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
1
vote
1 answer

How do I make AB+(A'C')+(B'C) use only nand or not gates?

I know that I have to use the rule where (A+B)' is A'*B' in order to solve it but I don't know how to approach the problem. I tried messing around with the equation and making the complement and then just adding but it doesn't turn out to be the…
Aryan Gupta
  • 11
1
vote
0 answers

Simplify boolean expression using other boolean expressions

There are ways of minimising boolean expressions, such as the Quine-McKluskey algorithm I was wondering if there is a way to minimise expressions that takes into account also other available expressions. For example, let's say I want to simplify (A…
Juan Felipe Coirolo
  • 11
1
vote
1 answer

How to simplify $\overline{A\oplus \overline{AB}}$ correctly?

My attempt: $$\overline{A\oplus \overline{AB}}$$ Let $\overline{AB}=C$. Now, $$\overline{A\oplus…
tryingtobeastoic
  • 3,176
1
vote
1 answer

Monomorphisms of Boolean Algebras

Suppose $f: B->C$ is a homomorphism between Boolean Algebras such that $ker(f)=0$. Is $f$ necessarily a monomorphism? Thank You.
Paul Epstein
  • 355
1
vote
0 answers

Boolean algebra and infinite "bits"

I'm aware of the following elementary result on Boolean algebras. Let $\{p_1, ..., p_n\} \subseteq B$ be a finite subset of $B$. Let $2^n$ be the set of all functions from $\{1,..., n\}$ to $\{0, 1\}$. For any $\sigma \in 2^n$, let $p^\sigma =…
Severine Climacus
  • 391
1
vote
1 answer

boolean algebra: simplify $ a* b *d + \tilde a *\tilde c*d + b* \tilde c* d$

Simplify the following function(algebraically): $$y = a*b*d + \tilde a *\tilde c*d + b *\tilde c *d$$ the solution is: $$a*b*d + \tilde a * \tilde c * d$$ which i checked via karnaugh and also wolfram. my "solution" so far: $b*d*(a + \tilde c) +…
ChrizZz
  • 39
1
vote
1 answer

How to show these two Boolean expressions are the same?

How do I show using the laws of Boolean algebra that: $$ (a \wedge c) \, \vee \, (a \wedge b) \, \vee \, (b \wedge c) \equiv (\bar{a} \wedge b \wedge c) \, \vee \, (a \wedge \bar{b} \wedge c) \, \vee \, (a \wedge b \wedge \bar{c}) \, \vee \, (a…
user348382
  • 447
1
vote
2 answers

Is it possible to convert boolean operations to addition/multiplication modulo $2$?

I mean, is it possible to convert AND OR NOT to simple algebraic expressions? If it's not possible, how do you prove it?
user3698446
  • 27
1
vote
1 answer

How can I prove that A or ((not A) and B) is equivalent to A or B?

I tried AND'ing the leftmost A with itself to keep the expression balanced, but then I wasn't able to put A and (not A) in evidence. I was also trying to apply De Morgan's Theorem, but I cannot AND the leftmost A with (not B).
yellow_submarine
  • 568
1
vote
2 answers

How do I finish the last few steps of this simplification of a boolean expression?

Say I want to expand the following $$\overline{(x' + y)(x + y')},$$ The correct answer is given through for example the following procedure $$ \overline{(x' + y)(x + y')} = \overline{x'+y} + \overline{x + y'} = x''y'+ x'y'' = xy' + x'y.$$ I should…
NoName123
  • 407
1
vote
2 answers

Infinite distributive laws

Let $A$ be a $\sigma$-algebra. A family $\{A_{m,n}\}$ of elements of $A$ satisfies the infinite distributive law in case $$\bigwedge_{m\in\omega}\bigvee_{n\in\omega}…
Ben
  • 57
  • 6
1
vote
1 answer

Binomial expansion through combinations.

If you have $(a+b)(c+d)(e+f)$ how can you expand this? Someone was mentioning that you get different combinations so that you get $adf+ade+acf+ace+bdf+bde+bce+bcf$? Is that the full expansion? As an example, how can it be applied to the boolean…
user1766888
  • 623
  • 3
  • 12
  • 24
1
vote
0 answers

Example of a completion of a Boolean Algebra that is not minimal

Let $A$ be a Boolean Algebra. The pair, $(B, \rho:A \to B)$, is a completion of $A$ if $\rho$ is an injective homomorphism, preserves arbitrary meets and joins, and the complete Boolean Algebra generated by $\rho(A)$ is all of $B$. We call a…
Good Morning Captain
  • 1,733
1
vote
1 answer

$F=ABCD'+AB'CD'+A'BC'D+AB'D'+A'B'CD+A'C'D+AB'CD$

Need to solve by Boolean algebra. I tried but didn't solve it and the answer on K-map is $\;A'C'D+B'CD+ACD'+A'BC'$. The question is $\;ABCD'+AB'CD'+A'BC'D+AB'D'+A'B'CD+A'C'D+AB'CD$.
Omar Mamdouh Zake
  • 13
1
vote
2 answers

Boolean simplification, 5 variables

I'm currently learning for my maths exam, and in the part about boolean algebra I came across an exercise that I can't seem to solve. I probably only need the first few steps to get started. $$ (xyz + uv)(x+\overline{y}+\overline{z}+uv) $$ Usually,…
Simon Verbeke
  • 627
1 2 3
…
64 65