Dear Micromass,
Are you sure that I have been arguing over this thing all the time since then? :smile:
The thing of publishing is a complicated matter, you know ))
The goal was not to enjoy myself though. I wanted a discussion over my point...
Yours,
Dan
Dear Micromass,
It is your way of thinking that one can not give an example of a logical error in math paper with things that are not well-formed formula. Sorry ))
Yours,
Dan
Dear Micromass,
Are you sure that when we taik about logical errors we have to talk about well-formed formula (otherwise we can not do it) !? That is a thing I can not agree with))
Yours,
Dan
Dear Micromass,
This is an example of the logical error.
I could not invent the example of this logical error in math different from Russell's paradox-like things...
Yours,
Dan
Dear Fredric,
Nor do I! My point is that each of them separately is a fallacious, that is, containing a logical error, argument.
Well it depends on what you mean by valid...
As to me it contains logical errors but at the same time it is a theorem in predicate logic (that is quite...
Dear Fredrik,
No! You missed the point, sorry. Please try to read my message#20 ))
I thought you missed it - that is why I wanted to put it here again...
Yours,
Dan
Dear Rubi,
It is a logical error to make conclusion such a way )) Which is not a good thing )))
Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to...
Dear Micromass,
Hi! Remember me? ))
The reference is in the reference section of the paper)) It is a book by a Russian logician Ivin. He mentions this. By the way I am not quoting.
Yours,
Dan
Dear Rubi,
Yes, in predicate logic Argument 1 is a theorem. I know it ))
My point is that assumption R ∈ R contradicts the definition of R because if R ∈ R, R includes a member that is included in itself (R itself is such a member).
Or, symbolically: R ∈ R → ∃ y: y ∈ R ∧ y ∈ y → R ≠ {x: x∉x}...
Dear Rubi,
I said you correctly used the axiom of predicate logic ))
Actually, I suspect you did not even try to read my message#20 after the sentence "Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of...
Dear Rubi,
Russell's paradox is like:
Let R be the set of all sets that are not members of themselves. Then it is a member of itself if and only if it is not a member of itself. - paradoxical incoherence.
The usual conclusion that we make from the paradox is that there is no such R.
"Let...
Dear Rubi,
I answered your first question as well as I could.
"∃R∀x(x∈R↔x∉x) is false" - this is correct, of course :) There is no such R :). I am not making any alternative "my" axiomatic system )))).
What exactly (which line) have you failed to understand in my message#20?
Yours,
Dan
Dear Rubi,
You correctly applied the axiom of predicate calculus.
Have you read my message #20 above? Have you understood everyithing there?
Yours,
Dan