The Grades of Initiation

[whocoulditbe?]

This still makes 0 sense to me. If {not A, not B} means anything at all, it would be a set of both not-As and not-Bs, since {A, B} is a set of both A and B.

[Daniel]

11 minutes ago, whocoulditbe? said:

... therefore {} ∩ whatever must equal {}, even {} ∩ {}.

 

How about this?

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

Pick one!  Is it true or false?

[Daniel]

8 hours ago, whocoulditbe? said:

This still makes 0 sense to me. If {not A, not B} means anything at all, it would be a set of both not-As and not-Bs, since {A, B} is a set of both A and B.

 

The intersection / union of a disjunction produces infinite negations, "mu" or "wu"... "無"

 

It's the partnership of the disjointed sets which produces the action of nullification.

 

From the wiki-monster:

 

 

Causing to be non-existent.  Original non-being.  Those are both verbs.  Actions.

 

I'm bringing a mental construct ( many actually, infinitely many ) which systematically produces a realization of 無.  That's the point.

 

 

Edited September 15, 2023 by Daniel

[whocoulditbe?]

5 minutes ago, Daniel said:

How about this?

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

Pick one!  Is it true or false?

Please show that A ⊆ B => A ∩ B ≠ {}.

[Daniel]

1 minute ago, whocoulditbe? said:

Please show that A ⊆ B => A ∩ B ≠ {}.

 

Let A = { 1 }

Let B = { 1 , 2 , 3 }

 

A ⊆ B = { 1 }

A ∩ B = { 1 }
{ 1 } ≠ {}

 

Any mistakes?

[whocoulditbe?]

Just now, Daniel said:

Any mistakes?

I should have been more specific:
Please show that ∀A, B: A ⊆ B => A ∩ B ≠ {}.

[whocoulditbe?]

4 minutes ago, Daniel said:

A ⊆ B = { 1 }

I also have no idea what this means.

[Daniel]

1 hour ago, whocoulditbe? said:

You mean disjunct. {1, 2} and {2} are non-equal but have {2} as their intersection.

 

Disjunction is boolean "OR" isn't it?

 

Edit:  confirmed:

 

https://en.m.wikipedia.org/wiki/Logical_disjunction

 

 

I really think the best choice is "disjoint".

 

Edited September 14, 2023 by Daniel

[Cobie]

Delete and forget about it.

 

 

Edited September 14, 2023 by Cobie

[Daniel]

18 minutes ago, whocoulditbe? said:

I also have no idea what this means.

 

Please be honest.  You're being stubborn. I think you "have an an idea of what it means"  ~eye-rolls~

 

Let A = { 1 }

Let B = { 1 , 2 , 3 }

 

A ⊆ B = 1 

A ∩ B = 1
1  ≠ {}

 

Every element or group of elements can be represented as a set. The set is represtented as the element or elements enclosed in curly brackets.

 

1 = { 1 }.

 

Better?

 

[Daniel]

31 minutes ago, Cobie said:

Delete and forget about it.

 

... deleted and stricken from the record, your honor

Edited September 14, 2023 by Daniel

[Daniel]

@who

31 minutes ago, whocoulditbe? said:

I also have no idea what this means.

 

why didn't you answer the question?  I answered your question.  Fair is fair.

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

Pick one!  Is it true or false?

[Daniel]

duplicate.

Edited September 14, 2023 by Daniel

[Daniel]

Looks like confidence is not the metric to undermine competence.  This appears to be an effort to demonize those with non-standard points of view.

 

 

 

[whocoulditbe?]

Ahhhhh DisJOINT not Disjunct, I am above the peak!

[whocoulditbe?]

30 minutes ago, Daniel said:

 

Please be honest.  You're being stubborn. I think you "have an an idea of what it means"  ~eye-rolls~

 

Let A = { 1 }

Let B = { 1 , 2 , 3 }

 

A ⊆ B = 1 

A ∩ B = 1
1  ≠ {}

 

Every element or group of elements can be represented as a set. The set is represtented as the element or elements enclosed in curly brackets.

 

1 = { 1 }.

 

Better?

 

You just did ∃A, B: A ⊆ B => A ∩ B ≠ {} again. A ⊆ B = 1 makes no more sense to me.

PS, Sorry if I'm being a bit aggressive in my replies. It's kind of fun to be able to talk about math on here.

Edited September 14, 2023 by whocoulditbe?

[Daniel]

5 minutes ago, whocoulditbe? said:

Ahhhhh DisJOINT not Disjunct, I am above the peak!

 

Not really sure what you're saying here.  You corrected me.  I accepted it.  But, then I remembered, no... disjuction is OR.  This should be a welcome correction, just as I welcomed your correction.  Are we allies, or enemies?  Do we have a common goal, or are we in some sort of competition?

 

It seems like there is an effort to discredit what I'm saying, but, there is no reason to do so. I *actually* know what I'm taking about.  I said disjoint was ideal earlier in the thread.

 

4 hours ago, Daniel said:

regarding being disconnected, I prefer the word disjointed, never coinciding.

 

this was 4 hours ago.

[whocoulditbe?]

1 minute ago, Daniel said:

Not really sure what you're saying here.  You corrected me.  I accepted it.  But, then I remembered, no... disjuction is OR.  This should be a welcome correction, just as I welcomed your correction.  Are we allies, or enemies?  Do we have a common goal, or are we in some sort of competition?

I'm just lamenting how wrong I was! I missed your previous correction.

Edited September 14, 2023 by whocoulditbe?

[Daniel]

4 minutes ago, whocoulditbe? said:

You just did ∃A, B: A ⊆ B => A ∩ B ≠ {} again. A ⊆ B = 1 makes no more sense to me.

PS, Sorry if I'm being a bit aggressive in my replies. It's kind of fun to be able to talk about math on here.

 

It's OK to be agressive.  It's not OK to be dishonest.  

 

56 minutes ago, whocoulditbe? said:

I also have no idea what this means.

 

This ^^ is dishonest.

[whocoulditbe?]

1 minute ago, Daniel said:

58 minutes ago, whocoulditbe? said:

I also have no idea what this means.

 

This ^^ is dishonest.

Ok. I could kinda guess that "A ⊆ B = 1" just means "A ⊆ B," but I have no idea why you added the "= {1}" on, or if there's any significance to it that I'm missing.

[Daniel]

50 minutes ago, whocoulditbe? said:

You just did ∃A, B: A ⊆ B => A ∩ B ≠ {}

 

If that's true, then I answered your question correctly.  All I need is one example which saticfies the condition.  

 

"∃A, B: A ⊆ B => A ∩ B ≠ {}" = there exists at least 1 pair of sets, A, B such that A ⊆ B => A ∩ B ≠ {}"

 

Example:  A = { 1 } B = { 1 , 2 , 3 )

 

Fair is fair, please answer my question:

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

Pick one!  Is it true or false?

 

42 minutes ago, whocoulditbe? said:

Ok. I could kinda guess that "A ⊆ B = 1" just means "A ⊆ B," but I have no idea why you added the "= {1}" on, or if there's any significance to it that I'm missing.

 

OK.. Step-by-Step , no AI assist, no online research, purely from memory  

 

? A ⊆ B => A ∩ B ≠ {} ?

 

Premise 'P':  A ⊆ B

Premise "Q": A ∩ B ≠ {}

 

Truth Table for P "=>" Q

 

P | Q | True/False

T | T | T

T | F | F

F | T | T

F | F | T

 

Let A = { 1 }

Let B = { 1 , 2 , 3 }

 

{ 1 } ⊆ { 1 , 2 , 3 } is true

A ⊆ B is true

P is true

 

{ 1 } ∩ { 1 , 2 , 3 } = 1 

1 ≠ {}

{ 1 } ∩ { 1 , 2 , 3 } ≠ {}

A ∩ B ≠ {}

Q is true

 

P is true

Q is true

 

P | Q | True/False

T | T | T

 

P => Q is true!

 

? A ⊆ B => A ∩ B ≠ {} ?

 

If A = { 1 } AND B = { 1 , 2 , 3 } Then A ⊆ B => A ∩ B ≠ {} is True!

 

Edited September 14, 2023 by Daniel

[Daniel]

48 minutes ago, whocoulditbe? said:

Ok. I could kinda guess that "A ⊆ B = 1" just means "A ⊆ B," but I have no idea why you added the "= {1}" on, or if there's any significance to it that I'm missing.

 

I said Let A = { 1 }.

 

A = { 1 }.

 

{ 1 } => { 1 } =/= {} is true.  This seemed obvious to me.  Path of least resistance... backfired.

 

Edited September 14, 2023 by Daniel

[whocoulditbe?]

5 minutes ago, Daniel said:

 

If that's true, then I answered your question correctly.  All I need is one example which saticfies the condition.  

 

"∃A, B: A ⊆ B => A ∩ B ≠ {}" = there exists at least 1 pair of sets, A, B such that A ⊆ B => A ∩ B ≠ {}"

 

Example:  A = { 1 } B = { 1 , 2 , 3 )

 

Fair is fair, please answer my question:

 

If {} is disjoint from itself, {} ⊆ {} is true or false?  If it's true then it cannot be disjoint from itself.  If it's false it cannot be s subset of every set.

 

Pick one!  Is it true or false?

 

 

OK.. Step-by-Step , no AI assist, no online research, purely from memory  

 

? A ⊆ B => A ∩ B ≠ {} ?

 

Premise 'P':  A ⊆ B

Premise "Q": A ∩ B ≠ {}

 

Truth Table for P "=>" Q

 

P | Q | True/False

T | T | T

T | F | F

F | T | T

F | F | T

 

Let A = { 1 }

Let B = { 1 , 2 , 3 }

 

{ 1 } ⊆ { 1 , 2 , 3 } is true

A ⊆ B is true

P is true

 

{ 1 } ∩ { 1 , 2 , 3 } = 1 

1 ≠ {}

{ 1 } ∩ { 1 , 2 , 3 } ≠ {}

A ∩ B ≠ {}

Q is true

 

P is true

Q is true

 

P | Q | True/False

T | T | T

 

P => Q is true!

 

? A ⊆ B => A ∩ B ≠ {} ?

 

If A = { 1 } AND B = { 1 , 2 , 3 } Then A ⊆ B => A ∩ B ≠ {} is True!

 

I'm saying ∃ isn't enough. You need ∀ to cover A={} B={}. If you prove that, then I'm forced to pick between {} ⊆ {} and {} ∩ {} = {}. For now I maintain that both are true. It now occurs to me that proving ∀A: A ⊆ A => A ∩ A ≠ {} would also be enough to do the trick for what you're arguing.

[Daniel]

2 hours ago, whocoulditbe? said:

Please show that A ⊆ B => A ∩ B ≠ {}.

 

This is what you asked ^^. 

 

This is an answer:

 

If A = { 1 } AND B = { 1 , 2 , 3 } Then A ⊆ B => A ∩ B ≠ {} is True!

 

10 minutes ago, whocoulditbe? said:

I'm saying ∃ isn't enough. You need ∀ to cover A={} B={}. If you prove that, then I'm forced to pick between {} ⊆ {} and {} ∩ {} = {}. For now I maintain that both are true. It now occurs to me that proving ∀A: A ⊆ A => A ∩ A ≠ {} would also be enough to do the trick for what you're arguing.

 

Sounds like you wanted to ask:

 

For any A and any B show that A ⊆ B => A ∩ B ≠ {}.

[whocoulditbe?]

Just now, Daniel said:

For any

Thats what ∀ means...