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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Simplify $F=MNO+Q'P'N'+PRM+Q'OMP'+MR$

How can we simplify $$F=MNO+Q'P'N'+PRM+Q'OMP'+MR$$ using the theorems of boolean algebra, not Karnaugh or anything else? Well, I can obviously simplify $MR(P+1)=MR$, so the expression becomes $$MNO+Q'P'N'+MR+Q'OMP'$$ But from here, I tried to use De…
Iulia
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Simplify the expression: $\sim ((p\rightarrow (\sim q \vee r))\wedge (\sim p \wedge \sim q \wedge \sim r ))$

Simplify the expression: $\sim ((p\rightarrow (\sim q \vee r))\wedge (\sim p \wedge \sim q \wedge \sim r ))$ Correct answer: $p\vee q\vee r$ I don't know how to start on this problem. Are there some useful laws to use here or is there something…
Daniel Andersson
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problem with simplifying the boolean function

Can someone explain me why $ AB \vee A \bar B C = AB \vee AC\\$ ? How do we get rid of $ \bar B \\$?
bvl
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How can I simplify my boolean expression further

I have the following boolean expression that I want to simplify $$B\cdot D+ \overline{A\cdot B\cdot D} + \overline{B}\cdot C\cdot \overline{D}$$ Here is what I have been able to due so far $$B\cdot D+ \overline{A} + \overline{B} + \overline{D} +…
J.C
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Understanding the simplification of a boolean expression

$ABC+AB'C+A'BC+A'BC'= AB'C+A'BC+A'BC$ I was reading a book about digital systems and it said that i could supress $ABC$ because it belongs to $BC$ but i can't just look to the expression and have a clear understanding of how this works. I can look…
Wilson Lima
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Complementing a Boolean Function from Shannon's Expansion

Given a Boolean function $f: B^n \longrightarrow B$, using Shannon's expansion across variable $x$, I can write it as $f = xf_x + x^{'}f_{x^{'}}$. I want to complement $f$. Using Shannon's expansion of $f^{'}$ directly, I can write $$f^{'} =…
RayaneCTX
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Can any boolean expression with OR operators be converted to only AND operators?

I'm fairly new to Boolean algebra and I was wondering, using Boolean theorems,can any Boolean expression with an OR operators in it be converted to an equivalent expression using only AND operators? Do some expressions come to a point where you…
sparpo
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How to automatically create proofs in Boolean Algebra

Given a complete set of axioms (for example, associativity, communtative, distributivity, identity, annihilator, idepotence, and the "complementation" laws) for boolean algebra, I know any other true statement follows logically from these axioms. As…
Chris Jefferson
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How do I get Z' + X'Y' + XY from Z' + ZX'Y' + ZXY?

So I'm trying to simplify an equation using boolean algebra. I'm still very confused as to how all the rules work and such, but I feel like I'm very close to the answer. The original equation is (in canonical SOP form): Z'X'Y' + Z'X'Y + Z'XY' + Z'XY…
ESM
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Which properties of boolean algebra are used to prove the DeMorgan Laws?

Given a proof, I tried to analyze it and identify the properties of boolean algebra used at each step. However, I am stumped with the first line: $(a+b) + \sim(a) \cdot \sim(b) = (a + b + \sim(a))(a+b+\sim(b))$. Which property makes those two things…
Jonathan Komar
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Prove the identity in this boolean equation

$$AD'+A'B+C'D+B'C=(A'+B'+C'+D')(A+B+C+D)$$ Don't know where to begin with this.
Aaron
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Simplifying a Boolean algebra expression

~ABC + ~BCD + ~A~BC~D + ABCD = ? The solution book gives this answer: ~AC + CD I arrived this one: ACD My question is if the method used to simplify can affect the result or if should it be equal? Note: ~ stands for not ( ~A = Not A) My steps: ~ABC…
Wilson Lima
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Finding the simplest circuit formula for this boolean logic statement

I have a boolean logic statement: boolean logic statement I get the truth table to derive the statement: truth table derived statement: (-A^-B^C) V (-A^B^-C) V (-A^B^C) V (A^-B^C) V (A^B^C) But when I apply de Morgan's to simplify it, I end up with…
Jakub Greluwski
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Boolean algebra and closure axiom

A Boolean algebra is an algebraic system (B,$∨$,$∧$,$¬$), where $∨$ and $∧$ are binary, and $¬$ is a unary operation. One of the Boolean algebra axiom is: If $a$ and $b$ are elements of $B$, then $(a ∨ b)$ and $(a ∧ b)$ are in $B$. i.e. the set $B =…
V. Gai
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Help with boolean algebra simplification and equivalent

I have the following boolean expression: See number 5 and 7 What I have trouble with are the actual steps of simplification using the boolean algebra laws for number 5 and 7 . I'm probably missing something really really obvious, because I'm trying…
Ahmad Aryanto
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