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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Simplify (W + Y) (W' + X' + Y) (X' + Y' + Z')

Okay, so I started from ( (W + Y + Z')' (W' + X' + Y)' (X' + Y' + Z')' (W + Y + Z)' )' then simplified further to: ( W'Y'Z + WXY' + XYZ + W'Y'Z' )' then: (W + Y + Z')(W' + X' + Y)(X' + Y' + Z')(W + Y + Z) to: (W + Y)(W' + X' + Y)(X' + Y' + Z') The…
ESM
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Can there be prime implicants or essential prime implicants of SOP form?

I had got a following question let $f(A,B,C,D)=Π(2,3,5,9,11,12,13)$. The total number of prime implicants and essential prime implicants are denoted by P and Q respectively. What is the value Q%P where ′%′ denotes the modulo operator? I solved it…
Mr. Sigma.
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Solving this equation to be 3 literals

I am trying to solve this to be 3 literals, but I keep getting errors: Question:$(x'y' + z)' + z + xy + wz$ My answer: $(x+y)z' + xy + z(1+w)\implies xz' + yz' + xy + z\implies zx + zz' + yz' + xy$ $\implies zx + yz' + xy \implies xyz + xzz' + xy…
Maysa Sarsour
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Boolean Algebra, 4-variable Expression Simplification

I have the following Boolean expression: $$w'x'y'z + wx'y'z + xz + xyz'\tag{1}$$ Upon doing my own work, I can only get as far as: $$zx + xy + zy'\tag{2}$$ Now, when I put the original equation into the following webpage…
skippr
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I need help in simplifying a Boolean expression.

My starting point was (A+D)*(A+B+C)*(~A+C+~D) And I should end at ~A*B*D +A*~D +C*D (according to online solvers.) But when I do it by hand on a paper I end up with this: ~A*B*D +A*~D +C*D +A*C And for the life of me I can't figure out how to…
Andres Eelma
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1+AB=1 using boolean algebra ? Am I right or not?

As 1+AB Now if I put A=0 & B=1 then the above expression gives the answer 1 Conversely if I put A=1 & B=0 then again the answer of above expression is 1 I've seen manly rules or laws to solve boolean algebra problems but I didn't see anyone like…
Diana Joseph
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Using the replacement laws to prove that ($a \to $b) $\vee$ ($a \to $c) = $a \to ($b $\vee$ c)

I have been asked to prove that ($a \to $b) $\vee$ ($a \to $c) = $a \to ($b $\vee$ c). I believe it is just the simple case of using the distributive law: $a \wedge ($b $\vee$ c)= (a $\wedge c) \vee ($a $\wedge$ b). But I am not sure.
Math-Man
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Simplify the following expression using Boolean Algebra into sum-of-products (SOP) expressions

Simplify the following expression using Boolean Algebra into sum-of-products (SOP) expressions $Q.S.U + (Q' + S').(R + V) + U.(R + V) + Q' + S.T.U$ $.$ = AND $+$ = OR This is what I have so far $Q.S.U + (Q' + S').(R + V) + U.(R + V) + Q' +…
Lawrence Wong
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how do I simplify this particular boolean expression?

so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how…
ali farhad
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Which one is the right answer in this boolean expression?

(A+B)' (A'+B')' =(A'B')+(AB)= A'B'+AB (Ans) OR (A+B)' (A'+B')' =(A'+B')(A+B) = A'A+ A'B+ AB'+ BB' = 0+ A'B+ AB'+ 0 = A'B+ AB' (Ans) I get two different answers. I think the procedures I used…
Kurapika
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Simplifying $(A'+ D) \times [(B \times C) + D']$ in Boolean algebra

$(A'+ D) \times [(B \times C) + D']$ I can rewrite as $(A' + D) \times (D' + B) \times (D' + C)$. Is there a way of simplifying to get to the step of $D' + D = 1$ or $D' \times D = 0$ or any other simplifying that I can do? Thank you.
Brian
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Boolean Identity: (x + yx) = (x + y)(x + z) what is z

In the boolean Identity: (x + yx) = (x + y)(x + z) Where does z come from and what does it mean?
Alabaster
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Can anyone Simplify this Boolean expression?

The expression is: [AB {C+(BD)'} + (AB)']CD
Ali
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2 answers

Boolean Algebra simplify problem

I am reading a section on Boolean algebra in a text book and trying to understand a solution to simplifying problem they have presented.the expression $$(¬p ∧ ¬q ∨ p)$$ is simplified to $$(¬q ∧ p).$$ I have not been able to work out what law was…
Stephen
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Boolean Algebra and Karnaugh

I'm starting learning this year, Boolean algebra, truth table and Karnaugh map, on the expression below, and if any one can explain, how to pass it to table and Karnaugh map? And what are the values of 1 I get and…
David E.
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