Boolean algebras are structures which behave similar to a power set with complement, intersection and union. Questions regarding Boolean algebras as structures, or regarding functions defined from/to Boolean algebras fit into this tag very nicely. For Boolean logic use the tag propositional logic

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Simplifying Boolean Algebra law

I've got a problem here that I could use help solving. I have simplified it to this point. Using Wolfram Alpha, I know it is still possible. My lecturer did it but I didn't catch all of it. It is ...
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Boolean algebra generated by value sets of polynomials over $\mathbb{N}$

Update For each polynomial $P \in \mathbb{N}[X]$, let $S_P = \{ P(n) \mid n \in \mathbb{N}\}$. Does the Boolean algebra generated by the subsets $S_P$ of $\mathcal{P}(\mathbb{N})$ such that $P$ is ...
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Is it logically valid to prove DeMorgan's laws using the duality of boolean algebra?

I'm taking an introductory course in boolean algebra, and have been assigned the task of proving DeMorgan's Laws (so, disclaimer, this is homework). One line of reasoning that I came up with would be ...
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353 views

Boolean formula over 64 Boolean variables X

This question comes from this homework assignment from ECS20 at UC Davis. Chess is played on an 8 x 8 board. A knight placed on one square can move to any unoccupied square that is at a distance ...
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Prove the following with equivalence statements.

I need to prove the following statement with equivalence statements. $\exists x \in D,(P(x) \Rightarrow Q(x)) \ \text{is equivalent to} \ (\forall x \in D, P(x)) \Rightarrow (\exists x \in D, ...
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Boolean Algebra A + AB = A

Hi I have a question about the following algebra rule A + AB = A My textbook explains this as follows A + AB = A This rule can be proved as such: Step 1: Dustributive Law: A + AB = A*1 = A(1+B) ...
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4answers
164 views

Proving logical equivalence: $P \Leftrightarrow P \vee (P \wedge Q)$

I'm a first year CS student about to write his first term test and this question is part of our practice package. I have not been successful in writing a sequence of equivalences to justify this ...
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2answers
511 views

Disjunctive normal form expansion

I do not understand this at all. Find the sum-of-products expansions of these Boolean functions. $F(x, y, z) = x + y + z$ $F(x, y, z) = (x + z)y$ $F(x, y, z) = x$ $F(x, y, z) = x y$ ...
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190 views

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 + ...
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Convert boolean expression into SOP and POS

Convert the following expression into SOP (sum of products) and POS (product of sums) canonical forms using boolean algebra method: $(ac + b)(a + b'c) + ac$ Attempt at solution: $(ac + b)(a + b'c) ...
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94 views

Boolean Simplification Problem

I have to simplify the figure I came up with the equation $$(ac+\bar b)\cdot (a+b+\bar c) \cdot (b+ac).$$ Help me out Attempted Answer: \begin{align*} (ac+\bar b)\cdot (a+b+c)\cdot (b+ac) ...
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90 views

Simplify the boolean expression to two literals

Expression: $$[AB'(C+BD) + A'B']C$$ I start off using the distributive law, and then nowhere to go. I need help.
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485 views

Trying to simplify the expression $A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$

So far I've got: $A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$ $= A'C'D'(B' + B) + C'D(A'B + AB')$ $= A'C'D'(1) + C'D(A \;\text{ XOR }\; B)$ $= C'[A'D' + D(A \;\text{ XOR }\; B)]$ Did I do this ...
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Trying to simplify $A'B'C'D + A'B'CD + A'BC'D + AB'CD + ABCD$

My solution so far: $A'(B'C'D + B'CD + BC'D) + A(B'CD + BCD)$ $= A'(C'D(B' + B) + B'CD) + A(CD(B'+B))$ $= A'(C'D(1) + B'CD) + A(CD(1))$ $= A'C'D + A'B'CD + ACD$ $= D(A'C' + AC) + A'B'CD$ $= ...
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Boolean Algebra - Demorgan Laws

I am given the problem: !((!A * B) * !((!B + C) * (!C * !D))) where ! = NOT, * = AND, and + = OR and I tried simplifying it using only Demorgan Laws (no absorption) and I got: (A + !B) + ((!B + ...
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338 views

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 ...
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2answers
220 views

Parity function proofing for every n>=1 using only AND, OR, 0, and 1

Consider the parity function: $F_n$($x_1$, $...$ ,$x_n$) $=$ $\oplus_{i=1}^n$$x_i$ where each $x_i$ is boolean. Prove that, for every $n \ge 1$, there is no way to compute $F_n$ using only ...
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987 views

Functionally complete sets of boolean functions

A (finite) set of boolean functions $S=\{f_1,\ldots,f_n\}$ is called functionally complete if every boolean function (of a finite number of variables) can be presented as a finite composition of ...
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63 views

For a valid chain of deduction in boolean algebra, does the chain of its dual hold as well?

For example, xy + x'z + yz = xy + x'z + yz(x+x') = xy + x'z + yzx + yzx' = xy + xyz + x'z + x'zy = xy(1+z) + x'z(1+y) = xy + x'z Hence xy + x'z + yz = xy + x'z. Also ...
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510 views

Writing a boolean formula and logic circuit that computes mux

Let $mux(p_{11}, p_{10}, p_{01}, p_{00}, x_1, x_0) = P_{x1x0}$ (with all variables bits). Write a boolean formula, and then draw a circuit, that computes mux. For ...
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Simplifying Boolean Function

I am in a computer class with Karnaugh Maps and one of the questions is X 'Y Z + X 'Y 'Z + 'X Y 'Z + X Y Z and I need to simplify it where ' means not so 'x means not x. I think the answer is X 'Y ...
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Is the following sentence a tautology: $(p\Rightarrow q)\vee(r \Rightarrow p)\vee(r\Rightarrow s)\vee(r\Rightarrow q)$?

If both $p$ and $q$ are false then ($p\Rightarrow q$) is true. If either $p$ or $q$ is true then one of ($r\Rightarrow p$) or ($r\Rightarrow q$) is true. If both $p$ and $q$ are true then all are ...
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172 views

Proving a Logic Equation

I have two information. $x+y = 1$ and $xy = 0$. Now,I need to prove this equation : $xz + x'y + yz = y + z$ What I tried: $z(x+y) + x'y = z + x'y$ Thats all What do you think?
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Simplifying Boolean expression: $F(a,b,c,d)$

I have a Boolean function below and I need to simplify it. F(a,b,c,d) = !a&&!b&&d || !a&&c&&!d || !a&&b&&!c || a&&!c&&!d || ...
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318 views

Simplify $(A+C)(AD+AD) + AC + C$ using Boolean algebra

I have solved the equation like this: ...
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144 views

Negation bar meaning?

I know that the horizontal bar on top means it's a negation. But I've never encountered one over more than one term like this one: $\overline{\bar{x} + \bar{y}x}(y + \overline{xy})$ Is that ...
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Commutative ring addition where $a^2 = a$

I'm trying to solve following question: If $a^2=a$ for all $a \in R$ where $R$ is a commutative ring, then $a+a=0$. I have tried to solve this problem for a while now and I'm more or less stuck. I ...
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Boolean logic simplification

To simplify $$ A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AABC'D + ABCD' + AB'C'D' + AB'CD $$ I have no idea how to start the first step. Thanks in advance!!
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boolean logic simplify

To prove: $(X+Y)(X'+Z) = XZ + X'Y$ I try to simply $(X+Y)(X'+Z)$ to $XZ + X'Y + YZ$ then I have no idea how to simply further. Thanks in advance!!
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Simplify the expression below by using the Algebra laws:

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

I have the equation (CD)+(BC'D'+A'B'C'D)+(C'D') and I can't seem to simplify it enough for my homework. I've tried multiple ways of simplifying the equation and I keep ending with an extra variable D. ...
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87 views

How to find/generate a 6 variable Bent Function?

I want to find a Bent Function with 6 variables. I read some papers about how to generate Boolean Functions, but I don't want to implement an algorithm from zero just to find one function. It is also ...
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Three Boolean Algebra Proofs - I just don't get it!

I'm having a very difficult time proving the following 3 expressions: $$\begin{align*} &x\cdot y\cdot z+x'\cdot z=y\cdot z+x'\cdot z\\ &x\cdot y+y\cdot z+x'\cdot z=x\cdot y+x'\cdot z\\ ...
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Discrete: Boolean Function

~(pV~q) v (~p^~q) is equal to ~p? I know the answer is yes and I've been using DeMorgans initially then distributive law after. However I keep messing up on the algebra. Help is appreciated so I can ...
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Simplify Boolean Expression ABC' + A'BC + A'B'C'

Can anyone help me simplify this boolean expression? ABC' + A'BC + A'B'C' EDIT: (only using AND gates (multiply [(x)(x)]) and OR gates (addition [+]))
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Steps to simplify a Boolean Expression

Simplify: (x ∧ y) ∨ (x ∧ ¬y) ∨ (¬x ∧ y) I need to simplify this using the using properties going step by step. I keep ending up with (x ∧ y) as the answer but when I map is out I get that is should ...
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sum-of-products expansions of these Boolean functions

Find the sum-of-products expansions of these Boolean functions. $a)$ $F(x, y) = \text{~}x + y$ $b)$ $F(x, y) = x \text{~}y$ $c)$ $F(x, y) = 1$ $d)$ $F(x, y) = \text{~}y$
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687 views

Conclude the premise using rules of inference

First question I have solved I belive... show, s -> (q -> r) <-> (s ^ q) -> r using the defintion of implication and Boolean algebra. s ->(~q V r) <-> ~ (s ^ q ) V r ~s V (~q V r ) <-> ~s ...
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how many semantically different boolean functions are there for n boolean variables?

In short, this is an assignment question for a course I am taking - the exact wording is this: "Given n Boolean variables, how many 'semantically' different Boolean functions can you construct?" ...
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Expressing boolean functions using the not or operator

I need to express these with $\downarrow$ $x+ y + z$ This one I think I can do, I guess at it and copy the wikipedia page since my book has no explanation on how to do this I get $(x \downarrow y ...
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Given a finite set of points construct a polynomial that meets the points.

Say I have a set of points in $\mathbb{Z}^3 \times \mathbb{Z}_2$ each of which represent part of a mapping $(z_1, z_2, z_3) \mapsto z_4 \in \mathbb{Z}_2$. How do I find the the simplest polynomial ...
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The set of all polynomial functions from $\mathbb{Z}^3 \rightarrow \mathbb{Z}/(2)$

Let $f:\mathbb{Z}^3 \rightarrow \mathbb{Z}_2$ be a polynomial function in $\mathbb{Z}[x_1, x_2, x_3]$. Then $f$ has the form $f(x_1, x_2, x_3) = c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_1 x_2 + c_5 x_1 ...
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Prove that S is a Boolean Algebra

Let $n\ge1\in\Bbb N$, we define the set of binary boolean vectors with $\varphi^n .$ Prove that $\varphi^n$ is a boolean algebra. So (...) I know that: Let $\varphi=\{0,1\}, \mathrm ...
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Is “(p AND q) OR r” logically equivalent to “p AND (q OR r)” ??

In the context of discreet math / boolean algebra / logic, is "(p AND q) OR r" logically equivalent to "p AND (q OR r)"? I believe so, but my professor said: ...
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Proving boolean algebra property

Let $S$ a boolean algebra; $a,b,c \in S$ prove that $(a'+b)'+(a'+b')'=a$ Then: $(a'+b)'+(a'+b')'= (a')'+b'+(a')'+(b')'=a+b'+a+b = (a+a)+(b+b')=a+1=1$ Maybe i'm wrong but I think that the problem is ...
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Simplification of boolean expressions

Question 1 $$\begin{align*}A'BC + AB'C + ABC + A'B'C' &=A'B C + B' C (A +A') + ABC\\ &=A'B + C + B'C + ABC \end{align*}$$ Question 2 $$\begin{align*}A'B + B'C + CB &=A'B + C(B +B')\\ ...
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Boolean-expression simplification $F = AB'C + (A'B' + ABC'D)'$

Here are my solutions. Hence: I am stuck on where or what path I am going to take Problem: F = AB'C + (A'B' + ABC'D)' Solution 1 --------------- F = AB'C + (A'B' + ABC'D)' = AB'C + (A'B')' (ABC'D)' ...
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Proving a boolean algebra question [duplicate]

Let $\sqsubseteq$ be a boolean ordering of the boolean algebra $X$, which means that for each $x$ and $y$ the following applies: $x \sqsubseteq y$ if $x \sqcap y = x$. Let $v, w, a, b \in X$ with $v ...
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101 views

Matrix of integers to boolean matrix

My Question is about converting a matrix of numbers, say each row is an item and each column is a feature of the item. The features are currently integers but I want to convert the feature ...
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276 views

Check if tautology (w/o truth table)

$(A+B)(A+C)(B+C) = AB + AC + BC$ is a tautology (checked with Wolfram Alpha) and not hard to see if you apply duality principle $(invert + * 0 1)$ But how to prove with simplyfication It'S not much ...