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0 answers

Number of canonical expressions

There is a question: What is the number of canonical expressions that can be developed over a 3-valued boolean algebra? I was trying to solve this. Canonical expression is the combination of minterms. In a 3 valued boolean algebra, the number of…
Rebooting
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Simplifying a four term Boolean expression using Boolean algebra

I have the following Boolean expression that I would like to simplify. I know that I could use a Karnaugh map to simplify, but I need to use Boolean Algebra and simplify step by step. $'$ is NOT $+$ is OR $$AB'C+A'BC+ABD+CD'+D'$$ I was able to only…
cdignam
  • 575
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Number of possible functions using minterms that can be formed using n boolean variables.

Consider 3 boolean variables $x, y$ and $z$. Then you can form a total of 8 expressions using each variable or its complements exactly once in each expression i.e. $xyz$, $xyz′$, $xy′z$, $xy′z′$, $x′yz$, $x′yz′$, $x′y′z$, $x′y′z′$ where $x′$…
Aditya Naidu
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Need help with Boolean Algebra simplification.

I need to simplify the following expression: $$ P = AC + A\bar{B} + \bar{A}BC + \bar{A}\bar{B}\bar{C} $$ Using a K-Map I get the correct answer of: $$ P =AC + A\bar{B} + BC + \bar{B}\bar{C} $$ My problem is that I don't understand how $$ \bar{A}BC…
Movers
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If $[0,a] \cong [0,b]$ and $a \wedge b = 0$, does it follow that $[0,a'] \cong [0,b']$?

Let $\mathbf{B} = \langle B, \wedge, \vee, ', 0, 1 \rangle$ be a Boolean Algebra. Then $[0,a] = \{ x \in B : 0 \leq x \leq a \}$, and this can be made into a Boolean Algebra $\mathbf{B}_a = \langle [0,a], \wedge, \vee, ^*, 0, a \rangle$, where $x^*…
amrsa
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Boolean function simplification

I'm having problems with the following expression: (A'+B)'+B(A'+AC)+ABC' And here is what I tried to simplify: AB' + B(A'+AC) + ABC' (De Morgan's) AB' + B(A'+C) + ABC' (Identity) AB' + A'B + BC + ABC' (Distribuitive) Here I convert to Standard SOP…
Joao Ferreira
  • 165
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Simplify semi-boolean expression

I'm trying to simplify the following expression: (A == B) OR ( (A > B) AND (A < C) ) Given that B <= C, this is my bet: B <= A < C (also (B <= A) AND (A < C)) Explanation: (A == B) OR ( (A > B) AND (A < C) ) ((A == B) OR (A > B)) AND ((A ==…
user141170
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1 answer

Stuck on boolean algebra problem

Could someone please explain me why $x.y+x.z+y'.z$ Is equal to $x.y+y'.z$? I just can't simplificate it..
Sinan Samet
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Consensus Theorem and Boolean algebra

I am trying to prove the following boolean equality. $$bc + abc + bcd + a’(d+c) = abc + a’c + a’d$$ I have simplified the left side to $bc + a'd + a'c$ by factoring out a $bc(1)$. However, I am not able to figure out where that extra $a$…
Jonathan
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Simplify $AC'+A'C+BCD'=AC'+A'C+ABD'$

How to prove that $$AC'+A'C+BCD'=AC'+A'C+ABD'$$ approch: a way to demonstrate is expressed in its canonical form. Any hint would be appreciated.
felipeuni
  • 5,080
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Boolean functions that are not too far from all linear functions.

Suppose I have a Boolean function $ f:\mathbb{F}_2^n\rightarrow \mathbb{F}_2 $ which satisfies the following property: $$d(f, \ell)\leq 2^{n-1}\quad\forall\; \text{linear functions}\; \ell .$$ where $d(.,.)$ is the Hamming distance between two…
JQX
  • 159
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boolean algebra simplification to remove extra term

how do i simplify this equation using boolean algebra: AB + ¬AC + BC to be equal to AB + ¬AC the BC is unneeded, but how do i remove that term using boolean algebra?
Ken L
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