Boolean algebra Definition and 146 Threads

Hello! I’m an assistant of a mathematical scientific researcher, and my research programme evolves around finding and developing all the (possible) solutions regarding all unsolved mathematical, logic, exact, and IQ puzzles ever created. If you search on the internet for: “The hardest unsolved...

Hey! 😊 I want to prove by using the rules of boolean algebra that the following statement is always true $$\{b\land [\neg a\Rightarrow \neg b]\} \Rightarrow a$$ Since we have to use the rules of boolean algebra, we cannot use the truth table, right? Could you give me a hint how we could...

Hello, can anyone help how i prove this logic rule? I am not sure whether i have to draw a digital circuit or something. if someone could help me solve it showing the steps they took i'd appreciate thanks

I was asked to use De Morgans law to find the complement of a particular equation, I applied the law correctly and simplified my solution down to A'B'CD'+CA+CB'+D'A+D'B' I ran the problem through a boolean simplifier to check my work...

Can anyone tell me if I'm doing this right so far? My question really is... does this mean that the next step for me is to expand the last statement shown?Asking because that seems like a lot of work.

Hey! :o We have the following equalities: \begin{align*}\left (\overline{y}\land z\lor x\land \left (\overline{z}\lor y\right )\right )\land \left (\overline{x}\lor \overline{y}\right )& \overset{(1)}{=}\left (\overline{y}\land z\lor x\land \overline{z}\lor x\land y\right )\land \left...

I have received (unasked) a digital edition of "Laws of Form" (1969) by G. Spencer-Brown; I have glanced at it, and also at the Wikipedia article https://en.wikipedia.org/wiki/Laws_of_Form. OK, another logical system; logical journals (e.g. by ASL) are full of them, and I am not sure whether...

Hello to everyone who's reading this. The problem I need help with is the following.: Homework Statement "Simplify to obtain minimum SOP. F(A, B, C, D) = A’B’CD’+AC’D’+ABC’+AB’C+AB’C+BC’D" The problem stated above has two provided solutions, the "main" one and the "alternate" one. I'm...

Homework Statement i'm viewing an example written in class. it looks like this: f(x1, x2, x3, x4) = [(not x1) * x2 * x4] ∨ [x2 * x3 * x4] what should be function after applying absorption law? Homework Equations i know how another option called "gluing" works: [x1 * x2 * x3] ∨ [(not x1) *...

Homework Statement My solution, is this correct? This is what I came up with. Y=A+((A*B)+B+C'+(B+C'*D)+D) Is it safe to say that it is correct or did I make a mistake?

Homework Statement The problem is given in the picture attached. It is a network of switches.Homework EquationsThe Attempt at a Solution I managed to simplify the expression to this: ## (S + x'(w+y) + xvz)(x'+y)(v+z') ## but I just can't find a way to simplify it to 9 literals. I've tried all...

Homework Statement given the function $F(a,b,c,d)=Σ(0,1,2,4,5,10,12)$ how can i know for sure if 8 is a don't care? and is it possible for 6 and 14 to be don't cares? Homework Equations it's part of a bigger exercise, so the boolean expression i got from previous parts is a'c'+c'd'+b'd' The...

Homework Statement [/B] How to simplife it? I would like to see a process of simplification. Y = A'B'C'D + A'B'CD + A'BC'D + A'BCD' + A'BCD + AB'C'D + AB'CD + ABC'D' The ' denotes a bar over the previous letter.My second question is: Y = BC + AB'D = (uploaded picture) Y = A'D + B'D + A'BC +...

Homework Statement This excresice is supposed to help you understand the basic operations of sets, later used in probability. I am given the following phrases and have to write them in using mathematics. Given three events A, B and C, which belong to sample space S, calculate the following...

For the expression: (B and C) or (not B and not C) or (A and (not B) and not C) I get two different answers depending on the order I do the simplification. 1. If I factor C out of the first and third terms I get (B and C) or (A and C) or (not B and not C) 2. If I factor not B out of the...

Homework Statement This is from a past exam paper for logic design in my course. I have an exam coming up and would love to know how to solve this one. [/B] Develop a circuit realization of the XOR function with three inputs. You may use AND, OR and NOT-gates with not more than two inputs...

Homework Statement Prove that $$(\bar{a} + b)(b+c) + a\bar{b}$$ where ##a,b## can be from the set ##B\in\{0, 1\}## equals $$a+b+c$$ Homework Equations Rules of Boolean Algebra 3. The Attempt at a Solution [/B] My attempt: ##\bar{a}b + \bar{a}c + bb + bc + a\bar{b}## ##b(\bar{a} + 1+c) +...

Homework Statement Express the function Y= (abd + c)' + ((acd)'+(b)')' as the complete disjunctive normal form: 2.1 by applying Boole's theorerm, Homework EquationsThe Attempt at a Solution I separated the equations to two terms (T1,T2) T1= (abd + c)' T2=((acd)'+(b)')' T1= (abd+c)'...

The problem I have been trying to solve a long problem but my answer differs from my books answer with just a few peculiar terms. My answer: ##x_1' \vee x_0'x_1x_2 \vee x_0x_1'x_2' \vee x_2## Book: ##x_1' \vee x_2 ## My question is: Is it possible to simplify ##x_0'x_1x_2 \vee...

Homework Statement Use the definition of exclusive or (XOR), the facts that XOR commutes and associates (if you need this) and all the non-XOR axioms and theorems you know from Boolean algebra to prove this distributive rule: A*(B (XOR) C) = (A*B) (XOR) (A*C) Homework Equations All the...

The problem This is not a complete homework problem. I am at the last step of the solution to a long problem and only interested to know whether these following expressions are equivalent. My answer: ## a \oplus ab \oplus ac ## Answer in my book: ## a \oplus b \oplus c ## The attempt I...

The problem I am trying to show that ##a'c' \vee c'd \vee ab'd ## is equivalent to ## (a \vee c')(b' \vee c')(a' \vee d) ## The attempt ## (a \vee c')(b' \vee c')(a' \vee d) \\ (c' \vee (ab'))(a' \vee d)## The following step is the step I am unsure about. I am distributing the left...

I am asking this question as I have found Boolean Algebra quite intriguing. I have a good understanding of high school level probability and statistics and also Algebra II. Is this enough or do I need more "mathematical maturity"? Anyway, thank you in advance.

Homework Statement (A OR C) AND NOT(C AND A AND B OR C AND A AND NOT B) or (A + C) (CAB + CAB')' Relevant Equations (A+B)' = A'B' A(B+C) = (AB) + (AC) (AB)' = A' + B' The attempt at a solution I'm not sure how I'm suppose to expand (CAB + CAB')' for simplifying. I keep arriving at false which...

I want to prove: $$A\cup B=A\cap B\Longleftrightarrow A=B$$ Forall A,B sets By using the axioms and theorems of the Boolean Algebra. Any hints ??

https://scontent-mxp1-1.xx.fbcdn.net/v/t35.0-12/13324008_10209600482268675_3779883_o.png?oh=abd4c815b9472f70cfbfbb1923a37b3c&oe=5752F454 Number 12 : I have simplified it and I have got AB but the model answer in red say A(comp(B)+c) I just want is that answer true or not , thanks in advance

Homework Statement Simplify the expression ##cb' + ca'b + cabd + cad'## Homework Equations All the properties of boolean algebra. The Attempt at a Solution Here's how I did it: $$cb' + ca'b + cabd + cad' = $$ $$c(b' + a'b + abd + ad') = $$ $$c(b' + a'b + a(bd + d')) = $$ $$c((a + a')b' + a'b...

Homework Statement I'm trying to show that the output of this XOR circuit is ##F=A'B+AB'##, Homework Equations ##(A+B)'=A'\cdot B'## ##(A\cdot B)'=A'+B'## The Attempt at a Solution From the gates the output is ##[(A\cdot B)+(A+B)']'##, using De Morgan's laws this becomes ##[(A\cdot...

This truth table that represents statement p v (q ^ r) is equivalent to (p v q) ^ (p v r)Showing that this statement is not equivalent to (p v q) ^ r.. Now I need to what property of Boolean Algebra is being demonstrated by the fact that the first two statements were equivalentp q r q ^ r...

Most of the results on google happily prove A+(B.C) = (A+B).(A+C), which is that OR is distributive (over AND). But as part of their proof, they use the law that AND is distributive (over OR), namely that A.(B+C) = (A.B)+(A.C) which I can't seem to find any algebraic proof for. So are there...

Homework Statement Am I doing this right? The question is... (ab) + (a' + b') = 1 Homework Equations (a) Commutative a · b = b · a a + b = b + a (b) Associative (a · b) · c = a · (b · c)...

So, I have an equation: ~A * B * C * ~D + ~A * B * C * D + A * ~B * ~C * D + A * ~B * C * D + A * B * ~C * D + A * B * C * ~D where * represents "AND" and + represents "OR", ~ being NOT. Part of the reason I'm having trouble is due to the length of the equation. So far, I've managed to use...

Homework Statement Minimize the following using boolean identities 1. AB'CD+(ABC')'+ABCD' Homework Equations Identity 1A=A 0+A = A Null (or Dominance) Law 0A = 0 1+A = 1 Idempotence Law AA = A...

1. Simplify the expression: F = xyz' + xy'z' + x'yz + xyz2. Postulates and theoremsThe Attempt at a Solution F = xyz' + xy'z' + x'yz + xyz = x(yz' + y'z') + yz(x' + x) (Distributive) = x(yz' + y'z') + yz.1 (Complement) = x(yz' + y'z') + yz (identity) This is where I need...

1. prove that: X'Y'Z + X'YZ' + XY'Z' + XYZ = (X⊕Y)⊕Z Homework Equations Use postulates and theorems. The Attempt at a Solution X'Y'Z + X'YZ' + XY'Z' + XYZ (original expression) X'Y'Z + X'YZ' + X(Y'Z' + YZ) (distributive) X'Y'Z + X'YZ' + X.1 (complement) X'Y'Z + X'YZ' + X (identity)[/B]...

I know De-Morgan's law that $$ -(p∧q) = -p∨-q $$ Also $$ -(p∨q) = -p∧-q $$ But for material implication and bi conditional operations there are also some transformation. What is the law or proof for it? Like $$ p⇒q = -p∨q $$ $$ p ↔q = (p∧q) ∨ (-p∧-q) $$ There may be other properties also that I...

ab'c + a'b + bc' + abc = ac + a'b + bc' (How to further reduce this?) Kmap gives B + AC

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color] (A+B)&(C+D) + (A+B)&(C+D)' + C (A+B)&(C+D) + (A+B)&(C'&D') + C by deMorgans (A+B)&[(C+D)+(C'&D')] + C by Distributive I'm just wondering if I did anything wrong in this simplification or...

Homework Statement In Game 6 of a 6-team double-elimination tournament, Team 1, the top-ranked team, faces the loser of a previous game involving Team 2, the second-ranked team, against one of either Team 3, the third-ranked team or Team 6, the sixth-ranked team. This exercise tests the limits...

C) How did they go from the first red line to the second? f) How did they go from the first green line to the second g) B + B(bar) = 1, so surely the answer should have a +1 ? 2) How did they go from the first purple line to the second? I have a list of the Boolean laws and I have used...

Homework Statement Simplify (A+B')(B+C) The Attempt at a Solution I first expanded it and got = AB + AC + B'B + B'C = AB + AC + B'C Turns out the solution is AB + B'C (according to an online source). How do we get rid of the AC term?

Homework Statement The Attempt at a Solution Is my solution correct or can I simplify it even further?

is x + xz = xz?

Homework Statement I need to obtain the minimum second order circuit of the function: Homework Equations The Attempt at a Solution I know that in order to get the minimum second order circuit, I need to use K-Map using both miniterms and maxiterm and compare them. My problem is to put the...

Homework Statement Homework Equations Boolean Algebra The Attempt at a Solution I use the distri to change the A+C'.D and Demorgan. Should i use dis and Demorgan firstly like below? Which property I should use in the next step? If the first part is wrong,which property i should...

According to wikipedia, absorption is an axiom for a boolean algebra. This seems incorrect to me, since I believe absorption can be proved from the other axioms (distributivity, associativity, commutativity, complement, identity). Thoughts? ## AB' + A = AB' + A*1 = A(B'+1) = A(1) = A ## BiP

Homework Statement Prove the following expression using Boolean algebra: 1. X'Y' + Y'Z + XZ + XY + Z'Y = X'Y' + XZ + YZ' Homework Equations Laws of Boolean algebra The Attempt at a Solution I tried to take Y common but failed. I did the same with X and Z, but the method did...

Homework Statement I'm studying function simplification in boolean algebra, and I didnt understand the following step: (NOT A)(NOT B)(C) + B = (NOT A)(C) + B What happened to the NOT B? Homework Equations The Attempt at a Solution

Hi I am not sure where to post this question but I am trying to simplify this expression: r*c'w+c (As in R AND NOT C AND W OR C) to c+wr (As in C OR W AND R) and I know that it simplifies to this and they are both equivalent; however my question is which boolean simplification property is...

is this a valid statement \overline{A\overline{B}} = \overline{A}B