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## DEV Community

Alex

Posted on • Updated on

# Learn \LaTeX

Here I'll share a bit of my experience about how to learn \LaTeX from zero, the best editor for \LaTeX (IMO) as well as typical mistakes.

## "Who Is LaTeX" Stage

### Typing

It is a requirement for any programming activity. Open any trainer for blind printing with ten fingers — and go ahead!

### How To Learn

• I recommend this free online introduction to \LaTeX in three parts, made by Overleaf. Instead of a dry book guide, they made it as a presentation. There are a lot of examples, with a focus only on the most important stuff. This introduction should take about 4-hours for you to read.

Also, Overleaf has great docs. Later, when you have some questions about \LaTeX, you can look it up in their docs.

• Videos are usually the easiest way to learn, and for a quick start in \LaTeX take a free course from the HSE (in Russian).

The basics could be learned in 4-5 hours, and two days off are enough for the entire course. Video lessons are better than thick books in the beginning.

### Where To TeX

There are two ways.

1. Online compiler like overleaf.
• Very simple.
• Real-time online teamwork.
2. Set up your environment.
• So much faster compiling!
• Takes time.
• Easier to integrate Git.

#### GUI / Text Editor for LaTeX

The best option, in my opinion, is LaTeX in Vim. There are gold articles in which Gilles Castel shows his Vim + LaTeX setup and how it works.

Otherwise — you can try TexStudio, but it looks like an ancient artifact.
Also, there is the texpad app for macOS, but I had some issues with buggy behavior (in 2020).

PS: The Vim + LaTeX setup can easily take 20+ hours to make it work if you are new.

## Life Hacks

1. Use \newcommand{command_name}{script_of_command} for commands you use very often.

Compare f: \mathbb{R} \to \mathbb{R} and f: \R\to\R.

$f: \mathbb{R} \to \mathbb{R}$
2. Learn how to TeX in Vim =)

## Typical Mistakes

1. Please, God, don't do 500+ lines main.tex...

• It will be an unmaintainable mess.
• You don’t have a universal preamble.

Instead -- use input{anotherFile.tex} or include{anotherFile.tex} to structure code.

2. Not using or defining LaTeX environments.
For every theorem, proof, note, etc. you can set special environment (\begin{theorem}, \begin{proof}).

• Now you can change the style of your theorems by changing one line.
3. Not using multiline or align environments for long formulas.

\begin{align*} a &= b + c + d - d \\ &= b + c \end{align*}
\begin{align*}
a &=  b + c + d - d \\
&= b + c
\end{align*}

4. Not using \cfrac or \displaystyle when necessary.

• Otherwise, the formula may get hard to read.

$\frac{2 + \frac{8}{4}}{6}$ -- without \displaystyle

$\cfrac{2 + \frac{8}{4}}{6}$ -- with \displaystyle

5. Not using \text inside formulas.

\begin{align*} a &= a + b + c + d , when\; sun\; is\; red \\ a &= b + c, \text{ when sun is white} \end{align*}
\begin{align*}
a &= a + b + c + d , when\; sun\; is\; red  \\
a &= b + c, \text{ when sun is white}
\end{align*}

6. All variables inside text should be math mode.

-a or $-a$

7. Avoid making new lines in text with \\.

• Using empty lines instead makes code much more readable.
8. Not using \ldots for ...

... or $\ldots$

9. Ignoring warnings and errors.

• You must not ignore erros.
• Warnings help you deepen your knowledge and improve code.
10. Not using Git for big projects.
Best of luck...

11. Using tutorials from ancient times.
You can meet outdated and deprecated practices.


text:
Expand description

Application-local prelude: conveniently import types/functions/macros which are generally useful and should be available everywhere.

## Re-exports

• pub use crate::application::app_config;
• pub use crate::application::app_reader;
• pub use crate::application::app_writer;

## Traits

• Commonly used Abscissa traits Application types implementing this trait own global application state, including configuration and arbitrary other values stored within application components.
• Commonly used Abscissa traits Subcommand of an application: derives or otherwise implements the Options trait, but also has a call() method which can be used to invoke the given (sub)command.
• Commonly used Abscissa traits Runnable is a common trait for things which can be run without any arguments.

format: plain


text:
# Linear Charge Density

1. Oct 4, 2015

### kahlua13

1. The problem statement, all variables and given/known data

This is a wire whose shape is given by y = acos(x/L). This wire has a linear charge density of +λ, and is it desired to determine the electric field at the point (0,y) where y > a.

a) If a=0, determine the amount of charge the wire has.

b)If a > 0, is the total charge on this wire greater than, less than, or equal to the charge when a = 0. Why?

c) If a > 0, is there any statement you can make about the electric field at the POI? Why or why not?

d) a > 0. Locate a point on the wire. What are its coordinates?

e) If I want to move along the wire a small distance, ds,away from this point. Give the x and y-components of ds.

f) Construct, but do not evaluate, the expression for the electric field at the POI. Give enough detail to explain your reasoning.

2. Relevant equations
linear charge density = charge/length

3. The attempt at a solution

I am able to answer a-d.
For part A, length * linear charge density = charge
For part B, because the wire would be longer if a > 0, the total charge would be greater than if a = 0.
For part C, I believe the answer is that the x-components of the E-field will cancel, allowing us to use only the y-component to calculate the electric field.

However I really don't know where to start with part E! My chosen point is (0,a)

2. Oct 5, 2015

### TSny

Hello. Welcome to PF.

I think that for (e) you are supposed to consider $ds$ as a little displacement that has x and y components. The length $ds$ can then be constructed from these components. You should express the x and y components of $ds$ in a way that will be helpful for part (f).


text:
The goal of sourmashconsumr is to parse, analyze, and visualize the outputs of the sourmash python package. The sourmashconsumr package is still under active development.

## Installation

You can install the development version of sourmashconsumr from GitHub with:

# install.packages("remotes")
remotes::install_github("Arcadia-Science/sourmashconsumr")

Eventually, we hope to release sourmashconsumr on CRAN and to provide a conda-forge package. We’ll update these instructions once we’ve done that.

## Usage

See the vignette for full instructions on how to run the sourmashconsumr package (coming soon!).

To access the functions in the sourmashconsumr package, you can load it with:

library(sourmashconsumr)

The sourmashconsumr package contains a variety of functions to work with the outputs of the sourmash python package. The table below summarizes which sourmash outputs the sourmashconsumr package operates on and the functions that are available. For a complete list of functions in the sourmashconsumr package, see the documentation.

## Developer documentation

The sourmashconsumr package follows package developer conventions laid out in https://r-pkgs.org/, and changes can be contributed to the code base using pull requests. For more information on how to contribute, see the developer documentation.


text:
## the end of Series B!

Posted in Statistics, University life, Books, pictures with tags , , , , on May 25, 2016 by xi'an

I received this news from the RSS today that all the RSS journals are turning 100% electronic. No paper version any longer! I deeply regret this move on which, as an RSS member, I would have appreciated to be consulted as I find much easier to browse through the current issue when it arrives in my mailbox, rather than being t best reminded by an email that I will most likely ignore and erase. And as I consider the production of the journals the prime goal of the Royal Statistical Society. And as I read that only 25% of the members had opted so far for the electronic format, which does not sound to me like a majority. In addition, moving to electronic-only journals does not bring the perks one would expect from electronic journals:

• no bonuses like supplementary material, code, open or edited comments
• no reduction in the subscription rate of the journals and penalty fees if one still wants a paper version, which amounts to a massive increase in the subscription price
• no disengagement from the commercial publisher, whose role become even less relevant
• no access to the issues of the years one has paid for, once one stops subscribing.

“The benefits of electronic publishing include: faster publishing speeds; increased content; instant access from a range of electronic devices; additional functionality; and of course, environmental sustainability.”

The move is sold with typical marketing noise. But I do not buy it: publishing speeds will remain the same as driven by the reviewing part, I do not see where the contents are increased, and I cannot seriously read a journal article from my phone, so this range of electronic devices remains a gadget. Not happy!

## RSS statistical analytics challenge 2014

Posted in Kids, R, Statistics, University life, Wines with tags , , , , on May 2, 2014 by xi'an

Great news! The RSS is setting a data analysis challenge this year, sponsored by the Young Statisticians Section and Research Section of the Royal Statistical Society: Details are available on the wordpress website of the Challenge. Registration is open and the Challenge goes live on Tuesday 6 May 2014 for an exciting 6 weeks competition. (A wee bit of an unfortunate timing for those of us considering submitting a paper to NIPS!) Truly terrific, I have been looking for this kind of event to happen for many years (without finding the momentum to set it rolling…)  and hope it will generate a lot of exciting activity and replicas in other societies.

## chain event graphs [RSS Midlands seminar]

Posted in pictures, Statistics, University life with tags , , , , , , , , , , on October 16, 2013 by xi'an

Last evening, I attended the RSS Midlands seminar here in Warwick. The theme was chain event graphs (CEG), As I knew nothing about them, it was worth my time listening to both speakers and discussing with Jim Smith afterwards. CEGs are extensions of Bayes nets with originally many more nodes since they start with the probability tree involving all modalities of all variables. Intensive Bayesian model comparison is then used to reduce the number of nodes by merging modalities having the same children or removing variables with no impact on the variable of interest. So this is not exactly a new Bayes net based on modality dummies as nodes (my original question). This is quite interesting, esp. in the first talk illustration of using missing value indicators as a supplementary variable (to determine whether or not data is missing at random). I also wonder how much of a connection there is with variable length Markov chains (either as a model or as a way to prune the tree). A last vague idea is a potential connection with lumpable Markov chains, a concept I learned from Kemeny & Snell (1960): a finite Markov chain is lumpable if by merging two or more of its states it remains a Markov chain. I do not know if this has ever been studied from a statistical point of view, i.e. testing for lumpability, but this sounds related to the idea of merging modalities of some variables in the probability tree…

Posted in Books, pictures, Running, Statistics, Travel, University life with tags , , , , , , , on September 5, 2013 by xi'an

Although I could not stay at the RSS Annual Conference for the three days, I would have liked to do so, as there were several interesting sessions, from MCMC talks by Axel Finke, Din-Houn Lau, Anthony Lee and Michael Betancourt, to the session on Anti-fragility, the concept produced by Nassim Taleb in his latest book (reviewed before completion by Larry Wasserman). I find it rather surprising that the RSS is dedicating a whole session to this, but the usually anti-statistic stance of Taleb (esp. in The Black Swan) may explain for it (and the equally surprising debate between a “pro-Taleb” and a “pro-Silver”. I will also miss Sharon McGrayne‘s talk on the Bayesian revolution, but look forward to hear it at the Bayes-250 day in Duke next December. And I could have certainly benefited from the training session about building a package in R. It seemed, however, that one-day attendance was a choice made by many participants to the conference, judging from the ability to register for one or two days and from the (biased) sample of my friends.

Incidentally, the conference gave me the opportunity to discover Newcastle and Tynemouth, enjoying the architecture of Grey Street and running on the huge meadows almost at the city centre, among herds of cows in the morning fog. (I wish I had had more time to reach the neighbourly Hadrian wall and Durham, that I only spotted from the train to B’ham!)

Posted in Books, Statistics, Travel, University life with tags , , , , , , , on September 3, 2013 by xi'an

Today, I attended the RSS Annual Conference in Newcastle-upon-Tyne. For one thing, I ran a Memorial session in memory of George Casella, with my (and his) friends Jim Hobert and Elias Moreno as speakers. (The session was well-attended if not overwhelmingly so.) For another thing, the RSS decided to have the DIC Read Paper by David Spiegelhalter, Nicky Best, Brad Carlin and Angelika van der Linde Bayesian measures of model complexity and fit re-Read, and I was asked to re-discuss the 2002 paper. Here are the slides of my discussion, borrowing from the 2006 Bayesian Analysis paper with Gilles Celeux, Florence Forbes, and Mike Titterington where we examined eight different versions of DIC for mixture models. (I refrained from using the title “snow white and the seven DICs” for a slide…) I also borrowed from our recent discussion of Murray Aitkin’s (2009) book. The other discussant was Elias Moreno, who focussed on consistency issues. (More on this and David Spiegelhalter’s defence in a few posts!) This was the first time I was giving a talk on a basketball court (I once gave an exam there!)

## Dennis Lindley and Tony O’Hagan on YouTube

Posted in Statistics, University life with tags , , , on August 20, 2013 by xi'an

The great discussion Tony O’Hagan had with Dennis Lindley last March for the Bayes 250 meeting at the RSS is now available on line.

Since this is still close to Dennis’s birthday, I take the opportunity to wish him the best for his 90th birthday.

## a general framework for updating belief functions [reply from the authors]

Posted in Statistics, University life with tags , , , , , , on July 16, 2013 by xi'an

Thanks to Christian for the comments and feedback on our paper “A General Framework for Updating Belief Distributions“. We agree with Christian that starting with a summary statistic, or statistics, is an anchor for inference or learning, providing direction and guidance for models, avoiding the alternative vague notion of attempting to model a complete data set. The latter idea has dominated the Bayesian methodology for decades, but with the advent of large and complex data sets, this is becoming increasingly challenging, if not impossible.

However, in order to do work with statistics of interest, we need to find a framework in which this direct approach can be supported by a learning strategy when the formal use of Bayes theorem is not applicable. We achieve this in the paper for a general class of loss functions, which connect observations with a target of interest. A point raised by Christian is how arbitrary these loss functions are. We do not see this at all; for if a target has been properly identified then the most primitive construct between observations informing about a target and the target would come in the form of a loss function. One should always be able to assess the loss of ascertaining a value of $\theta$ as an action and providing the loss in the presence of observation x. The question to be discussed is whether loss functions are objective, as in the case of the median loss,

$l(\theta,x)=|\theta-x|$

or subjective, in the case of the choice between loss functions for estimating a location of a distribution; mean, median or mode? But our work is situated in the former position.

Previous work on loss functions, mostly in the classical literature, has spent a lot of space working out what are optimal loss functions for targets of interest. We are not really dealing with novel targets and so we can draw on the classic literature here. The work can be thought of as the Bayesian version of the M-estimator and associated ideas. In this respect we are dealing with two loss functions for updating belief distributions, one for the data, which we have just discussed, and one for the prior information, which, due to coherence principles, must be the Kullback-Leibler divergence. This raises the thorny issue of how to calibrate the two loss functions. We discuss this in the paper.

To then deal with the statistic problem, mentioned at the start of this discussion, we have found a nice way to proceed by using the loss function $l(\theta,x)=-\log f(x|\theta)$. How this loss function, combined with the use of the exponential family, can be used to estimate functionals of the type

$I=\int g(x)\,f_0(x)\, dx$

is provided in the Walker talk at Bayes 250 in London, titled “The Misspecified Bayesian”, since the “model” $f(x|\theta)$ is designed to be misspecified, a tool to estimate and learn about I only. The basic idea is to evaluate I by ensuring that we learn about the $\theta_0$ for which

$I=\int g(x)\,f(x|\theta_0)\, dx.$

This is the story of the background, we would now like to pick up in more detail on three important points that you raise in your post:

1. The arbitrariness in selecting the loss function.
2. The relative weighting of loss-to-data vs. loss-to-prior.
3. The selection of the loss in the M-free case.

In the absence of complete knowledge of the data generating mechanism, i.e. outside of M-closed,

1. We believe the statistician should weigh up the relative arbitrariness in selecting a loss function targeting the statistic of interest versus the arbitrariness of selecting a misspecified model, known not to be true, for the complete data generating mechanism. There is a wealth of literature on how to select optimal loss functions that target specific statistics, e.g. Hüber (2009) provides a comprehensive overview of how this should be done. As far as we are aware, we know of no formal procedures (that do not rely on loss functions) to select a false sampling distribution $f(x|\theta)$ for the whole of x; see Key, Pericchi and Smith (1999).
2. The relative weighting of loss-to-data vs. loss-to-prior. This is an interesting open problem. Our framework shows in the absence of M-closed or the use of self-information loss that the analyst must select this weighting. In our paper we suggest some default procedures. We have nowhere claimed these were “correct”. You raise concerns regards parameterisation and we agree with you that care is needed, but many of these issues equally hold for existing “Objective” or “Default” Bayes procedures, such as unit-information priors.
3. The selection of the loss in M-free. You say “….there is no optimal choice for the substitute to the loss function…”. We disagree. Our approach is to select an established loss function that directly targets the statistic of interest, and elicit prior beliefs directly on the unknown value of this statistic. There is no notion here of a a pseudo-likelihood or where this converges to.

Thank you again to Christian for his critical observations!


text:
I request someone to clear the following doubt of mine:

Consider a closed room of volume 100 litres at STP. 1 mole of a gas (say oxygen) is introduced inside the room. I can conclude the following by my knowledge:

1) Avogadro's Law states that 1 mole of any gas at STP will occupy 22.4 litres. Hence, V[gas] = 22.4 litres.

2) Due to diffusion, the gas will move from a region of high concentration to a region of low concentration. Hence, the gas will occupy the volume of the entire room .Therefore, V[gas] = 100 litres.

Since, 22.4 litres $$\neq$$ 100 litres, something mus be wrong with my line of reasoning. Could anyone kindly point out the fallacy?

Note by Nilabha Saha
2 years, 3 months ago

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Yes, you see Avogadro stated that 1 mole of any gas occupies 22.4 L at STP. When you diffuse it out, the conditions no longer will be that of STP , as temperature will increase as volume has increased. Incidentally, V is proportional to T, so accordingly Volume becomes 100 L and there is no STP. So, Avogadro's Law doesnot work here directly. So, that's the fallcy

- 2 years, 3 months ago

Thank you very much for providing me with this answer. You have relieved me of a painful sensation which I just couldn't get over. Perhaps it stands proof to the fact that laws of chemistry are amicable (in this case, the Ideal Gas Law with Molar Volume Of Gases).

- 2 years, 3 months ago

You are welcome.

- 2 years, 3 months ago

Suppose that the closed room is a basketball (balloon with rigid-ish walls). What happens as we pump air into it?.

Staff - 2 years, 3 months ago

I thank you a lot for guiding me to a brilliant exploration here on Brilliant.

- 2 years, 3 months ago

The first thing is you have applied ( pv=mRT) to find the volume that will be occupied by a certain amount of oxygen which is (1 mole ) at STP conditions ، you have obtained that it will be 22.4 liters ، then you have put that amount of oxygen in a bigger room (100 liters volume) so you should apply another law which is (P1V1=P2V2), The value of P2 is what you will have as you kept the T is constant ........ let's approximate that you have 1 bar at STP so the P2 will be decreased around 0.2 bar in your 100 litter room which means that you have create vacuum state ..

Thank you ...

- 2 years, 3 months ago


text:
# [XeTeX] Fontspec Bug

Heba Soliman Heba.soliman at hindawi.com
Wed Nov 19 08:21:49 CET 2008

Dear Mr Will:

First of all thank you for your hard efforts in improving the fontspec
package.

Second I have a big problem that is related to my work so I need please

The problem is that I am writing an arabic article and the diacritics in
some places are collapsed with the letters and I don't know why this
problem generated and how to fix it, the only thing I am sure of is that
the font has no problem as I checked with other programs as Adobe
illustrator and this problem doesn't exist:

This is my sample file:

\documentclass{amsart}

\usepackage{fontspec}
\usepackage{xltxtra}
\usepackage{xunicode}
\usepackage{bidi}

\setmainfont[Script=Arabic]{YakoutLinotypeLight-Regularxx.ttf}

\begin{document}

لمّا ألّا لمًّا آلّلًة لمُا

\end{document}

Please find attached the generated PDF file.

Best regards,

Heba
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text:
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For every integer k from 1 to 10 inclusive the kth terem of

Author Message
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Joined: 17 Sep 2011
Posts: 209
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Kudos [?]: 55 [2] , given: 8

For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  31 Jan 2012, 16:58
2
KUDOS
19
This post was
BOOKMARKED
For every integer k from 1 to 10 inclusive the kth term of a certain sequence is given by (-1)^(k+1)*(1/2^k). If T is the sum of the first 10 terms in the sequence then T is
1) greater than 2
2)between 1 & 2
3) between 0.5 and 1
4)between 0.25 and 0.5
5)less than 0.25

Could someone please provide a solution to this problem ?
_________________

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Kudos [?]: 8371 [28] , given: 192

Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  01 Feb 2012, 22:11
28
KUDOS
Expert's post
12
This post was
BOOKMARKED
gpkk wrote:
For every integer k from 1 to 10 inclusive the kth term of a certain sequence is given by (-1)^(k+1)*(1/2^k). If T is the sum of the first 10 terms in the sequence then T is
1) greater than 2
2)between 1 & 2
3) between 0.5 and 1
4)between 0.25 and 0.5
5)less than 0.25

Could someone please provide a solution to this problem ?

To get a hang of what the question is asking, put values for k right away. Say, k = 1, k = 2 etc
You get terms such as (1/2) when k = 1, (-1/4) when k = 2 etc

T = 1/2 - 1/4 + 1/8 - 1/16 +.... + 1/512 - 1/1024 (Sum of first 10 terms)

Of course GMAT doesn't expect us to calculate but figure out the answer using some shrewdness.

We have 10 terms. If we couple them up, two terms each, we get 5 groups:
T = (1/2 - 1/4) + (1/8 - 1/16) ...+ (1/512 - 1/1024)

Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. e.g. 1/2 is larger than 1/4 so 1/2 - 1/4 = 1/4 i.e. a positive number
1/8 - 1/16 = 1/16, again a positive number.

We will get something similar to this: T = 1/4 + 1/16 +.... (all positives)
Definitely this sum, T, is greater than 1/4 i.e. 0.25

Now, let's group them in another way.

T = 1/2 + (- 1/4 + 1/8) + (- 1/16 + 1/32) ... - 1/1024
You will be able to make 4 groups since you left the first term out. The last term will also be left out.
Each bracket will give you a negative term -1/4 + 1/8 = -1/8 etc
Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative.

So the sum, T, must be more than 0.25 but less than 0.5

_________________

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Verbal Forum Moderator Status: Preparing for the another shot...! Joined: 03 Feb 2011 Posts: 1422 Location: India Concentration: Finance, Marketing GPA: 3.75 Followers: 154 Kudos [?]: 905 [13] , given: 62 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 15 Jan 2013, 06:16 13 This post received KUDOS Expert's post 1 This post was BOOKMARKED an alternative: The series can be written as $$1/2 - 1/4 + 1/8 - 1/16.............1/1024$$ Since the terms beyond $$1/16$$ are too small, so I am not considering those terms. Now on adding $$1/2 - 1/4 +1/8 - 1/16$$, we get $$5/16$$ which is slightly more than $$4/16$$. Hence its value would be around $$0.3$$. The answer choice which includes this number is D. _________________ Intern Joined: 21 Dec 2011 Posts: 14 Schools: Tepper '17 (S) Followers: 0 Kudos [?]: 10 [4] , given: 7 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 24 Feb 2014, 17:23 4 This post received KUDOS I did it the following way. K=(1/2)-(1/4)+(1/8)-(1/16)+..... ----- 1 Multiply K by 2 2K=1-(1/2)+(1/4)-.....-(1/512) ------- 2 Adding 1 and 2 3K = 1 -(1/1024) K= (1/3) -{1/(3*1024)} Now 1/(3*1024) will be very small So K= 1/3 = .3333 Ans Option D Director Joined: 03 Feb 2013 Posts: 840 Location: India Concentration: Operations, Strategy Schools: Tepper '17 (M) GMAT 1: 760 Q49 V44 GPA: 3.88 WE: Engineering (Computer Software) Followers: 72 Kudos [?]: 482 [4] , given: 536 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 12 Jul 2014, 06:54 4 This post received KUDOS This is GP. The terms will be 1/2-1/4+1/8-.... Common ratio is (-1/4)/(1/2) = -1/2 So the sum of terms = 1/2 [1- (-1/2)^10]/(1-(-1/2)) = 1/2 *[1-1/1024]/3/2 = 1023/(1024*3) close to 1/3 so Option D _________________ Thanks, Kinjal Struggling with GMAT ? Experience http://www.gmatify.com/ My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961 Please click on Kudos, if you think the post is helpful EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 3906 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 167 Kudos [?]: 1045 [1] , given: 58 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 03 Jun 2015, 14:14 1 This post received KUDOS Expert's post Hi vijaydoli, This question comes up every so often in this Forum. There are a couple of different ways of thinking about this problem, but they all require a certain degree of "math." Without too much effort, you can deduce what the sequence is: +1/2, -1/4, +1/8, -1/16, etc. The "key" to solving this question quickly is to think about the terms in "sets of 2"… 1/2 - 1/4 = 1/4 Since the first term in each "set of 2" is greater than the second (negative) term, we now know that each set of 2 will be positive. 1/8 - 1/16 = 1/16 Now we know that each additional set of 2 will be significantly smaller than the prior set of 2. 1/4....1/16....1/64....etc. Without doing all of the calculations, we know…. We have 1/4 and we'll be adding tinier and tinier fractions to it. Since there are only 10 terms in the sequence, there are only 5 sets of 2, so we won't be adding much to 1/4. Based on the answer choices, only one answer makes any sense… Final Answer: [Reveal] Spoiler: D GMAT assassins aren't born, they're made, Rich _________________ Intern Joined: 27 Jul 2010 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 11 Feb 2012, 17:44 Perfect explanation! Thanks Karishma! Intern Joined: 01 Dec 2010 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 34 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 13 Feb 2012, 06:30 VeritasPrepKarishma wrote: gpkk wrote: For every integer k from 1 to 10 inclusive the kth term of a certain sequence is given by (-1)^(k+1)*(1/2^k). If T is the sum of the first 10 terms in the sequence then T is 1) greater than 2 2)between 1 & 2 3) between 0.5 and 1 4)between 0.25 and 0.5 5)less than 0.25 Could someone please provide a solution to this problem ? To get a hang of what the question is asking, put values for k right away. Say, k = 1, k = 2 etc You get terms such as (1/2) when k = 1, (-1/4) when k = 2 etc T = 1/2 - 1/4 + 1/8 - 1/16 +.... + 1/512 - 1/1024 (Sum of first 10 terms) Of course GMAT doesn't expect us to calculate but figure out the answer using some shrewdness. We have 10 terms. If we couple them up, two terms each, we get 5 groups: T = (1/2 - 1/4) + (1/8 - 1/16) ...+ (1/512 - 1/1024) Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. e.g. 1/2 is larger than 1/4 so 1/2 - 1/4 = 1/4 i.e. a positive number 1/8 - 1/16 = 1/16, again a positive number. We will get something similar to this: T = 1/4 + 1/16 +.... (all positives) Definitely this sum, T, is greater than 1/4 i.e. 0.25 Now, let's group them in another way. T = 1/2 + (- 1/4 + 1/8) + (- 1/16 + 1/32) ... - 1/1024 You will be able to make 4 groups since you left the first term out. The last term will also be left out. Each bracket will give you a negative term -1/4 + 1/8 = -1/8 etc Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative. So the sum, T, must be more than 0.25 but less than 0.5 Answer (D) Intern Joined: 28 Dec 2013 Posts: 38 Followers: 0 Kudos [?]: 1 [0], given: 3 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 30 Jun 2014, 18:37 Marcab wrote: an alternative: The series can be written as $$1/2 - 1/4 + 1/8 - 1/16.............1/1024$$ Since the terms beyond $$1/16$$ are too small, so I am not considering those terms. Now on adding $$1/2 - 1/4 +1/8 - 1/16$$, we get $$5/16$$ which is slightly more than $$4/16$$. Hence its value would be around $$0.3$$. The answer choice which includes this number is D. QUESTION : WHERE IS 1/1024 coming from? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5953 Location: Pune, India Followers: 1522 Kudos [?]: 8371 [0], given: 192 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] 30 Jun 2014, 18:47 Expert's post sagnik242 wrote: Marcab wrote: an alternative: The series can be written as $$1/2 - 1/4 + 1/8 - 1/16.............1/1024$$ Since the terms beyond $$1/16$$ are too small, so I am not considering those terms. Now on adding $$1/2 - 1/4 +1/8 - 1/16$$, we get $$5/16$$ which is slightly more than $$4/16$$. Hence its value would be around `$$0.3$$. The answer choice which includes this number is D. QUESTION : WHERE IS 1/1024 coming from? When you put k = 10 (to get the last term) in the given expression $$(-1)^{k+1}*(1/2^k)$$, you get -1/1024 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  17 Jul 2014, 21:03
Can we some how solve this question using sum of a series formula- S = (n/2) × (2a + (n-1)d)??
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  17 Jul 2014, 21:14
Expert's post
sumitaries wrote:
Can we some how solve this question using sum of a series formula- S = (n/2) × (2a + (n-1)d)??

The formula you are talking about is used for sum of terms of an Arithmetic Progression only.

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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  04 Aug 2014, 05:45
1
This post was
BOOKMARKED
http://tinypic.com/r/2lvk4z5/8

Please check a simpler solution to the above problem in the above image link.

Cheers.
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For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  07 Oct 2014, 14:19
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Rohitesh,
Where did u get this 1/6?
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  13 Dec 2014, 20:00
taleesh wrote:
Rohitesh,
Where did u get this 1/6?

He used the summation of geometric series (base 1/4), I think.
Beautiful solution.
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]  03 Jun 2015, 12:56
I used the same calculation as above, which will probably take little more than 2 minutes. Is there a simple version to solve this problem?
Re: For every integer k from 1 to 10 inclusive the kth terem of   [#permalink] 03 Jun 2015, 12:56
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## OnceAWriter one year ago 72 is what percent of 480? Write and solve an equation to solve the problem

• This Question is Open
1. exploringphysics

480*x/100=72; x=15

2. HDNightster

I think its 15 percet do 72/480 * x/100 yo times 72 by 100 then divide by to get x

3. HDNightster

72/480=15/100

4. robtobey

$\frac{72}{480}*100=15\%$

5. Not the answer you are looking for?
Search for more explanations.

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• Corpus ID: 218581575

# A New Proof of Newman's Conjecture and a Generalization

@article{Dobner2020ANP,
title={A New Proof of Newman's Conjecture and a Generalization},
author={Alexander Dobner},
journal={arXiv: Number Theory},
year={2020}
}
Newman's conjecture (proved by Rodgers and Tao in 2018) concerns a certain family of deformations $\{\xi_t(s)\}_{t \in \mathbb{R}}$ of the Riemann xi function for which there exists an associated constant $\Lambda \in \mathbb{R}$ (called the de Bruijn-Newman constant) such that all the zeros of $\xi_t$ lie on the critical line if and only if $t \geq \Lambda$. The Riemann hypothesis is equivalent to the statement that $\Lambda \leq 0$, and Newman's conjecture states that $\Lambda \geq 0$. In…

## Figures from this paper

Jensen polynomials are not a viable route to proving the Riemann Hypothesis
Recent work on the Jensen polynomials of the Riemann xi-function and its derivatives found a connection to the Hermite polynomials. Those results have been suggested to give evidence for the Riemann
THE DE BRUIJN–NEWMAN CONSTANT IS NON-NEGATIVE
• Mathematics, Physics
Forum of Mathematics, Pi
• 2020
For each $t\in \mathbb{R}$, we define the entire function $$\begin{eqnarray}H_{t}(z):=\int _{0}^{\infty }e^{tu^{2}}\unicode[STIX]{x1D6F7}(u)\cos (zu)\,du,\end{eqnarray}$$ where

## References

SHOWING 1-10 OF 22 REFERENCES
The Laguerre inequalities with applications to a problem associated with the Riemann hypothesis
• Mathematics, Computer Science
Numerical Algorithms
• 2005
A new numerical method is investigated, base on the Laguerre inequalities, for determining lower bounds for the de Bruijn-Newman constant ∧, which is related to the Riemann Hypothesis, to obtain the new lower bound, -0.0991 < ∧ which improves all previously published lower bound for ∧.
An improved bound for the de Bruijn–Newman constant
• A. Odlyzko
• Mathematics, Computer Science
Numerical Algorithms
• 2004
Improve previous lower bounds and prove that −2.7⋅10−9<Λ satisfies Λ≤0.7, providing yet more evidence that the Riemann hypothesis, if true, is just barely true.
Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann Hypothesis
• Mathematics
• 1994
AbstractWe give here a rigorous formulation for a pair of consecutive simple positive zeros of the functionH0 (which is closely related to the Riemann ξ-function) to be a “Lehmer pair” of zeros ofH0.
NOTES ON LOW DISCRIMINANTS AND THE GENERALIZED NEWMAN CONJECTURE
Generalizing work of Polya, de Bruijn and Newman, we allow the backward heat equation to deform the zeros of qua- dratic Dirichlet L-functions. There is a real constant LKr (gen- eralizing the de
Fourier transforms with only real zeros
The class of even, nonnegative, finite measures p on the real line such that for any b > 0 the Fourier transform of exp(bt2) dp(t) has only real zeros is completely determined. This result is then
An improved lower bound for the de Bruijn-Newman constant
• Computer Science, Mathematics
Math. Comput.
• 2011
This article reports on computations that led to the discovery of a new Lehmer pair of zeros for the Riemann Ϛ function and improves the known lower bound for de Bruijn-Newman constant A.
A NEW LEHMER PAIR OF ZEROS AND A NEW LOWER BOUND FOR THE DE BRUIJN-NEWMAN CONSTANT
• Mathematics
• 1993
The de Bruijn-Newman constant has been investigated extensively because the truth of the Riemann Hypothesis is equivalent to the assertion that 0. On the other hand, C. M. Newman conjectured that 0.
On the structure of the Selberg class, I: 0≤d≤1
• Mathematics
• 1999
The Selberg class S is a rather general class of Dirichlet series with functional equation and Euler product and can be regarded as an axiomatic model for the global L-functions arising from number
A new lower bound for the de Bruijn-Newman constant
SummaryStrong numerical evidence is presented for a new lower bound for the so-called de Bruijn-Newman constant. This constant is related to the Riemann hypothesis. The new bound, −5, is suggested by


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# On open problems based on fuzzy filters of pseudo BCK-algebras

#### Abstract

We study the properties and relations of fuzzy pseudo-filters of pseudo-BCK algebras. After we discuss the equivalent conditions of fuzzy normal pseudo-filter of pseudo-BCK algebra (pP), we propose fuzzy implicative pseudo-filter and its relation with fuzzy Boolean filter of (bounded) pseudo-BCK algebras (pP). Then two open problems: “In pseudo-BCK algebra or bounded pseudo-BCK algebra, Is the notion of implicative pseudo-filter equivalent to the notion of Boolean filter?” and “Prove or negate the following conclusion: A pseudo-BCK algebra is an implicative pseudo-BCK algebra if and only if every pseudo-filter of it is a Boolean filter(or an implicative pseudo-filter)” are partly solved.

## 1Introduction

The rapid development of mathematical logic, technology and computing science put forward many new requirements, thus contributing to the non-classical logic and the rapid development of modern logic. The study of fuzzy logic has become a hot topic with scientific information and artificial intelligence, which make fuzzy logic study of algebra and logic inseparable. Historically, the logic of algebra initiated in the early Leibniz period, he made use of symbols proposition to establish a two-valued logic calculus theory. Modern classical mathematical logic has been formed, and characterized as any proposition are “true” and “false” binary judgments in the 20th century.

However, science and technology is constantly in progress and development, so in real life, not everything can be described by absolutely true and absolutely false binary logic to deal with the phenomenon around. The classic two-valued logic can not meet the needs of new types of reasoning. Therefore, the classical two-valued logic must be improved and new promotion and reasoning should occur to meet the needs of real life. One way to improve and extend the classical two-valued logic is to expand the domain of assignment, which formed a multi-valued logic systems and fuzzy logic system. This system consists of a well-known Lukasiewicz logic, Godel logic, Gainse-Resche logic systems, basic logic BL, fuzzy propositional calculus formal deductive system, and so on. Fuzzy logic is both the mathematical basis of the artificial intelligence and fuzzy reasoning. Based on the actual background, different forms of fuzzy logic system is proposed.

The logical algebras are the algebraic counterpart of the the non-classical logic. For example, BCK-algebra, BL-algebras, pseudo MTL-algebras and non-commutative residuated lattice are algebraic counterparts of BCK Logic, Basic Logic, monoidalt-norm-based logic and monoidal logic respectively[2, 18, 19].

In 1966, BCK-algebra was introduced by Iséki and Imai from BCK/BCI Logic [11]. From then on, numerous research work have been done on this, including ordered structure, algebraic structure, ideal, filter and the relations between them and logic. Afterwards, pseudo-BCK algebras, an extension of BCK-algebras, was proposed to express the non-commutative reasoning by Georgescu and Iorgulescu [6].

Filters theory plays a vital role not only in studying of algebraic structure, but also in non-classical logic and Computer Science [23]. From the aspect of logical point, filters correspond to various provable formulae sets [23]. For example, based on filters and prime filters in BL-algebras, Hájek proved the completeness of Basic Logic BL [18]. Literatures [1, 4, 5, 8, 12, 17, 24, 30], further studied filters of BL-algebras, lattice implication algebras, pseudo BL-algebras, pseudo effect-algebras, residuated lattices, triangle algebras and the corresponding algebraicstructures.

Fuzzy sets were introduced in 1965 by Zadeh [14]. At present, fuzzy ideas have been applied to algebraic structures. Furthermore, fuzzy filter ideas function well in studying of algebraic structures. In lattice implication algebras fuzzy positive implicative filters was introduced [30]. [27, 28] characterized fuzzy positive implicative filters of lattice implication algebras. [15, 16] further studied fuzzy filters of BL-algebras. Wang and Xin solved an open problem in pseudo BL-algebras between fuzzy normal and fuzzy Boolean filter [20]. Literatures [10, 26, 29] studied interval valued-fuzzy filters in pseudo BL-algebras and MTL-algebras. Therefore, fuzzy filters are a useful tool to obtain results on algebraic structures of logic algebras.

In [22], there is an open problem: “In pseudo-BCK algebra or bounded pseudo-BCK algebra, Is the notion of implicative pseudo-filter equivalent to the notion of Boolean filter?” In [8], there is an open problem: “Prove or negate that pseudo-BCK algebras is implicative BCK algebras if and only if every pseudo-filters of them is implicative pseudo-filters (or Boolean filter).” The two kinds of filters are important to deep study of the algebraic structure of pseudo-BCK algebra, thus the two open problems are interesting and meaningful topics for deep research in pseudo-BCK algebra and are the motivation of this paper. We found it difficult to directly discuss the relation between the two filters, and we also found that fuzzy idea will be helpful in study of the algebraic structure, so we focused on the fuzzy filters of the two kinds andtheir relation.

We explore the properties and relations of fuzzy pseudo-filters of pseudo-BCK algebras. After we discuss the equivalent conditions of fuzzy normal pseudo-filter of pseudo-BCK algebra (pP), we propose fuzzy implicative pseudo-filter and its relation with fuzzy Boolean filter of (bounded) pseudo-BCK algebras (pP). Then two open problems are partly solved.

## 2Preliminaries

Here we recall some basic results.

Definition 2.1. (Nola, Georgescu and Iorgulescu [2]) An algebraic structure (A, ∨ , ∧ , ⊙ , ⇀ , ⇝ , 0, 1) is called a pseudo-BL algebra if for all x, y, z ∈ A

(1) (A, ∨ , ∧ , 0, 1) is a bounded lattice,

(2) (A, ⊙ , 1) is a monoid,

(3) x ⊙ y ≤ z ⇔ x ≤ y ⇀ z ⇔ y ≤ x ⇝ z,

(4) x ∧ y = (x ⇀ y) ⊙ x = x ⊙ (x ⇝ y),

(5) (x ⇀ y) ∨ (y ⇀ x) = (x ⇝ y) ∨ (y ⇝ x) =1.

Definition 2.2. (Meng and Jun [9]) An algebraic structure (A, ⇀ , 1) is called an BCK-algebra if for all x, y, z ∈ A

(1) (z ⇀ x) ⇀ (y ⇀ x)) ≥ (y ⇀ z),

(2) (y ⇀ x) ⇀ x ≥ y,

(3) x ≥ x,

(4) x ≥ y and y ≥ x imply x = y,

(5) x ⇀ 1 =1,

where x ≤ y means x ⇀ y = 1.

Definition 2.3. (Georgescu and Iorgulescu [6]) An algebraic structure (A, ≥ , ⇀ , ⇝ , 1) is called a (reversed left-) pseudo-BCK algebra, such that for all x, y, z ∈ A, the axioms hold

(1)(z  ⇀  x)   ⇝   (y  ⇀  x))   ≥  y  ⇀  z,   (z ⇝ x)   ⇀ (y ⇝  x))≥y ⇝ z,

(2) (y ⇀ x) ⇝ x ≥ y, (y ⇝ x) ⇀ x ≥ y,

(3) x ≥ x,

(4) 1 ≥ x,

(5) x ≥ y and y ≥ x imply x = y,

(6) x ≥ y ⇔ y ⇀ x = 1 ⇔ y ⇝ x = 1.

Example 2.4 (Lee and Park [13]) Let X = [0, ∞] and let ≤ be the usual order on X. Define binary operations ⇀ and ⇝ on X by x ⇀ y = 0 (ifx ≤ y), xy=2xπarctan(ln(xy))(ify<x); x ⇝ y = 0 (ifx ≤ y), xy=xe-tan(πy2x)(ify<x). for all x, y ∈ X. Then (X, ≤ , ⇀ , ⇝ , 0) is a pseudo-BCK algebra.

Definition 2.5. (Georgescu and Iorgulescu [6]) A pseudo-BCK algebra (A, ≥ , ⇀ , ⇝ , 1) is called bounded if there exists unique element 0 such that 0 ⇀ x = 1 or 0 ⇝ x = 1 for any x ∈ A.

In a bounded pseudo-BCK algebra A we define: x - = x ⇀ 0, x  = x ⇝ 0 for any x ∈ A.

Proposition 2.6. (Georgescu and Iorgulescu [6]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebra. Then the following properties hold for any x, y, z ∈ A

(1) x ≤ y ⇀ y ⇀ z ≤ x ⇀ z and y ⇝ z ≤ x ⇝ z,

(2) x ≤ y ⇀ z ⇀ x ≤ z ⇀ y and z ⇝ x ≤ z ⇝ y,

(3)z ⇀ x ≤ (y ⇀ z) ⇀ (y ⇀ x), z ⇝ x ≤ (y ⇝ z)⇝ (y ⇝ x),

(4) z ⇝ (y ⇀ x) = y ⇀ (z ⇝ x).

Definition 2.7. (Georgescu and Iorgulescu [6]) A pseudo-BCK algebra (pP) (i.e., with pseudo-product) is a pseudo-BCK algebra (A, ≥ , ⇀ , ⇝ , 1) satisfying the condition (pP)

(pP) for all x, y ∈ A, there exists x ⊙ y = min {z|x ≤ y ⇀ z} = min {z|y ≤ x ⇝ z}.

Theorem 2.8. (Zhang [21]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebras (pP), x ⊙ y is defined as min {z|x ≤ y ⇀ z} or min {z|y ≤ x ⇝ z}, then the followings hold in A

(1) (x ⊙ y) ⇀ z = x ⇀ (y ⇀ z),

(2) (y ⊙ x) ⇝ z = x ⇝ (y ⇝ z),

(3) (x ⇀ y) ⊙ x ≤ x, y, x ⊙ (x ⇝ y) ≤ x, y,

(4) x ⊙ y ≤ x ∧ y ≤ x, y.

Proposition 2.9. (Iorgulescu [3]) Any pseudo-BL algebra is a pseudo-BCK algebra (pP).

Lemma 2.10. (Zhang [21]) pseudo-BCK algebra (pP) is categorically equivalent to partially ordered residuated integral monoid.

Proposition 2.11. (Georgescu [7]) In a pseudo-BL algebra A, for all x, y, z ∈ A the followings hold

(1) x ∨ y = ((x ⇀ y) ⇝ y) ∧ ((y ⇀ x) ⇝ x),

(2) x ∨ y = ((x ⇝ y) ⇀ y) ∧ ((y ⇝ x) ⇀ x),

(3) x ⇀ y = x ⇀ x ∧ y, x ⇝ y = x ⇝ x ∧ y.

Definition 2.12. (Meng and Jun [9]) A BCK algebra A is called an implicative BCK-algebra if for any x, y ∈ A, (x ⇀ y) ⇀ x = x.

Definition 2.13. (Zhang [22]) A pseudo-BCK algebra (A, ≥ , ⇀ , ⇝ , 1) is called a 1-type implicative pseudo-BCK algebra if for any x, y ∈ A, (x ⇀ y) ⇀ x = (x ⇝ y) ⇝ x = x.

Theorem 2.14. (Zhang and Gong [22]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebras. Then A is a 1-type implicative pseudo-BCK algebra if and only if A is an implicative BCK-algebra.

Definition 2.15. (Zhang and Gong [22]) A pseudo-BCK algebra (A, ≥ , ⇀ , ⇝ , 1) is called a 2-type implicative pseudo-BCK algebra if for any x, y ∈ A, (x ⇝ y) ⇀ x = (x ⇀ y) ⇝ x = x.

Theorem 2.16. (Zhang and Gong [22]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebras. Then A is a 2-type implicative pseudo-BCK algebra if and only if A is an implicative BCK-algebra.

## 3Some kinds of pseudo-filters and fuzzy pseudo-filters in pseudo BCK-algebras

First, we recall the definitions of pseudo-filter, Boolean filter, normal pseudo-filter and implicative pseudo-filter of pseudo-BCK algebras.

Definition 3.1. (Zhang and Gong [22]) A nonempty subset F of a pseudo-BCK algebra A is called a pseudo-filter if

(F1) x ∈ F, y ∈ A, x ≤ y ⇒ y ∈ F,

(F2) x ∈ F, x ⇀ y ∈ F or x ⇝ y ∈ F ⇒ y ∈ F.

We could get the equivalent conditions of pseudo-filter in pseudo-BCK algebra as follows.

Theorem 3.2. A nonempty subset F of a pseudo-BCK algebra A is a pseudo-filter if and only if it satisfies

(F3) 1 ∈ F,

(F4) x ∈ F, x ⇀ y ∈ F or x ⇝ y ∈ F impliesy ∈ F.

Proof. Let F be a nonempty subset of A. It is easy to see that 1 ∈ F, i.e., (F3) and (F4) hold if F is a pseudo-filter of A. Conversely, suppose F satisfies (F3) and (F4). Let x ∈ F, x ≤ y ∈ A, then x ⇀ y = x ⇝ y = 1 ∈ F, thus y ∈ F, so F is a pseudo-filter of A.

Theorem 3.3. A nonempty subset F of a pseudo-BCK algebra(pP) A is a pseudo-filter if and only if

(F5) x, y ∈ F implies x ⊙ y ∈ F,

(F6) x ∈ F, x ≤ y ∈ A implies y ∈ F.

Proof. Let F be a pseudo-filter of A and x, y ∈ F, we know x ⇀ (y ⇀ (x ⊙ y)) = (x ⊙ y) ⇀ (x ⊙ y) =1 ∈ F, then y ⇀ (x ⊙ y) ∈ F, so x ⊙ y ∈ F.

Conversely, suppose F satisfies (F5) and (F6). If x, x ⇀ y (orx ⇝ y) ∈ F, then (x ⇀ y) ⊙ x, x ⊙ (x ⇝ y) ≤ y ∈ F.

Definition 3.4. (Zhang and Gong [22]) A pseudo-filter F of a pseudo-BCK algebra A is normal, if

(NF) x ⇀ y ∈ F iff x ⇝ y ∈ F.

It’s obvious that any filter is normal in a BCK-algebra.

Definition 3.5. (Zhang and Gong [22]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebra. A pseudo-filter F of A is called an implicative pseudo-filter if for any x, y ∈ A

(1) (x ⇀ y) ⇀ x ∈ F implies x ∈ F,

(2) (x ⇝ y) ⇝ x ∈ F implies x ∈ F.

Definition 3.6. (Zhang and Gong [22]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebra. A pseudo-filter F is called Boolean if for any x, y ∈ A

(1) (x ⇀ y) ⇝ x ∈ F implies x ∈ F,

(2) (x ⇝ y) ⇀ x ∈ F implies x ∈ F.

Theorem 3.7. (Zhang and Gong [22]) Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebra and F a normal pseudo-filter of A. Then F is implicative if and only if F is Boolean.

Theorem 3.8. (Zhang and Gong [22]) Let (A ; ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra and F implicative pseudo-filter. The following properties hold for ∀x, y ∈ A

(1) ((x ⇀ 0) ⇀ x) ⇝ x ∈ F, i.e., (x - ⇀ x) ⇝ x ∈ F,

(2) ((x ⇝ 0) ⇝ x) ⇀ x ∈ F, i.e., (x  ⇝ x) ⇀ x ∈ F,

(3) ((x ⇀ y) ⇀ x) ⇝ x ∈ F,

(4) ((x ⇝ y) ⇝ x) ⇀ x ∈ F,

(5) if (x ⇀ y) ⇀ y ∈ F, then (y ⇀ x) ⇝ x ∈ F,

(6) if (x ⇝ y) ⇝ y ∈ F, then (y ⇝ x) ⇀ x ∈ F,

(7) if x ⇝ y ∈ F, then ((y ⇝ x) ⇀ x) ⇝ y ∈ F,

(8) if x ⇀ y ∈ F, then ((y ⇀ x) ⇝ x) ⇀ y ∈ F.

Definition 3.9. (Zhang and Jun [25]) A nonempty subset F of a pseudo-BCK algebra A is a positive implicative pseudo-filter of A if for all x, y ∈ A, (F1) and the followings hold

(F7) x ⇝ (y ⇀ z) , x ⇝ y ∈ F implies x ⇝ z ∈ F,

(F8) x ⇀ (y ⇝ z) , x ⇀ y ∈ F implies x ⇀ z ∈ F.

Theorem 3.10. (Zhang and Jun [25])A positive implicative pseudo-filter F of a pseudo-BCK algebra A is a pseudo-filter satisfying for all x, y ∈ A, x ⇀ (x ⇝ y) ∈ F implies x ⇀ y, x ⇝ y ∈ F.

Definition 3.11. A non-constant fuzzy subset f of a pseudo-BCK algebra A is called a fuzzy pseudo-filter if f t is either empty or a pseudo-filter for any t ∈ [0, 1].

It’s easy to find that F is a pseudo-filter iff χ F is a fuzzy pseudo-filter, where χ F stands for the characteristic function of F.

For pseudo-BCK algebras or pseudo-BCK algebras(pP), inspired by [20], the following results of fuzzy pseudo-filters are similar with the ones of pseudo BL-algebras and reader could refer to it for the proofs.

Proposition 3.12. A fuzzy set f is a fuzzy pseudo-filter of a pseudo-BCK algebra (or pseudo-BCK algebra(pP)) A if and only if for all x, y, z ∈ A, one the following conditions holds

(1) f (1) ≥ f (x) and f (y) ≥ f (x) ∧ f (x ⇀ y),

(2) f (1) ≥ f (x) and f (y) ≥ f (x) ∧ f (x ⇝ y),

(3) x ⇀ (y ⇀ z) =1 or x ⇝ (y ⇝ z)) =1 ⇒ f (z) ≥ f (x) ∧ f (y),

(4) x ⊙ y ≤ z or y ⊙ x ≤ z ⇒ f (z) ≥ f (x) ∧ f (y),

(5) f is order-preserving, f (x ⊙ y) ≥ f (x) ∧ f (y).

Corollary 3.13. An order-preserving fuzzy set fof a pseudo-BCK algebra(pP) A is a fuzzy pseudo-filter if and only if for any x, y ∈ A, f (x ⊙ y) = f (x) ∧ f (y).

Corollary 3.14. Let f, f i ((i ∈ τ)) be fuzzy pseudo-filters of a pseudo-BCK algebra(pP) A, then for any x, y ∈ A, the following properties hold

(1) f (x) = f (x ⊙ x) = f (x ⊙ x ⊙ … ⊙ x),

(2) f (x ⊙ y) = f (y ⊙ x),

(3) f (x ⇀ y) = f (1) or f (x ⇝ y) = f (1) implies f (x) ≤ f (y),

(4) f 1 ∧ f 2 is a fuzzy pseudo-filter of A,

(5) ∧ iτ f i is a fuzzy pseudo-filter of A.

In order to solve the mentioned open problems, we fuzzilize the several kinds of pseudo-filters of pseudo-BCK algebras and characterize them respectively.

Definition 3.15. Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebra. A fuzzy pseudo-filter f of A is called implicative if for any x, y ∈ A, the following conditions hold

(1) f ((x ⇀ y) ⇀ x) = f (x),

(2) f ((x ⇝ y) ⇝ x) = f (x).

Theorem 3.16. Let (A ; ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra and f fuzzy implicative pseudo-filter of A. Then ∀x ∈ A,

(1) f ((x - ⇀ x) ⇝ x) = f (1),

(2) f ((x  ⇝ x) ⇀ x) = f (1),

(3) f ((x - ⇝ x) ⇀ x) = f (1),

(4) f ((x  ⇀ x) ⇝ x) = f (1).

Proof. (1) From x ≤ (x - ⇀ x) ⇝ x, so ((x - ⇀ x) ⇝ x- ≤ x - and ((x - ⇀ x) ⇝ x- ⇀ x - = 1, then f (((x - ⇀ x) ⇝ x- ⇀ x -) = f (1). And x - ≤ (x - ⇀ x) ⇝ x, so we get ((x - ⇀ x) ⇝ x- ⇀ x - ≤ ((x - ⇀ x) ⇝ x- ⇀ ((x - ⇀ x) ⇝ x). Then f (((x - ⇀ x) ⇝ x- ⇀ ((x - ⇀ x) ⇝ x)) = f ((x - ⇀ x) ⇝ x) = f (1).

(2) It is similar to (1).

(3) From x ≤ (x - ⇝ x) ⇀ x, so ((x - ⇝ x) ⇀ x- ≤ x - and ((x - ⇝ x) ⇀ x- ⇀ x - = 1, then f (((x - ⇝ x) ⇀ x- ⇀ x -) = f (1). And x - ≤ (x - ⇝ x) ⇀ x, so we get ((x - ⇝ x) ⇀ x- ⇀ x - ≤ ((x - ⇝ x) ⇀ x- ⇀ ((x - ⇝ x) ⇀ x). Then f (((x - ⇝ x) ⇀ x- ⇀ ((x - ⇝ x) ⇀ x)) = f ((x - ⇝ x) ⇀ x) = f (1).

(4) It is similar to (3).

Theorem 3.17. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is implicative if and only if f t is either empty or an implicative pseudo-filter for each t ∈ [0, 1].

Proof. Let f be a fuzzy implicative pseudo-filter of A. For any t ∈ [0, 1], if f t  ≠ φ, then suppose (x ⇀ y) ⇀ x ∈ f t , then from f ((x ⇀ y) ⇀ x) = f (x), we get x ∈ f t . Dually we can get x ∈ f t if (x ⇝ y) ⇝ x ∈ f t . So f t is an implicative pseudo-filter of A.

Conversely, let f ((x ⇀ y) ⇀ x) = t, we get (x ⇀ y) ⇀ x ∈ f t , and f t is an implicative pseudo-filter of A, then x ∈ f t . That is, f ((x ⇀ y) ⇀ x) = t ≤ f (x). Since x ≤ (x ⇀ y) ⇀ x and f is isotone, then f ((x ⇀ y) ⇀ x) ≥ f (x), we get f ((x ⇀ y) ⇀ x) = f (x). Dually we can obtain f ((x ⇝ y) ⇝ x) = f (x), thus proves f is implicative.

Theorem 3.18. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is implicative if and only if f f(1) is an implicative pseudo-filter.

Proof. Necessity is obvious. Suppose f f(1) is an implicative pseudo-filter, then for all x, y ∈ A, if (x ⇀ y) ⇀ x ∈ f f(1), then x ∈ f f(1), so f ((x ⇀ y) ⇀ x) = f (x) = f (1). Dually we can obtain f ((x ⇝ y) ⇝ x) = f (x) = f (1). Thus we prove f is implicative.

Corollary 3.19. A nonempty subset F of a pseudo-BCK algebra A is an implicative pseudo-filter if and only if χ F is implicative.

Corollary 3.20. Let (A, ≥ , ⇀ , ⇝ , 1) be a pseudo-BCK algebra. A pseudo-filter F of A is called Boolean if for any x, y ∈ A, the following conditions hold

(1) f ((x ⇀ y) ⇝ x) = f (x),

(2) f ((x ⇝ y) ⇀ x) = f (x).

Similar to the fuzzy implicative pseudo-filter, we have the following results, so we omit the proof.

Theorem 3.21. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is Boolean if and only if f t is either empty or a Boolean filter for each t ∈ [0, 1].

Theorem 3.22. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is Boolean if and only if f f(1) is a Boolean filter of A.

Corollary 3.23. A nonempty subset F of a pseudo-BCK algebra A is Boolean if and only if χ F is Boolean.

Theorem 3.24. A fuzzy pseudo-filter f of a bounded pseudo-BCK algebra A is Boolean if and only if f (x ⇀ y) = f (x ⇀ (y - ⇝ y)), f (x ⇝ y) = f (x ⇝ (y  ⇀ y)) for all x, y ∈ A.

Proof. Suppose f is Boolean. Let f (x ⇀ (y - ⇝ y)) = t, then f t is a Boolean filter,x ⇀ (y - ⇝ y) ∈ f t . And (y - ⇝ y) ⇀ y ∈ f t . (y - ⇝ y) ⇀ y ≤ (x ⇀ (y - ⇝ y)) ⇀ (x ⇀ y), then x ⇀ y ∈ f t , i.e., f (x ⇀ y) ≥ t = f (x ⇀ (y - ⇝ y)). Since x ⇀ y ≤ x ⇀ (y - ⇝ y), then f (x ⇀ y) ≤ f (x ⇀ (y - ⇝ y)), so f (x ⇀ y) = f (x ⇀ (y - ⇝ y)). Dually, we can obtain f (x ⇝ y) = f (x ⇝ (y  ⇀ y)).

Conversely, let f ((x ⇀ y) ⇝ x) = t, then f t is a pseudo-filter and (x ⇀ y) ⇝ x ∈ f t . (x ⇀ y) ⇝ x ≤ x - ⇝ x, then x - ⇝ x ∈ f t . 1 ⇀ (x - ⇝ x) = x - ⇝ x ∈ f t , we get f (x) = f (1 ⇀ x) = f (1 ⇀ (x - ⇝ x)) ≥ t = f ((x ⇀ y) ⇝ x), and f (x) ≤ f ((x ⇀ y) ⇝ x) is obvious, then f (x) = f ((x ⇀ y) ⇝ x). We can also get f (x) = f ((x ⇝ y) ⇀ x). Then f is a fuzzy Boolean filter of A.

Theorem 3.25. A fuzzy pseudo-filter f of a bounded pseudo-BCK algebra A is Boolean filter if and only if f ((x  ⇀ x) ⇝ x) = f ((x - ⇝ x) ⇀ x) = f (1) for all x ∈ A.

Proof. (1) From x ≤ (x  ⇀ x) ⇝ x, we get ((x  ⇀ x) ⇝ x ≤ x and ((x  ⇀ x) ⇝ x ⇀ x  = 1, then f (((x  ⇀ x) ⇝ x ⇀ x ) = f (1). And x  ≤ (x  ⇀ x) ⇝ x, so we get ((x  ⇀ x) ⇝ x ⇀ x  ≤ ((x  ⇀ x) ⇝ x ⇀ ((x  ⇀ x) ⇝ x). Then f (((x  ⇀ x) ⇝ x ⇀ ((x  ⇀ x) ⇝ x)) = f ((x  ⇀ x) ⇝ x) = f (1).

Similarly we can prove f ((x - ⇝ x) ⇀ x) = f (1).

Conversely, assume f (x ⇀ (y - ⇝ y)) = t, then x ⇀ (y - ⇝ y) ∈ f t . And we have (y - ⇝ y) ⇀ y ≤ (x ⇀ (y - ⇝ y)) ⇀ (x ⇀ y) ∈ f t . Then x ⇀ y ∈ f t , i.e., f (x ⇀ y) ≥ t = f (x ⇀ (y - ⇝ y)). Since y - ⇝ y ≥ y and x ⇀ (y - ⇝ y) ≥ x ⇀ y, we get f (x ⇀ y) ≤ f (x ⇀ (y - ⇝ y)), then f (x ⇀ y) = f (x ⇀ (y - ⇝ y)). So f (x ⇝ y) = f (x ⇝ (y  ⇀ y)). Then f is a fuzzy Boolean filter of A.

Corollary 3.26. In a bounded pseudo-BCK algebra, every fuzzy implicative pseudo-filter is a fuzzy Boolean filter.

Theorem 3.27. Let (A, ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra and f a fuzzy Boolean filter. Then f ((y ⇝ x) ⇀ x) = f ((x ⇀ y) ⇝ y) for all x, y ∈ A.

Proof. Let f ((x ⇀ y) ⇝ y) = t, then (x ⇀ y) ⇝ y ∈ f t . (x ⇀ y) ⇝ y ≤ (y ⇝ x) ⇀ ((x ⇀ y) ⇝ x) = (x ⇀ y) ⇝ ((y ⇝ x) ⇀ x). And, we have x ⇀ y ≥ ((y ⇝ x) ⇀ x) ⇀ y and (x ⇀ y) ⇝ ((y ⇝ x) ⇀ x) ≤ (((y ⇝ x) ⇀ x) ⇀ y) ⇝ ((y ⇝ x) ⇀ x). Combining the above results we get (x ⇀ y) ⇝ y ≤ (((y ⇝ x) ⇀ x) ⇀ y) ⇝ ((y ⇝ x) ⇀ x). Thus, (((y ⇝ x) ⇀ x) ⇀ y) ⇝ ((y ⇝ x) ⇀ x) ∈ f t . Then we get (y ⇝ x) ⇀ x ∈ f t , i.e., f ((y ⇝ x) ⇀ x) ≥ f ((x ⇀ y) ⇝ y). Similarly, we can prove that f ((y ⇝ x) ⇀ x) ≤ f ((x ⇀ y) ⇝ y).

Corollary 3.28. Let (A, ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra and f a fuzzy Boolean filter. Then f (x -∼) = f (x ∼-) = f (x) for all x ∈ A.

Theorem 3.29. Let (A, ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra(pP) and f a fuzzy Boolean filter. Then f (x ⇀ y) = f (x ⊙ x ⇀ y) = f ((x ⊙ x ⊙ … ⊙ x) ⇀ y), f (x ⇝ y) = f (x ⊙ x ⇝ y) = f ((x ⊙ x ⊙ … ⊙ x) ⇝ y) for all x, y ∈ A.

Proof. Assume f (x ⇝ (x ⇝ y)) = t, then x ⇝ (x ⇝ y) ∈ f t . And x ⇝ (x ⇝ y) ≤ ((x ⇝ y) ⇝ y) ⇀ (x ⇝ y). It follows that ((x ⇝ y) ⇝ y) ⇀ (x ⇝ y) ∈ f t . So f (x ⇝ y) ≥ t = f (x ⇝ (x ⇝ y)). The inverse inequation is obvious, then f (x ⇝ y) = f (x ⇝ (x ⇝ y)) = f (x ⊙ x ⇝ y). Similarly, we can prove that f (x ⇀ y) = f (x ⇀ (x ⇀ y)) = f (x ⊙ x ⇀ y). And by induction, we can get the result.

Definition 3.30. A fuzzy filter f of a pseudo-BCK algebra is called normal if f t is either empty or a normal pseudo-filter for each t ∈ [0, 1].

Theorem 3.31. Suppose f is a fuzzy set of a pseudo-BCK algebra A. Then the followings are equivalent for any x, y, z ∈ A

(1) f is normal,

(2) f is a fuzzy pseudo-filter satisfying f (x ⇀ y) = f (x ⇝ y),

(3) f (1) ≥ f (x), f (x ⇝ z) ≥ f (y) ∧ f (x ⇀ (y ⇝ z)) and f (x ⇀ z) ≥ f (y) ∧ f (x ⇝ (y ⇀ z)),

(4) f (1) ≥ f (x), f (x ⇀ z) ≥ f (y) ∧ f (x ⇀ (y ⇀ z)) and f (x ⇝ z) ≥ f (y) ∧ f (x ⇝ (y ⇝ z)).

Proof. (1) ⇒ (2). For all x, y ∈ A, let f (x ⇀ y) = t 1, so x ⇀ y ∈ f t 1 , then x ⇝ y ∈ f t 1 , f (x ⇝ y) ≥ t 1 = f (x ⇀ y). Dually, we can obtain f (x ⇀ y) ≥ f (x ⇝ y).

(2) ⇒ (3). It is obvious.

(2) ⇒ (4). It is obvious.

(3) ⇒ (1). Let x = 1, we find f is a fuzzy pseudo-filter. For any t ∈ [0, 1], if f t  ≠ φ and x ⇀ y ∈ f t , so f (x ⇀ y) ≥ t. x ⇀ ((x ⇀ y) ⇝ y) =1 ∈ f t , then f (x ⇀ ((x ⇀ y) ⇝ y)) = f (1) ≥ t and f (x ⇝ y) ≥ f (x ⇀ y) ∧ f (x ⇀ ((x ⇀ y) ⇝ y)) ≥ t, so x ⇝ y ∈ f t . Also, we can prove that if x ⇝ y ∈ f t , then x ⇀ y ∈ f t . So f t is a normal pseudo-filter and f is normal.

(4) ⇒ (1). In the same way of (3) ⇀ (1).

Similar to the proof of fuzzy implicative pseudo-filter, we have the following results.

Theorem 3.32. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is normal if and only if f f(1) is a normal pseudo-filter of A.

Corollary 3.33. A nonempty subset F of a pseudo-BCK algebra A is a normal pseudo-filter if and only if χ F is normal.

Definition 3.34. A fuzzy pseudo-filter f of pseudo-BCK algebra A is called positive implicative if for any x, y ∈ A, the followings hold

(1) f (x ⇝ z) ≥ f (x ⇝ (y ⇀ z)) ∧ f (x ⇝ y),

(2) f (x ⇀ z) ≥ f (x ⇀ (y ⇝ z)) ∧ f (x ⇀ y).

Theorem 3.35. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is positive implicative if and only if for each t ∈ [0, 1], f t is either empty or a positive implicative pseudo-filter.

Theorem 3.36. A fuzzy pseudo-filter f of a pseudo-BCK algebra A is positive implicative if and only if f f(1) is a positive implicative pseudo-filter.

Corollary 3.37. A nonempty subset F of a pseudo-BCK algebra A is a positive implicative pseudo-filter if and only if χ F is positive implicative.

Theorem 3.38. Suppose f is a fuzzy normal pseudo-filter of a pseudo-BCK algebra A. Then the followings are equivalent

(1) f is positive implicative.

(2) f (x ⇀ z) ≥ f (x ⇀ (y ⇀ z)) ∧ f (x ⇀ y).

(3) f (x ⇝ y) = f (x ⇝ (x ⇀ y)).

(4) f (x ⇀ y) = f (x ⇀ (x ⇀ y)).

Proof. (1) ⇒ (2). For any x, y, z ∈ A, we have f (x ⇝ z) ≥ f (x ⇝ (y ⇀ z)) ∧ f (x ⇝ y). Since f is fuzzy normal, then we can get (2) holds.

(2) ⇒ (3). When we let y = x and z = y in (2), we have f (x ⇀ y) ≥ f (x ⇀ (x ⇀ y)) ∧ f (x ⇀ x) = f (x ⇀ (x ⇀ y)) ∧ f (1) = f (x ⇀ (x ⇀ y)). It is obvious to get the inverse inequation by the isotonic of fuzzy filter. Thus (3) holds.

(3) ⇒ (4). It is obvious.

(4) ⇒ (1). Let f (x ⇀ (y ⇀ z)) ∧ f (x ⇀ y) = t, then x ⇀ (y ⇀ z) , x ⇀ y ∈ f t . Since y ≤ (y ⇀ z) ⇝ z, we have x ⇀ y ≤ x ⇀ ((y ⇀ z) ⇝ z) = (y ⇀ z) ⇝ (x ⇀ z), then (y ⇀ z) ⇝ (x ⇀ z) ∈ f t , further (y ⇀ z) ⇀ (x ⇀ z) ∈ f t . And (y ⇀ z) ⇀ (x ⇀ z) ≤ (x ⇀ (y ⇀ z)) ⇀ (x ⇀ (x ⇀ z)) ∈ f t , then we get x ⇀ (x ⇀ z) ∈ f t . By (4), we can prove that x ⇀ z ∈ f t , i.e., if x ⇝ (y ⇀ z) , x ⇝ y ∈ f t , then x ⇝ z ∈ f t . Dually if x ⇀ (y ⇝ z) , x ⇀ y ∈ f t , then x ⇀ z ∈ f t . So f t is a positive implicative pseudo-filter and f is positive implicative.

## 4On open problems of pseudo-filters in pseudo-BCK algebra

In [22], there is an open problem that “In pseudo-BCK algebra or bounded pseudo-BCK algebra, Is the notion of implicative pseudo-filter equivalent to the notion of Boolean filter?” In [8], there is an open problem that “Prove or negate that pseudo-BCK algebras is implicative BCK algebras if and only if every pseudo-filters of them is implicative pseudo-filters (or Boolean filter).” In this section, inspired by [20], with the help of the similar equivalent conditions of fuzzy normal filter of pseudo-BCK algebras(pP), we present the relation between the two filters and get some results for the open problems in pseudo-BCK algebra.

Theorem 4.1. A fuzzy pseudo-filter f of a pseudo-BCK algebra(pP) A is a fuzzy normal pseudo-filter if and only if for any x, y ∈ A, f ((x ⊙ y) ⇝ (y ⊙ x)) ∧ f ((y ⊙ x) ⇀ (x ⊙ y)) ≥ f (y).

Proof. Let f be a fuzzy normal pseudo-filter. For any y ∈ A, let t = f (y), then y ∈ f t and f t is a normal pseudo-filter of A. For any x ∈ A, (y ⊙ x) ⊙ y = y ⊙ (x ⊙ y) ≤ x ⊙ y, since y ≤ (y ⊙ x) ⇝ (x ⊙ y), then (y ⊙ x) ⇝ (x ⊙ y) ∈ f t and (y ⊙ x) ⇀ (x ⊙ y) ∈ f t , thus f ((y ⊙ x) ⇀ (x ⊙ y)) ≥ t = f (y). Dually, we have f ((x ⊙ y) ⇝ (y ⊙ x)) ≥ t = f (y). From above, then we obtain f ((x ⊙ y) ⇝ (y ⊙ x)) ∧ f ((y ⊙ x) ⇀ (x ⊙ y)) ≥ f (y) for any x, y ∈ A. Conversely, for any t ∈ [0, 1], if f t  ≠ φ, then f t is a pseudo-filter and there exists y ∈ f t . f ((y ⊙ x) ⇀ (x ⊙ y)) ≥ f (y) ≥ t, then (y ⊙ x) ⇀ (x ⊙ y) ∈ f t . ((y ⊙ x) ⇀ (x ⊙ y)) ⊙ y = (y ⇀ (x ⇀ (x ⊙ y)) ⊙ y ≤ x ⇀ (x ⊙ y), then we have x ⇀ (x ⊙ y) ∈ f t , for any x ∈ A, y ∈ f t . For any a, b ∈ A, if a ⇝ b ∈ f t , from above, we get a ⇀ a ⊙ (a ⇝ b) = a ⇀ b ∈ f t . Dually, we can obtain x ⇝ (y ⊙ x) ∈ f t for any x ∈ A, y ∈ f t . For any a, b ∈ A, if a ⇀ b ∈ f t , from above, we can obtain a ⇝ (a ⇀ b) ⊙ a = a ⇝ b ∈ f t . Thus f t is a normal pseudo-filter, and f is normal.

Corollary 4.2. Every fuzzy Boolean filter of pseudo-BCK algebras (pP) is a fuzzy normal filter.

Corollary 4.3. Every fuzzy Boolean filter of a pseudo-BCK algebra (pP) is implicative.

Proof. Let f be a fuzzy Boolean filter. By Definition, for any x, y ∈ A, f ((x ⇀ y) ⇝ x) = f (x), f ((x ⇝ y) ⇀ x) = f (x). And we obtain f ((x ⇀ y) ⇀ x) = f (x), f ((x ⇝ y) ⇝ x) = f (x). Thus by Definition, f is implicative.

Corollary 4.4. Let X be a pseudo-BCK algebra (pP) and f be a fuzzy normal pseudo-filter. Then f Boolean if and only if it is implicative.

Based on the above Theorems, we can get the following result.

Theorem 4.5. In bounded pseudo-BCK algebra, every fuzzy implicative pseudo-filter is a fuzzy Boolean filter. In pseudo-BCK algebras(pP), every fuzzy Boolean filter is a fuzzy implicative pseudo-filter.

For the first open problem, we can get the following theorem.

Theorem 4.6. In bounded pseudo-BCK algebra, every implicative pseudo-filter is a Boolean filter. In pseudo-BCK algebras(pP), every Boolean filter is an implicative pseudo-filter.

For the second open problem, we have the following results.

Theorem 4.7. (Zhang and Gong [22]) Let (A ; ⇀ , 1) be a BCK-algebra. Then A is an implicative BCK-algebra if and only if its every filter is an implicative filter.

Theorem 4.8. (Zhang and Gong [22]) Let (A ; ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra. Then A is a bounded implicative BCK-algebra if and only if its every pseudo-filter is an implicative pseudo-filter.

Theorem 4.9. (Zhang and Gong [22]) Let (A ; ≤ , ⇀ , ⇝ , 0, 1) be a bounded pseudo-BCK algebra. Then A is a bounded implicative BCK-algebra if and only if its every pseudo-filter is a Boolean filter.

By the above results, we could get

Theorem 4.10. Pseudo-BCK algebras(pP) is implicative BCK algebras if and only if every pseudo-filters of them is Boolean filter.

Proof. Based on the results of [9], pseudo-BCK algebras is implicative BCK algebras if and only if x ⇀ y = x ⇝ y for all x, y ∈ A and every pseudo-filters of them is implicative pseudo-filters (Boolean filter is the same), so necessity is obvious.

Now suppose every pseudo-filter F of a pseudo-BCK algebra (pP) A is a Boolean filter, then every fuzzy pseudo-filter χ F of A is a fuzzy Boolean filter and χ F is normal. Further we get that χ F is positive implicative. So we get F is a positive implicative pseudo-filter. And by Theorem 4.5 [25] we get (A, ⇀ , 1) is a positive implicative BCK-algebra and for all x, y ∈ A, x ⇀ y = x ⇝ y. So we get all pseudo-filters of A is positive implicative, then implicative, so we get that A is an implicative BCK-algebras by the result of [9].

Corollary 4.11. Pseudo-BCK algebras (pP) is implicative BCK algebras, then every pseudo-filters of them is an implicative pseudo-filter.

Thus the following theorem in pseudo-BCK algebra(pP) could answer the second problem in [8].

Theorem 4.12. Pseudo-BCK algebras (pP) is implicative BCK algebras if and only if every pseudo-filters of them is Boolean filter (meanwhile they are implicative pseudo-filters).

## 5Conclusion

After we discuss the equivalent conditions of fuzzy normal pseudo-filter of pseudo-BCK algebra (pP), we propose fuzzy implicative pseudo-filter and its relation with fuzzy Boolean filter of (bounded) pseudo-BCK algebras (pP). Then we proposed solutions for the open problems in pseudo-BCK algebras. In the future, we will make use of the relations among the fuzzy pseudo-filters to solve the open problems in the common pseudo-BCK algebras.

## Acknowledgments

This work is partially supported by the National Natural Science Foundation of China (Grant No. 11571281,61175055); the project TIN-2009-0828; Sichuan Key Technology Research and Development Program of China (Grant No. 2011FZ0051); Wireless Administration of Ministry of Industry and Information Technology of China ([2011]146); Foundation project of the National Key Laboratory of Aerospace Dynamics(Grant No. 202300006) and doctor initial fund of Xi’an Shiyou University of China (Grant No. 2011BS017).

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# Difference between conventional Eye-diagram voltage plot and “Eye-density” plot in ADS Keysight Simulation

I know what is eye diagram. It is the sampled voltage plot at the clock frequency (Usually) of my circuit and superimposed. If the logic 0 corresponds 0V and logic 1 corresponds 1V then for the ideal case, the eye diagram Amplitude will also plot the superimposed samples of the signal which corresponds to 0V and 1V. For the lossy channel, the amplitude moves away from the 0V and 1V level for logic 0 and logic 1 respectively. And that movement depends on the attenuation and loss of the channel. Generally I have seen so far to be around 0.8V or 0.9V for logic 1. Like below.

This eye diagram is actually PAM4 level eye diagram. I generated in Cadence Virtuoso using its eye diagram tool using its embedded calculator. I am fine with that.

Now, I have experienced in ADS Keysight simulation that they have Eye density which also produces Eye diagram. but I see it always plot the eye diagram amplitude into half. Like my below eye diagram that I generated in ADS using its generic TX, RX driver and low pass filter like channel. I have driven the TX with V_low = -1V and V_high = 1V PRBS signal. So, the difference is 2V. At the receiver (in simulation), it produces the eye with V_high around 0.4V to 0.6V. This is the case most of the time I experienced in ADS. So I was confused that Eye density is not exactly the conventional Eye diagram plot (Maybe or my conception of Eye density is not clear).

Eye density Summary:

In the help of ADS it is written like below:

Density: A temperature view of the 3-dimensional eye histogram. The eye is automatically centered and displayed over 2UI on the horizontal axis. Color indicates the number of crossings of a segment in the time-voltage plane. Blue areas are cold, whereas red indicates comparatively more crossings.

In statistical mode, the density histogram is plotted as if 1e6 bits are run.

I am confused about what exactly then it produces if it is temperature view...because it is also showing the voltage amplitude of the signal. And the Amplitude (as it is showing in the summary is around 0.698V whereas my input signal difference is 2V. My channel was normal passive low pass filter-like channel.

So, is the Eye density plot in ADS different from conventional Eye diagram? If it is what is then? Because the definition in ADS help is not enough for me to understand density.


text:
With fractions, the relationship between the numbers, not the actual numbers themselves, is most important. Understanding how to multiply and divide fractions can give you a deeper understanding of why you can increase or decrease the numbers within a fraction without changing the value of the whole fraction.

When you multiply or divide any number by 1, the answer is the same number. This rule also goes for fractions, so

When a fraction has the same number in both the numerator and the denominator, its value is 1. In other words, the fractions

are all equal to 1. Look what happens you multiply the fraction

The net effect is that you’ve increased the terms of the original fraction by 2. But all you’ve done is multiply the fraction by 1, so the value of the fraction hasn’t changed. The fraction 6/8 is equal to 3/4.

Similarly, reducing the fraction 6/9 by a factor of 3 is the same as dividing that fraction by 3/3 (which is equal to 1):

So 6/9 is equal to 2/3.