## Resolution of Russell's and Cantor's paradoxes

 Quote by micromass 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

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 Quote by DanTeplitskiy 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.

 Quote by micromass 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

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 Quote by DanTeplitskiy 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.

 Quote by micromass 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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 Quote by DanTeplitskiy 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.

 Quote by micromass 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

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 Quote by DanTeplitskiy 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.

 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

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 Quote by DanTeplitskiy 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.

 Quote by micromass 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

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 Quote by DanTeplitskiy 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.

 Quote by micromass 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

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 Quote by DanTeplitskiy 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.

 Quote by micromass 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

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 Quote by DanTeplitskiy 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?

 Quote by micromass 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