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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simplifying following function

F(A, B, C, D) = A'B'C' + AC' + ACD + ACD' + A'B'D' I first did AC(D+D') which got me just AC, and then A(C+C') which left me with just A. The function is now A'B'C'+ A + A'B'D'. I am unsure if I can do things like pulling the A out of A'B'D' to do…
Kytex
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I need help writing the correct boolean function for a circuit

I was given a pre-made circuit. I was tasked with determining the input/output table and also the correct boolean function. The results of the input/output table is: ~x,~y = 0 ~x,y = 0 x,~y = 0 x,y = 1 From the results, I think the boolean…
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Simplifying with boolean algebra.

I have two parts to my question, the first being how do I use Boolean algebra to reduce A'C'+ABC+AC' to have only 3 literals. I sort of understood the concept where you can factor things out but for some reason it won't come to mind with this one.…
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Simplifying Boolean expressions with XOR

$$p=a\land b$$ pb = pb=true ⊕ pb=false $$=(a\land(1\lor c))\oplus(a\land(0\lor c))$$ $$=(a\land1)\oplus(a\land c)$$ $$=a\oplus(a\land c)\tag1$$ $$=a\land\neg c\tag2$$ Can someone point me out how $(1)$ has simplified to $(2)$?
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How would I approach this proof by cases question?

Using proof cases with cases $p$ and $~\neg p$, carry out a deductive sequence proof: $$[(p\to q)\land(r\to q)\land(p\to r)]\to\neg(p\land r\land\neg q)$$ I have been trying for about 30 minutes now, but I do not know how to go about this. I start…
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Inverse law Vs Complementary law

I being confused with A + A̅ = 1 / A . A̅ = 0 law. What is this law? Is this Complementary law or Inverse law or other law?
Bachi Nirosh
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How to Solve The Redundant Literal Rule (OR)?

I found on the internet this rule (A + B = A + A̅ B) and its called Redundant Literal Rule. this is it's OR form. But I want to know is this really form of the Redundant Literal Rule?. I saw in a website there are 2 Types in this Law. What do you…
Bachi Nirosh
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Boolean algebra - dual function and self dual function

I'm having a problem understanding what influences the dual function has on the original function output. For example, if I have a self dual function - for the input 11 the function out put will be 1, and the output of the dual function will be…
Guy
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Boolean Simplification (1.2)

I'm going to attempt to ask a question again. I'm new to this site and I think that this site could be useful to me through out the 4 years in University for me. Can you tell me whether I'm going the right direction with this simplification I came…
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Is my truth table correct?

My table on this link I am not really sure about the symbols. does the over line mean negation?does the asterisk mean conjunction?if there's mistake, i'd be grateful if someone could upload the corrected table. thanks
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boolean algebra with finite elements

I need to define a boolean algebra with 8 elements. I know all the Axioms to define a binary boolean algebra but I don't know how to do that with 8 elements. Someone can guide me please? Thanks.
sami
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Boolean algebra; what does <-> mean?

Expression :$$(p\rightarrow q)\leftrightarrow(\neg q\rightarrow \neg p)$$ What does the symbol $\leftrightarrow$ mean ? Please explain by drawing the truth table for this expression and also with other examples if possible. I'm in a desperate…
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Boolean Algebra Problem

$ab+(ac)'+ab'c(ab+c) = 1$? how ?
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Simpifying boolean algebra

I need some help simplifying a boolean algebra expression. $(~a*b*c*~d) + (a*~b*~c*~d) + (a*~b*c*~d) +(a*b*~c*d) + (a*b*c*~d)$ I have managed to simplify to $(~c*~d)*(~a+~b)+(a*b)*(~c*~d)+(a*~b*c*~d)$ but after this step, I manage to get a couple of…
E-may
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trying to simplify equation to fullest

I've got an equation down to $F=BA+BC +\bar{A}\bar{C}$ and according to Wolfram Alpha it can simplified to $(\bar{A} + \bar{C}) + B$. What's the next steps? I tried using de Morgan's law and not sure it helped.
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