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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Boolean Algebra - Why does (x'y' + x'y + xy' + xy) = 1

Have the answers to my Design Fundamentals homework but I do not know how they got the answer they did without $(x'y' + x'y + xy' + xy)$ equaling $1$. Thanks
Blake Thomson
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Follow-Up Help with Truth Tables

I've been trying to solve this circuit problem(and understand it frankly), and I wanted to double check my thought process with the community helpfully. After running the circuit out, I have $A+ \bar A B$ for the top half, with the bottom half…
tim belk
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Boolean Simplification of A'B'C'+AB'C'+ABC'

My question is how do I reduce $\bar A\bar B\bar C+A\bar B\bar C+AB\bar C$ To get $(A+\bar B)\bar C$. I'm so lost just been trying to get it for awhile only using the 10 boolean simplification rules.
noname
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2 answers

How to simplify this Boolean expression

F=(A+B+C)(A+B+C')(A+B'+C') I used sop method and I am left with A+BC', so the above expression should leave me with (A+B)(A+C'). Iam not able to get to this answer. Help is appreciated.
turtle
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What to do with a hanging $1$ in a Karnaugh map?

I am learning about Karnaugh maps to simplify boolean algebra expressions. I have this: $$\begin{bmatrix} & bc & b'c & bc' & b'c' \\ a & 0 & 1 & 1 & 0\\ a' & 1 & 1 & 0 & 1 \end{bmatrix}$$ There are no groups of four, so I am now looking for groups…
Saturn
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Simplify Boolean Algebra Expression

The problem is to simplify: $$ xz+\bar{x}y+zy $$ I have an answer key that says the answer is: $$ xz + \bar{x}y $$ I have no idea how they got this expression, though. The first thing I tried was to find the dual. Setting the original expression…
David
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Simplify this boolean algebra?

$$ \begin{align} &\lnot x_1(x_2\land\lnot x_3\lor x_3)\lor x_1(\lnot x_2\land\lnot x_3\lor x_2\land x_3)\\ &=\lnot x_1\land x_2\land\lnot x_3\lor\lnot x_1\land x_3\lor x_1\land\lnot x_2\land\lnot x_3\lor x_1\land x_2\land x_3 \end{align} $$ Is it…
Luka
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Finding the atoms of a Boolean Algebra

I have a homework question that asks me to find the atoms of the Boolean Algebra that contains 256 Boolean functions "such as F1(x,y,z) = x + y +z, F2(x,y,z) = x + xz, F3(x,y,z) = xyz+ xyz and so on". He gives an example in his notes that for B^2,…
Will Nasby
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Boolean Algebra, Simplification: Don't know the method used

Here's the Karnaugh map: The answer I should be getting from the Karnaugh should be: T = R ∙ (CGM)' I'm really not seeing how this was arrived at through any simplification methods I've learned thus far. I can see the answers that are intended…
Hamster
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3 answers

Boolean algebra simplification hw

I'm given the equation $F=(a+b+c)(a'+b')(a+b'+c)$ and it's supposed to simplify into a sum of two product terms, each with two literals. I know the answer is $ab'+a'c$, but I'm unsure how to get there. I know the laws and postulates that I need to…
user104693
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Boolean Algebra, removal of redundant term

How do I simplify this boolean expression A¬B + A¬C + BC¬D + A¬D to A¬B + A¬C + BC¬D with boolean algebra? The A¬D is redundant, I can see why it is when I examine the truth table for this expression, but I can't see how to simplify it.
Ross
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Boolean simplification expression

In words my problem is NOT(p AND q) AND (NOT p OR q) AND (p OR p). I have rewritten it in symbols ¬(p ∧ q) ∧ (¬p ∨ q) ∧ (p ∨ p) I got this far: (¬p ∨ ¬q) ∧(¬p ∨ q) ∧ p Any help please?
user128891
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Is my answer for this truth table & boolean expression correct?

I was given the following boolean diagram: I had to come out with the truth table and the simplified expression. I need help to check if my answers are correct below.
user3296827
  • 21
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Boolean algebra simplification problem

I can't solve this equation: $$(xy + x'yz)(xz + x'y') = xyz$$ After applying distribution I got this: $$xyz + yz + x'z = xyz$$ I can't find the answer and have been thinking for hours now.
user3152069
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Prove something using the Algebraic Foundation of the Boolean Algebra

When asked to prove a specific equation for a boolean algebra by using the "Algebraic Foundation of Algebra Boole" (I don't know how accurate that translation is. In greek I found it as "αλγεβρική θεμελίωση της άλγεβρας Boole") i.e. prove the…
Eternal_Light
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