Resolution of Russell's and Cantor's paradoxesby DanTeplitskiy Tags: cantor, paradoxes, resolution, russell 

#37
Aug1911, 04:49 PM

P: 70

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 



#38
Aug1911, 05:09 PM

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#39
Aug1911, 05:23 PM

P: 70

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 



#41
Aug1911, 05:35 PM

P: 70

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 



#43
Aug1911, 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 



#44
Aug1911, 05:44 PM

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#45
Aug1911, 05:49 PM

P: 70

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



#46
Aug1911, 05:52 PM

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#47
Aug1911, 05:54 PM

P: 70

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 



#48
Aug1911, 05:57 PM

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#49
Aug1911, 06:03 PM

P: 70

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 



#50
Aug1911, 06:06 PM

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#51
Aug1911, 06:13 PM

P: 70

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 



#52
Aug1911, 06:20 PM

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#53
Aug2211, 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 


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