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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Not every boolean function is constructed from $\wedge$ (and) and $\vee$ (or)

Prove that not every boolean function is equal to a boolean function constructed by only using $\wedge$ and $\vee$. Here is my solution, can I ask for a feed back on my solution please? $p∧q$ $0 0 0 1 $ $p∨q$ $0 1 1 1$ Not…
Kit lai
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Search the OR of negation between boolean algebra

I have this formula $$(a\cdot b)+(\neg a\cdot \neg b)$$ At first I thought this kind of $a+\neg a = 1$ so the answer is 1, but then I realized $(\neg a\cdot \neg b) \neq \neg (a\cdot b)$. I try to do De Morgan for each $(a\cdot b)$ and $(\neg a\cdot…
giripp
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Simplifying Boolean Algebra Expression with 3 variables

Can someone help me simplify this in Boolean algebra? It should be one step at a time so I can understand it. The expression is: $(x+y+z)(x+z)(x'+y+z)$ I tried doing this: (it's probably wrong, because I think it should simplify to just…
xyz
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Discrete Mathematics - are Boolean functions equal?

Are the boolean functions $(p\wedge \neg q)\vee (\neg r\wedge q)$ and $(p\vee \neg q)\wedge (r \vee \neg q)$ equal? Explain your answer. Here my solution, please give me a feedback on this solution, I'm very eager to learn more on my mistakes,…
Kit lai
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Boolean Functions-Algebraic rules for Boolean functions-Associative Rule

Is the function $(p \wedge q) \vee r$ equal to the function $p \wedge (q \vee r)$? Let $a(p,q,r)=(p \wedge q) \vee r$ $b(p,q,r)= p \wedge (q \vee r)$ By associate law $a=b$, but using $a(0,0,1)=1$ and $b(0,0,1)=0$? Any suggestions?
Asnil Deo
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Can I factor out or statements on the other side of an equation in boolean?

I have this boolean equation: X'Y'+XY+X'Y=X'+Y I want to prove it. Now I was wondering if I can rearrange this equation, if I could, so I can factor out the other side; tell me if this is allowed. I haven't seen anything to say I could in my…
munchschair
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Simplify a Boolean Expression

I have to simplify this w′x′y′z + wx'yz' + w'xyz' I keep getting different answers depending on whether I start on the left or the right of the expression Any advice or help would be appreciated. Thanks, Mary
user127378
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Boolean Algebra / Digital Logic

I am trying to figure out how to simply a canonical sum of products expression that is from this expression: $$ f_1(x_1,x_2,x_3) = \sum m (2,3,4,6,7) $$ where m is canonical minterms I got: $$ \bar{x}_{1}x_2\bar{x}_3 +…
Nick
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need help simplifying boolean algebra exrpressions

Can someone walk me through simplifying the following expression? $$a\lnot b\lnot s + ab \lnot s + \lnot abs + abs$$ help and advice is appreciated!
eternalNoob
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boolean algebra question here very short?

We have the Boolean expression Y=A'BC' + ABC'+A'BC Simplify it. Now, this is what I did Y=BC'(A'+A) +A'BC. Now using idempotence Y=BC' + A'BC=B(C'+A'C) I just dont know how to continue this. The result in my book is Y=A'B+BC'
user3210269
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Boolean Law that proves theorem

What Boolean Law proves the following theorem: (a && b) || (b && c) || (a && c) = (a || b) && (b || c) && (a || c) I made a truth table for both of them and they are equal, but I'm not able to prove it. I'd prefer some hints instead of the full…
jviotti
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Must complete atomless Boolean algebras of the same cardinality be isomorphic?

More generally: must complete Boolean algebras of the same cardinality and with the same cardinality of atoms be isomorphic?
Jeremy
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How to simplify this Boolean algebra expression?

I have this question that I need some help with, I just can't get to grips with simplifying. I'm looking at the rules and such but I just can't see where to apply them. Can someone show me the simplification steps to this so I can answer the rest of…
MrTurvey
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Solution to ax+b=c in a Boolean algebra

I have a question. In another forum, a user asked if there is a solution to ax+b=c in a Boolean algebra, where "ax+b=c" is "$(A \wedge X) \vee B = C$". The idea is that, in a Boolean ring, this equation admits the solution x=c/a-b/a; however, I'm…
Nagase
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Atomic Boolean Algebras that are not canonical

I define a Boolean Algebra B to be canonical iff it is isomorphic to the powerset of some set S. (under union, intersection, and complement, of course). Any Boolean Algebra that is both atomic and complete is canonical. Can this theorem be made…
user107952
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