Resolution of Russell's and Cantor's paradoxes

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The discussion centers on a pre-print paper proposing a resolution to Russell's and Cantor's paradoxes through a non-axiomatic approach using classical logic. The author seeks feedback and clarifications on the paper, emphasizing that it does not advocate for the existence of a universal set but rather analyzes the formation of paradoxes. Participants debate the definitions of terms like "replica" and the implications of accepting naive set theory arguments until contradictions arise. The author clarifies that the goal is to understand where invalid set formations occur, rather than to confine reasoning to strict rules. Overall, the conversation highlights the complexities of set theory and the philosophical implications of resolving paradoxes.
  • #31
micromass said:
His point is that if we take the implication p\rightarrow q and if p is false, then p\rightarrow q is always true. It doesn't matter what q is.

So if we have a legless man then p="his right ankle is bleeding" is always false. So p\rightarrow q= "if his right ankle is bleeding, then we take him to the hospital" is true.

Dear Micromass,

If we are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” - we do not know what is actually true (whether he is actually legless or not) as all we are given is contradiction.

Yours,

Dan
 
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  • #32
DanTeplitskiy said:
Dear Micromass,

If we are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” - we do not know what is actually true (whether he is actually legless or not) as all we are given is contradiction.

Yours,

Dan

If we are given that the man is legless then, p="his right ankle is bleeding" is false. Right??

And thus p\rightarrow q is always true. This are the basic laws of logic.
 
  • #33
micromass said:
If we are given that the man is legless then, p="his right ankle is bleeding" is false. Right??

And thus p\rightarrow q is always true. This are the basic laws of logic.

Dear Micromass,

We are not given that the man is legless.:smile:

We are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” which is logically equivalent to "IF (Dan is a completely legless man And his right ankle is bleeding) THEN he should be taken to the nearest hospital for legless people."

Do you consider the conclusion "...he should be taken to the nearest hospital for legless people" can be drawn from “Dan is a completely legless man. If his right ankle is bleeding..."?

Yours,

Dan
 
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  • #34
DanTeplitskiy said:
Dear Micromass,

We are not given that the man is legless.:smile:

We are given this: “Dan is a completely legless man. If his right ankle is bleeding he should be taken to the nearest hospital for legless people.” which is logically equivalent to "IF (Dan is a completely legless man And his right ankle is bleeding) THEN he should be taken to the nearest hospital for legless people."

Do you consider the conclusion "...he should be taken to the nearest hospital for legless people" can be drawn from “Dan is a completely legless man. If his right ankle is bleeding..."?

Yours,

Dan

Ah, but the same conclusion holds. Let p="Dan is completely legless and his right ankle is bleeding", then p is false. Thus the implication p\rightarrow q holds true never the less.
Why is p false? Well, p is the conjuction p="Dan is completely legless" AND "His right ankle is bleeding". And this conjunction is clearly false.
 
  • #35
micromass said:
Ah, but the same conclusion holds. Let p="Dan is completely legless and his right ankle is bleeding", then p is false. Thus the implication p\rightarrow q holds true never the less.
Why is p false? Well, p is the conjuction p= "Dan is completely legless" AND "His right ankle is bleeding". And this conjunction is clearly false.

Dear Micromass,

If we can derive anything from a contradiction will the thing we derive this way be of any value for us? Will this thing be logically grounded if this could be anything?

For example, we can derive "he should be taken to the nearest hospital for legless people And he should not be taken to the nearest hospital for legless people" from any contradiction (including the above- "Dan is completely legless" AND "His right ankle is bleeding"). Right?

Yours,

Dan
 
  • #36
DanTeplitskiy said:
Dear Micromass,

If we can derive anything from a contradiction will the thing we derive this way be of any value for us? Will this thing be logically grounded if this could be anything?

Yes, it will be logically grounded. The truth table of the implication is logically sound. There are no problems with this.

For example, we can derive "he should be taken to the nearest hospital for legless people And he should not be taken to the nearest hospital for legless people" from any contradiction (including the above- "Dan is completely legless" AND "His right ankle is bleeding"). Right?

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.
 
  • #37
micromass said:
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
 
  • #38
DanTeplitskiy said:
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.
 
  • #39
micromass said:
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
 
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  • #40
DanTeplitskiy said:
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.
 
  • #41
micromass said:
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
 
  • #42
DanTeplitskiy said:
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.
 
  • #43
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
DanTeplitskiy said:
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.
 
  • #45
micromass said:
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
 
  • #46
DanTeplitskiy said:
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 R\in R~\leftrightarrow R\notin R can be proven from the axioms and the inference rules, means that it is true.
 
  • #47
micromass said:
If it can be proven with inference rules then it is a true premise. And since R\in R~\leftrightarrow R\notin R 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
 
  • #48
DanTeplitskiy said:
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 R=\{x~\vert~x\notin x\}. There is no need for (R=\{x~\vert~x\notin x\}~\text{and}~R\neq \{x~\vert~x\notin x\}). In fact, that is still true, but there's no need for it.
 
  • #49
micromass said:
Why can it be derived only from that?? It can also be derived from R=\{x~\vert~x\notin x\}. There is no need for (R=\{x~\vert~x\notin x\}~\text{and}~R\neq \{x~\vert~x\notin x\}). 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
 
  • #50
DanTeplitskiy said:
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 R\in R~\rightarrow~R\notin R can be derived from that. But why can it only be derived from that?
 
  • #51
micromass said:
Yes, I agree that R\in R~\rightarrow~R\notin R 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
 
  • #52
DanTeplitskiy said:
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 R\in R~\leftrightarrow~R\notin R from R=\{x~\vert~x\notin x\}. There is no need for R\neq \{x~\vert~x\notin x\}...
 
  • #53
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
 
  • #54
DanTeplitskiy said:
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.
 
  • #55
micromass said:
Yes, that looks OK.

Dear Micromass,

What if I take this: "Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number."

and change it to that: "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number."

Will the latter be the OK implication to you? (just in case: to me it definitely won't)

Yours,

Dan
 
  • #56
DanTeplitskiy said:
Dear Micromass,

What if I take this: "Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number."

and change it to that: "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number."

Will the latter be the OK implication to you? (just in case: to me it definitely won't)

Yours,

Dan

No, the latter is wrong. I don't see where you're taking me.
 
  • #57
micromass said:
No, the latter is wrong. I don't see where you're taking me.

Dear Micromass,

We are close to the point:smile:

Remember this? :wink: : Let p="Dan is completely legless and his right ankle is bleeding", then p is false. Thus the implication p\rightarrow q holds true never the less.

What if we apply the same reasoning to the case we are discussing now:

Let p = "N - the greatest natural number And N < 2", then p is false.
Thus the implication p\rightarrow q holds true never the less.

Does the implication "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number." seem to be OK now? (just in case: to me it still doesn't :smile:)

Yours,

Dan

P. S. Micromass, please do not take the above for any kind of personal attack. :shy: It is not that - you are a nice guy! I am just eager to explain my point to you now.
 
  • #58
DanTeplitskiy said:
Does the implication "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number." seem to be OK now? (just in case: to me it still doesn't :smile:)

Well, this is actually ok by me. Considering that there is no natural number, and thus considering that N<2 is actually false. Thus the implication

N<2 => N is the greatest natural number

is ok.
 
  • #59
micromass said:
Well, this is actually ok by me. Considering that there is no natural number, and thus considering that N<2 is actually false. Thus the implication

N<2 => N is the greatest natural number

is ok.

Dear Micromass,

Does it mean that you changed your mind and now you consider the reasoning
"Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number."
to be OK? :confused:

Yours,

Dan
 
  • #60
DanTeplitskiy said:
Dear Micromass,

Does it mean that you changed your mind and now you consider the reasoning
"Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number."
to be OK? :confused:

Yours,

Dan

I did not change my mind. Both

"If N<2, then N is the greatest natural number"

as

"If N<2, then N is not the greatest natural number"

are ok. This is of course a contradictory situation and this leads to the conclusion that N<2 is false.

I mean, if you define "N=the greatest natural number", then for every statement p holds that

If p, then N is the greatest natural number.

is always true! By definition.
 

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