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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Question about poset and boolean ordering, inf and sup

We have a poset $(X, \sqsubseteq)$, and we define operations $+$ and $\cdot$ by $x+y=inf(x, y)$ and $x\cdot y=sup(x, y)$ ($+$ can be seen as union in sets and $\cdot$ as intersection in sets). The question is: show that $+$ and $\cdot$ satisfy the…
user95514
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Boolean simplification of $AB'(B' + C)$

Simplifying $AB'(B'+ C)$, then using the distributive property I know I would get $AB'B' + AB'C$ I am just confused as to how to simplify $B'B'$
user2105982
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Boolean Algebra: Explain why (M AND (NOT N)) OR (X AND M AND N) = (M AND NOT N) OR (X AND M)?

I have no idea how this is true, by what theorem, and I literally have been thinking about this for 3 hours now. I know it's really simple, but I just must not be in the right mindset to discover this now. Here is (one of) the exact places where…
user2055216
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Boolean function with characters in the function value, not 0 and 1

I am currently studying the textbook Digital Design by Mano, and learned that the Boolean function can be expressed algebraically from a given truth table by forming a minterm for each combination of the variables that produces a 1 in the function…
user65452
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When does a unary version of a binary operator make sense?

Subtraction ($x - y$) is a binary operator that also has a useful unary version ($-x$), which can be seen as a simple shorthand for the binary version ($-x \Leftrightarrow 0-x$). A unary version of the addition operator ($x+y$) can also be defined…
Michel Schinz
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How to simplify ABC + A'B'C' into A XNOR B AND B XNOR C

I'm currently practicing on Boolean algebra and theorems, recently I found a question that asks to simplify the Boolean expression ABC+A'B'C'. During trying to simplify this when I was trying to construct a…
tritop21
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Simplify boolean expression using consensus

I am supposed to simplify this expression: = ⋅⋅' + ⋅⋅ + ′⋅(⋅) + ⋅⋅′. I have received a hint that states that the consensus theorem can be used. My thinking is that B is the common term and we get B⋅(A⋅C' + A⋅D + A'⋅C + C⋅D') but I get stuck here…
flacko
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Can an expression complement be equal to its dual?

Those are 3 questions I need to answer: a. Can the dual of a Boolean expression be equal to the expression itself? If yes, give an example. b. Can the complement of a Boolean expression be equal to the expression itself? If yes, give an example. c.…
Abdo Ismail
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Define Boolean operations in terms of inclusive denial

Define the binary operation of inclusive denial, denoted by $|$ on a Boolean algebra making $x|y = x' \vee y'$. Show that the binary operations of disjunction $\vee$, conjunction $\wedge$, the unary complement $'$ and the nullary constants $0$ and…
Hugo
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Converting from binary to RC(Radix Complement - Two's complement) and back from RC to binary

Assume we have a negative number X. To convert X from binary to Radix Complement we perform two's complement (Complement on the digits) + 1. Now to convert X from Radix Complement we can perform two's complement again to get back to binary. For…
Noodle
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Boolean Algebra Simplification - In sum of products form

How would you simplify this expression? I've been struggling with it for a while, but seem not to be getting anywhere near the right answer. Y = (A' + BD + C'D)' (B'CD')
Mathsstruggler
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Lindenbaum algebra complete/incomplete

Consider the Lindenbaum algebra for classical logic (defined below). I'm pretty sure that this algebra is not complete, but I can't seem to prove it or even find a proof of it. I can't even seem to find a statement of the incompleteness or…
Philip Kremer
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Simplify the boolean equation using boolean algebra rules

If I have the boolean equation: H = M'CD' + MC + MC' + CRD I think I can combine so that it's H = M'CD' + M(C + C') + CRD Does C + C' go to simplify to zero? So, I'm left with H = M'CD' + CRD
user1766888
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Which K-map to chose?

I got a K-map with the following boolean function: F(A,B,C,D) = ΠM[3,4,6,9,11,14]+ Σm[0,7,8,10,13,15] In the following K-map following prime-implicants are considered: But I can chose ($\bar{A}$+$\bar{D}$) instead of ($\bar{A}$+$B$) like: So now I…
crimsonKnight
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Full adder convert sum of products to XOR

I have this expression (which represents $C_{out}$ of a full adder): $$a \neg b \neg c + \neg a b \neg c + \neg a \neg b c + abc$$ which is generated by some program I wrote that converts a truth table into a list of expressions that represent each…
mochaccino
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