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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Simplifying this boolean function

How can I completely simplify this equation using algebraic simplification rules? $$x'y'z + x'yz + xyz$$
Brad
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Simplify Boolean Product of Sums Function

I've got a product of sums expression: F=(A'+B+C')&(A+D')(C+D') I need to show it as a sum of products and then simplify it. Right now I got: F=(A'&D')+(A&B&C)+(B&D')+(C&D') But the problem is that the values in the truth table are not the same. I…
Greg Reed
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How can I get a product-of-sums from this sum-of-products?

I have this function: $$f(A,B,C,D) = A'B' + CD' + ABC + A'B'CD' + ABCD'$$ I used a Karnaugh map to minimize the function to: $$Minimum SOP = A'B' + C D' + A B C$$ How can I turn this into a product-of-sums? My attempt was to first find the…
Søren
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Can this Boolean expression be simplified any further?

I have simplified a Boolean expression to $$(\lnot a \land \lnot b \land \lnot c) \lor (a \land (b \lor c)).$$ Is there any way to simplify this further, e.g. using De Morgan's or anything?
J0FVB
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Proving that a given operator is universal.

I've got the next operator: $L(W,X,Y)=(W+Y)X'$ I need to prove whether the operator is universal, And if it isn't is: {$L,1$} or {$L,0$} are universal. I know that what i need to do is either showing that i can create and NAND or NOR table. But i…
StationaryTraveller
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Demultiplexer Equation?

I am currently working on a project that requires a demultiplexer to be used. My problem is that I want to represent it in equation form, but no matter how much I try I cannot find the equation for a demultiplexer! I have tried to come up with a…
user110778
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Doubt in K-map rule (loops)

Is it possible to mark a loop in k-map as in second figure in above diagram (indicated in blue). According to my knowledge the first figure is correct. Is figure 3 or 4 correct? Note: Consider red bordered 4 cells as one loop and blue colored…
Marlon Abeykoon
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Homework: Conjunctive Normal Form

The way I understand CNF is as an expression containing AND's of OR's. So an AND-GATE with 3 inputs (A, B and C) should just be A AND B AND C. But apparently this is incorrect. Any guidance would be greatly appreciated, thank you.
Ogen
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Prove This Bool Expression

Prove $x'z+xyz+xy'z=z$ can you show how you solve this using Boolean Algebra. My main problem is when I do this $xz (y + y') = 1 $ So $1$ times $x =$ ?
MarkNinja
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Proving at boolean algebra

Must prove that $$(X+Y )=X+(X.Y')$$ i tried a lot of ways, using logic things and expanding this things, but cant reach the Y. $$(X+Y )=(X+X).(X+Y')$$ Whats the possible prove to this?
PlayMa256
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How to minimize $\bar{A}.\bar{C}+\bar{A}.B+A.C$ further?

$\bar{A}.\bar{C}+\bar{A}.B+A.\bar{B}.C+B.C$ $=>\bar{A}.\bar{C}+\bar{A}.B+A.\bar{B}.C+\color{Orchid…
user91187
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Interior algebras: an element need not be distinct from its interior?

There is a Wikipedia article about interior algebras. An interior algebra is a Boolean algebra with an additional unary operator, the interior operator, satisfying certain additional axioms. The axioms are dual to the Kuratowski closure axioms. A…
MikeC
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Minimize SOP and POS algebraically?

Is it possible to simplify an SOP (sum of products) or POS (product of sums) expression algebraically? I can only do it through k-maps. Example: $a'b'c'd' + a'b'c'd + a'b'cd' + a'b'cd + ab'c'd + abc'd' + abc'd + abcd' + abcd$
larry21
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Reducing a product-of-sums expression

f = ($x_1$ + $x_3$ + $x_4$) * ($x_1$ + $\overline x_2$ + $x_3$) * ($x_1$ + $\overline x_2$ + $\overline x_3$ + $x_4$) I've been working on this problem for a while but I cannot for the life of me figure out how to simplify the function without…
jeebface
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Simplify the expression below by using the Algebra laws:

Simplify the expression below by using the Algebra laws: $$ AB + \overline{(\bar AC + B)\cdot \overline{(\bar B \oplus C)}} $$
james
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