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Resolution of Russell's and Cantor's paradoxes

by DanTeplitskiy
Tags: cantor, paradoxes, resolution, russell
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DanTeplitskiy
#37
Aug19-11, 04:49 PM
P: 70
Quote Quote by micromass View Post
Yes, it will be logically grounded. The truth table of the implication is logically sound. There are no problems with this.



Yes, from a false hypothesis, we can derive everything. In latin: "ex falso sequitur quodlibet". This is not a flaw in logic, however it might be confusing to some. For example, the following is also true

"Dan is completely legless and his right ankle is bleeding" THEN "1+1=3"

This does not make 1+1=3 true. It only makes the implication true.
Dear Micromass,

I completely agree that the implication is true. My point is if we get something this way can this something (not the implication but what we get from it) be considered to be logically grounded?

Yours,

Dan
micromass
#38
Aug19-11, 05:09 PM
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Dear Micromass,

I completely agree that the implication is true. My point is if we get something this way can this something (not the implication but what we get from it) be considered to be logically grounded?

Yours,

Dan
Well, everything we get from following logical inference rules will be logically valid. So yes.
DanTeplitskiy
#39
Aug19-11, 05:23 PM
P: 70
Quote Quote by micromass View Post
Well, everything we get from following logical inference rules will be logically valid. So yes.
Dear Micromass,

Do you mean if we get "1+1=3" from some contradiction, "1+1=3" by itself is logically grounded?

As far as I understand if something can be derived only from some contradiction it is not logically valid at all.

Yours,

Dan
micromass
#40
Aug19-11, 05:24 PM
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Dear Micromass,

Do you mean if we get "1+1=3" from some contradiction "1+1=3" by itself is logically grounded?

Yours,

Dan
Yes, provided that the contradiction is logically grounded.
DanTeplitskiy
#41
Aug19-11, 05:35 PM
P: 70
Quote Quote by micromass View Post
Yes, provided that the contradiction is logically grounded.
Dear Micromass,

If contradiction is just given to you, like, say, the following way:
If (this forum exists and it does not exist) Then 1+1=3, will you consider "1+1=3" logically grounded?

Yours,

Dan
micromass
#42
Aug19-11, 05:37 PM
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Dear Micromass,

If contradiction is just given to you, like, say, the following way:
If (this forum exists and it does not exist) Then 1+1=3, will you consider "1+1=3" logically grounded?

Yours,

Dan
No, since the premise is false.
DanTeplitskiy
#43
Aug19-11, 05:42 PM
P: 70
Dear Micromass,

That is my point!

If we analyze the formula: R = {x: x∉x} ⇒ R ∈ R ↔ R ∉ R we can see that it is actually the following one: (R = {x: x∉x} And R ≠ {x: x∉x}) ⇒ R ∈ R ↔ R ∉ R.
That, in my opinion, makes "R ∈ R ↔ R ∉ R" logically ungrounded.

Yours,

Dan
micromass
#44
Aug19-11, 05:44 PM
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Dear Micromass,

That is my point!

If we analyze the formula: R = {x: x∉x} ⇒ R ∈ R ↔ R ∉ R we can see that it is actually the following one: (R = {x: x∉x} And R ≠ {x: x∉x}) ⇒ R ∈ R ↔ R ∉ R.
That, in my opinion, makes "R ∈ R ↔ R ∉ R" logically ungrounded.

Yours,

Dan
It's not logically ungrounded, since it can be logically proven. The premise is true, thus so must the conclusion.
DanTeplitskiy
#45
Aug19-11, 05:49 PM
P: 70
Quote Quote by micromass View Post
It's not logically ungrounded, since it can be logically proven. The premise is true, thus so must the conclusion.
Dear Micromass,

Sorry but how can this (R = {x: x∉x} And R ≠ {x: x∉x}) be a true premise?!

Yours,

Dan
micromass
#46
Aug19-11, 05:52 PM
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Dear Micromass,

Sorry but how can this (R = {x: x∉x} And R ≠ {x: x∉x}) be a true premise?!

Yours,

Dan
If it can be proven with inference rules then it is a true premise. And since [itex]R\in R~\leftrightarrow R\notin R[/itex] can be proven from the axioms and the inference rules, means that it is true.
DanTeplitskiy
#47
Aug19-11, 05:54 PM
P: 70
Quote Quote by micromass View Post
If it can be proven with inference rules then it is a true premise. And since [itex]R\in R~\leftrightarrow R\notin R[/itex] can be proven from the axioms and the inference rules, means that it is true.
Dear Micromass,

What I mean is "R ∈ R ↔ R ∉ R" can be derived only from (R = {x: x∉x} And R ≠ {x: x∉x}) .
Does not this alone make "R ∈ R ↔ R ∉ R" logically ungrounded?

Yours,

Dan
micromass
#48
Aug19-11, 05:57 PM
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Dear Micromass,

What I mean is "R ∈ R ↔ R ∉ R" can be derived only from (R = {x: x∉x} And R ≠ {x: x∉x}) .
Does not this alone make "R ∈ R ↔ R ∉ R" logically ungrounded?

Yours,

Dan
Why can it be derived only from that?? It can also be derived from [itex]R=\{x~\vert~x\notin x\}[/itex]. There is no need for [itex](R=\{x~\vert~x\notin x\}~\text{and}~R\neq \{x~\vert~x\notin x\})[/itex]. In fact, that is still true, but there's no need for it.
DanTeplitskiy
#49
Aug19-11, 06:03 PM
P: 70
Quote Quote by micromass View Post
Why can it be derived only from that?? It can also be derived from [itex]R=\{x~\vert~x\notin x\}[/itex]. There is no need for [itex](R=\{x~\vert~x\notin x\}~\text{and}~R\neq \{x~\vert~x\notin x\})[/itex]. In fact, that is still true, but there's no need for it.
Dear Micromass,

If R ∈ R R ≠ {x: x∉x}. That is so because under the assumption R ∈ R R includes a member that is included in itself (R itself is such a member). Do you agree?

That is, when we write
R = {x: x∉x} And R ∈ R -> R ∉ R
it is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R -> R ∉ R which is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} ⇒ R ∈ R -> R ∉ R

Yours,

Dan
micromass
#50
Aug19-11, 06:06 PM
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Quote Quote by DanTeplitskiy View Post
Dear Micromass,

If R ∈ R R ≠ {x: x∉x}. That is so because under the assumption R ∈ R R includes a member that is included in itself (R itself is such a member). Do you agree?

That is, when we write
R = {x: x∉x} And R ∈ R -> R ∉ R
it is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R -> R ∉ R which is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} ⇒ R ∈ R -> R ∉ R

Yours,

Dan
Yes, I agree that [itex]R\in R~\rightarrow~R\notin R[/itex] can be derived from that. But why can it only be derived from that?
DanTeplitskiy
#51
Aug19-11, 06:13 PM
P: 70
Quote Quote by micromass View Post
Yes, I agree that [itex]R\in R~\rightarrow~R\notin R[/itex] can be derived from that. But why can it only be derived from that?
Dear Micromass,

That is so because:
1. The only way to derive R ∉ R is (R = {x: x∉x} And R ∈ R). Right?
2. (R = {x: x∉x} And R ∈ R) is equivalent to (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). Right?
3. From 1. and 2. you can see that when you derive R ∉ R from (R = {x: x∉x} And R ∈ R) you actually derive R ∉ R from (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). You can not avoid it.

Like if you know that 2+2 = 4 you know that every time you use "2+2" you actually use "4". You can not avoid it either.

Yours,

Dan
micromass
#52
Aug19-11, 06:20 PM
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Quote Quote by DanTeplitskiy View Post
Dear Micromass,

That is so because:
1. The only way to derive R ∉ R is (R = {x: x∉x} And R ∈ R). Right?
2. (R = {x: x∉x} And R ∈ R) is equivalent to (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). Right?
3. From 1. and 2. you can see that when you derive R ∉ R from (R = {x: x∉x} And R ∈ R) you actually derive R ∉ R from (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). You can not avoid it.

Like if you know that 2+2 = 4 you know that every time you use "2+2" you actually use "4". You can not avoid it either.

Yours,

Dan
Well no, what you have said in (1) is that we can derive [itex]R\in R~\leftrightarrow~R\notin R[/itex] from [itex]R=\{x~\vert~x\notin x\}[/itex]. There is no need for [itex]R\neq \{x~\vert~x\notin x\}[/itex]...
DanTeplitskiy
#53
Aug22-11, 09:07 AM
P: 70
Dear Micromass,

Thanks for reply!

Let me explain my point another way.

"Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number.
If N >= 2 then N+1> N that is N is not the greatest natural number."

Do you consider the above proof that there is no greatest natural number to be OK? (just in case - I do)

Yours,

Dan
micromass
#54
Aug22-11, 09:10 AM
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Quote Quote by DanTeplitskiy View Post
Dear Micromass,

Thanks for reply!

Let me explain my point another way.

"Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number.
If N >= 2 then N+1> N that is N is not the greatest natural number."

Do you consider the above proof that there is no greatest natural number to be OK? (just in case - I do)

Yours,

Dan
Yes, that looks OK.


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