Is religion dying?

[viole]

Well. if we're using your so-called logic, if I don't know latin, caput is known to mean everything, anything and nothing.

I think you are confusing caput with kaputt.

Ciao

- viole

 

[viole]

Sure, a valid proof needs to address counter-examples and contradictions, otherwise those counter examples and contradictions disprove.

This is described in chapter 9 of the attached PDF. Your so-called proof ignores and omits the contradiction of the definition of Jew and Atheist. And this same contradiciton exists in any set, any property, and any propostion of vacuous truth about the empty-set excluding the identity.

Here's info on the author. But you should know this already. And you have already denied accepting contradictions as true.

Directory | VCU Department of Mathematics and Applied Mathematics

Full A to Z directory of administrators, faculty and staff in the VCU Department of Mathematics and Applied Mathematics

math.vcu.edu

View attachment 78032

Yes, very cool pdf. I think I will reference it quite often. Hope you have nothing against it, since it is material provided by you, and assumed authoritative by yourself.

Interestingly enough, at page 12 (Part 1, Fundamentals) it lists the following fact (fact 1.2)

Fact 1.2 The empty set is a subset of all sets

He also proves it. Exactly like I did. So, he really seems to understand logic.

So, that will settle it, right?

Wow. Faster than I thought

Ciao

- viole

 

[dybmh]

Yes, very cool pdf. I think I will reference it quite often. Hope you have nothing against it, since it is material provided by you, and assumed authoritative by yourself.

Interestingly enough, at page 12 (Part 1, Fundamentals) it lists the following fact (fact 1.2)

Fact 1.2 The empty set is a subset of all sets

He also proves it. Exactly like I did. So, he really seems to understand logic.

So, that will settle it, right?

Wow. Faster than I thought

Ciao

- viole

Not so fast. This is a definition, not a proof. He didn't *actually* prove anything. And because of the definition of Atheism, no set which can contain an atheist is ever empty. It will either contain atheism or theism, but not both. And this is true for any set, any property, and any propostion which asserts that the property is a member or belongs to that set, excluding its identity.

The fact that this is early in the book means that the concept of disproof hasn't been introduced. So the reader, if they are lacking that knowledge, as you appear to be lacking that knowledge, doesn't know that this can be easily disproven using the same logic applied here. I referrd you to chapter 9. The later chapter brings details which are not included in this earlier chapter.

This is normal when teaching anything to beginners. The teacher omits details that will be too complicated for the student, or will distract from the lesson being taught.

Here's what he said, and I will include the disproof, inline with what he has said. Try to refute any of it.

​

There is a special set that, although small, plays a big role. The empty set is the set {} that has no elements. We denote it as ∅ so ; ∅ = {}. Whenever you see the symbol ∅, it stands for {}. Observe that | ∅ | = 0. The empty set is the only set whose cardinality is zero.​

​

Definition 1.3 Suppose A and B are sets. If every element of A is also an element of B, then we say A is a subset of B, and we denote this as A ⊆ B. We write A ⊈ B if A is not a subset of B, that is, if it is not true that every element of A is also an element of B. Thus A ⊈ B means that there is at least one element of A that is not an element of B.​

​

This brings us to a significant fact: If B is any set whatsoever, then ∅ ⊆ B. To see why this is true, look at the last sentence of Definition 1.3. ( But what about the first sentence, the first sentence is ignored. ) It says that ∅ ⊈ B would mean that there is at least one element of 0 that is not an element of B. ( The first sentence says: If every element of A is also an element of B, then we say A is a subset of B. A has no elements ), But this cannot be so because ∅ contains no elements! ( True! Looking for elements in an empty-set always fails. ) Thus it is not the case that ∅ ⊈ B, ( No, the test is invalid, the definition given describes a set with elements ) so it must be that ∅ ⊆ B. ( No, the same reasoning can be applied in reverse. )​

​

No, looking at the first sentence of the defintion given, If every element of A is also an element of B, then we say A is a subset of B. But this cannot be so because ∅ contains no elements! Thus it is not the case that ∅ ⊆ B, so it must be that ∅ ⊈ B.​

​

Here is another way to look at it. Imagine a subset of B as a thing you make by starting with braces { }, then filling them with selections from B. For example, to make one particular subset of B = {a, b, c}, start with { }, select b and c from B and insert them into { to form the subset {b, c}. Alternatively, you could have chosen just a to make {a}, and so on. But one option is to simply select nothing from B. ( No, the definition above says a subset is made by "filling them with selections from B." Nothing was selected. Nothing was filled. ) This leaves you with the subset {}. ( But you didn't make a subset. You made nothing. You started with {} and did nothing to it. )Thus {} ⊆ B. More often we write it as ∅ ⊆ B. ( No, all that you did was show ∅ = ∅, nothing was done, no operation was attempted. )​

​

Unless you can refute any of what I have in bold-red, then this is just a basic intro in chapter 1. What I am talking about is brought in chapter 9. Ignoring chapter 9, is folly.

The earlier definition of the empty set says: it is a special set. These explanations consider the empty-set identical to non-empty sets. This is the same problem with considering unknown Jews like an empty-set, when in fact any set containing people contains either the property theist or atheist, but not both. Any set describing people is non-empty by definition.

 

[viole]

Not so fast. This is a definition, not a proof.

Nope. No definition, but demonstration. Let's see what he says. Again, page 12:

This brings us to a significant fact: If B is any set whatsoever, then the empty set is a subset of B.
To see why this is true, look at the last sentence of Definition 1.3. It says
that empty-set not subset of B would mean that there is at least one element of the empty set that is not an
element of B. But this cannot be so because the empty-set contains no elements! Thus
it is not the case that the entry-set is not a subset of B, so it must be that the empty set is a subset of B.

Exactly like my previous demonstration.
QED

I would suggest that you read, and possibly understand what you post, in order to avoid further embarrassments. And lies.

Ciao

- viole

 

[viole]

No, looking at the first sentence of the defintion given, If every element of A is also an element of B, then we say A is a subset of B. But this cannot be so because ∅ contains no elements! Thus it is not the case that ∅ ⊆ B, so it must be that ∅ ⊈ B.

You must be out of your mind

Of course there is no problem to say "every element of A" even if A has no element. That is still what you don't understand. Against the entire mathematical community.

Ciao

- viole

 

[dybmh]

Nope. No definition, but demonstration. Let's see what he says. Again, page 12:

Ummmmm, he refers to the defintion. But ONLY the last sentence, and ignores the first.

This brings us to a significant fact: If B is any set whatsoever, then the empty set is a subset of B.
To see why this is true, look at the last sentence of Definition 1.3. It says

There it is. It's not a proof.

that empty-set not subset of B would mean that there is at least one element of the empty set that is not an
element of B. But this cannot be so because the empty-set contains no elements! Thus
it is not the case that the entry-set is not a subset of B, so it must be that the empty set is a subset of B.

This ignores half the defintion.

Exactly like my previous demonstration.

Which also fails, because dispoof is described later.

Definition 1.3 Suppose A and B are sets. If every element of A is also an element of B, then we say A is a subset of B, and we denote this as A ⊆ B. We write A ⊈ B if A is not a subset of B, that is, if it is not true that every element of A is also an element of B. Thus A ⊈ B means that there is at least one element of A that is not an element of B.

Looking at the first sentence of the defintion given, If every element of A is also an element of B, then we say A is a subset of B. But this cannot be so because ∅ contains no elements! Thus it is not the case that ∅ ⊆ B, so it must be that ∅ ⊈ B.

QED

You have been disproven. Again.

I would suggest that you read, and possibly understand what you post, in order to avoid further embarrassments. And lies.

I would suggest reading past chapter 1.

The only person who apears to have been lying is you. You claimed knowledge, when you *actually* are ignorant.

 

[viole]

Looking at the first sentence of the defintion given, If every element of A is also an element of B, then we say A is a subset of B. But this cannot be so because ∅ contains no elements! Thus it is not the case that ∅ ⊆ B, so it must be that ∅ ⊈ B.

Nope. There is, again, no problem to say "every element of A..." even if A has no element. You make up definitions to adapt your conclusion that is in contradiction with the entire mathematical community. And that is form of lying.

So, at the end of the day, do you think that the professor, and all other professors, are totally mistaken, by making such huge error in the basic section? Or isn't more likely that you are not even able to comprehend the basics?

What is more likely?

CIao

- viole

 

[dybmh]

You must be out of your mind

Nope. You simply don't know how to *actually* evaluate things. Nor can you tell the difference between true and false. And you appear to be morally bankrupt.

Of course there is no problem to say "every element of A" even if A has no element.

Sure, there's no problem saying it, but nothing can be proven about those elements if they do not exist unless a contradiction is considered true.

That is still what you don't understand.

You are in denial of the simple fact that using a vacuous truth to prove anything is considering a contradiction true. And that violates the law of non-contradiction.

Against the entire mathematical community.

So what? All you're doing is admitting you're behaving like a sheep. I am not a sheep.

 

[viole]

So what? All you're doing is admitting you're behaving like a sheep. I am not a sheep.

LOL, nope. Flat earth advocates are not sheep, either.

I seriously hope that your job does not require analytical thinking.

Ciao

- viole

 

[viole]

Sure, there's no problem saying it, but nothing can be proven about those elements if they do not exist unless a contradiction is considered true.

Another fallacy. Nobody is proving anything about their elements.

Anyway, again, do you think the professor that you used as reference, is totally mistaken right at the basic section?

Ciao

- viole

 

[dybmh]

Nope. There is, again, no problem to say "every element of A..." even if A has no element. You make up definitions to adapt your conclusion that is in contradiction with the entire mathematical community. And that is form of lying.

No, it's not a lie to invalidate by contradiction. That's in Chapter 9. And here you are admitting that all you have is reliance on a defintion, no *actual* proof. You have FAITH.

So, at the end of the day, do you think that the professor, and all other professors, are totally mistaken, by making such huge error in the basic section? Or isn't more likely that you are not even able to comprehend the basics?

No, like I said, it's chapter 1. Chapter 9 has not been introduced.

What is more likely?

Again, since you cannot refute what I'm saying using *actual* logic, and can only protest and appeal to authority, it's likely you are wrong.

 

[dybmh]

LOL, nope. Flat earth advocates are not sheep, either.

They are if they are behaving like you.

I seriously hope that your job does not require analytical thinking.

I am extremely successful in my career, and yes it absolutely requires analytical thinking. And even if I wasn't, that would simply show how ignoant you are to have been defeated by someone like me.

 

[viole]

I am extremely successful in my career, and yes it absolutely requires analytical thinking. And even if I wasn't, that would simply show how ignoant you are to have been defeated by someone like me.

May I ask what you do?

ciao

- viole

 

[dybmh]

Another fallacy. Nobody is proving anything about their elements.

No one is proving anything. And yes, the definition is talking all about elements. Since I can't not find them.... they MUST be in there somewhere.

Here let me take this bag and turn it inside out, and shake it out. Now, I can't not find anything in it, it MUST be full.

And the other explanation, a subset is made by "taking the elements from one set and putting it in another". Ummmm, hello, that is workign with elements.

Anyway, again, do you think the professor that you used as reference, is totally mistaken right at the basic section?

B-A-S-I-C.

Basic.

 

[viole]

Again, since you cannot refute what I'm saying using *actual* logic, and can only protest and appeal to authority, it's likely you are wrong.

This is the best of all.

- you posted an article of a professor, to make your point. Which IS, in fact, appeal to authority
- that backfired comically, since the article actually made my point. as they always do. for obvious reasons.
- i made you notice that the professor is actually making my point, and how is that possible that he, and all professors, agree with me
- you accuse me of appeal to authority

what is more laughable than that?

and what on earth is actual logic? Something you just made up? Is the chapter on that book concerning logic, NOT actual logic?

ciao

- viole

 

[viole]

B-A-S-I-C.

Basic.

So, he was wrong About the basics, correct?

ciao

- viole

 

[dybmh]

May I ask what you do?

ciao

- viole

I'm a network administrator for a high volume dental office, I wrote the software that is used to monitor the health of the network, including all the backup logs, intrusion detection, HW health. I've been doing that for almost 20 years for the same office. The software needs to be updated periodically as new devices come on the network, and ther servers have come and gone and been replaced. But I did the same thing for local government, before that for 7ish years. I wrote their montoring software, I think they had around 2000 nodes and 30 servers. The software they used previously only was able to monitor around 75% of their nodes because the client software was buggy. My software was clientless, and I setup a system for their helpdesk to proactively seek out devices which had not reported. Within 2 weeks of implementation, 98% of the nodes were reporting successfully.

I know I'm successful, not just because of my bank account and assets, but because the office I work for stores terabytes of data, they do all digital records, very high-quality 3-d imaging of all their patients, I developed a custom off-site backup system for them, which is superior to any other that the owner has compared, I never have failed HW on the network, everything is monitored. I know for certain, no one is hacking into the system. I have never lost a single file out of millions that are stored, and the office has legal obligation for file retension.

 

[viole]

I'm a network administrator for a high volume dental office, I wrote the software that is used to monitor the health of the network, including all the backup logs, intrusion detection, HW health. I've been doing that for almost 20 years for the same office. The software needs to be updated periodically as new devices come on the network, and ther servers have come and gone and been replaced. But I did the same thing for local government, before that for 7ish years. I wrote their montoring software, I think they had around 2000 nodes and 30 servers. The software they used previously only was able to monitor around 75% of their nodes because the client software was buggy. My software was clientless, and I setup a system for their helpdesk to proactively seek out devices which had not reported. Within 2 weeks of implementation, 98% of the nodes were reporting successfully.

I know I'm successful, not just because of my bank account and assets, but because the office I work for stores terabytes of data, they do all digital records, very high-quality 3-d imaging of all their patients, I developed a custom off-site backup system for them, which is superior to any other that the owner has compared, I never have failed HW on the network, everything is monitored. I know for certain, no one is hacking into the system. I have never lost a single file out of millions that are stored, and the office has legal obligation for file retension.

of course no one is hacking the system. Because it would be someone, otherwise.
actually, how can no one hack anything? Are you attributing agency to nobody?

anyway, what programming language do you use?

ciao

- viole

 

[dybmh]

This is the best of all.

- you posted an article of a professor, to make your point. Which IS, in fact, appeal to authority

No, YOU asked for support for what I was saying. I brought YOU someone from your own community, that shows a proof is invalidated by contradiction. That is not an appeal to authority, that is answering your question.

- that backfired comically, since the article actually made my point. as they always do. for obvious reasons.

No... you need to refute chapter 9, since chapter 9 is what I referred to.

- i made you notice that the professor is actually making my point, and how is that possible that he, and all professors, agree with me

They all agree on the manner to teach a beginner. And you seem to be unable to move past the basics.

- you accuse me of appeal to authority

When you say "the whole math community agrees with me, so you must be wrong", yeah, that's ad pop, and an appeal to authority.

If you cannot *actually* refute what I'm saying, nor find a fault in contradiction I have identified, nor establish that anything that exists can have two mutually exclusive propoerties simultaneously, THEN

You have lost the debate.

what is more laughable than that?

Your claim to be a gnostic. That's pretty laughable.

and what on earth is actual logic?

It is logic which does not contradict itself when it claims that it rejects contradictions.

Actual logic follows it's own rules.

Something you just made up?

No, you have been declaring that your method is brainless and it is rule-following. But you are repeatedly breaking the law of non-contradiction. So, it's not *actual* logic, by your own definition.

Is the chapter on that book concerning logic, NOT actual logic?

What is described for the empty-set as a subset is not logic. No. It's a definition.

 

[dybmh]

of course no one is hacking the system. Because it would be someone, otherwise.
actually, how can no one hack anything? Are you attributing agency to nobody?

No, I'm not.

anyway, what programming language do you use?

Bash with some powershell helper scripts.