Re: [Haifux] My New Signature re-factored.

[Nadav Har'El]

On Tue, Oct 23, 2001, Shlomi Fish wrote about "[Haifux] My New Signature re-factored.":
> 
> Nadav, please tell me that at least you understood the joke. One righteous
> man in Sedom?

Shlomi, I'm not sure why you decided to test me in public, or whether this is
a test of my Mathematical abilities or of my sense of humor... (I have masters
degree in the former, and a failure degree in the latter), or why this test
was aimed at me, of all people.

But I'll bite...

> If:
> 1. A is A
> 2. A is not not-A
> does it imply that 
> 1. B is B 
> 2. B is not not-B

Why is this a "joke"? Is it supposed to be funny? How is it funny?

Or is it some sort of deep logic thing? I read your forwarded post too, but
it didn't help me understand what you're talking about. Here is what I can
say about this issue:

If a statement (call it S) is true for some "A", it isn't necessarily true for
another "B". Not unless you have another statement saying that the original
statement S is true for any object in the class Z, *and* that both A and
B are members of Z.

But since your wrote specific, not general, statements about A and B, (in
our case S(x)="(x is x) AND (x is not not-x)", and we looked at the two
statements S(A) and S(B)) it is conceivable that S(x) be right or wrong for
*all* objects x of any type. For the answer to your question to be positive,
we need S(x) to have the same truth-value (true or false) for any x of any
type because we have no idea what type A or B is...

But then a question arises - is your statement (S) at all defined for *all*
objects? Before you answer, remember that a "set of all objects" cannot
even be defined (it's easy to prove by paradox - ask me if you're interested).
Also remember that your statement is given in some sort of language (English,
logic, etc.) and it is not at all clear how it is defined for objects B
that cannot be expressed in that laguage at all...

Here are two ways to try to understand what you wrote by trying to define
how your statement S works on large classes of objects (first, all objects
expressible by English, and then all objects expressible by logics). Neither
sounds very funny to me, nor very insightful, so maybe I'm missing something??

If you define "is" and "not" for other classes of abstract objects, you
can get different answers....

======== The "natural language" way to understand S ============
In this case, the "B is not not-B" in the statement S(B) simply
means
        "B" and "not B" are different things.

This seems true for all English phrases B you can think of, but on second
thought it might not be (an example follows). If it can be true to A, and
false for B, then your question "does it imply" deserves the answer of no.

For example, for A="True" (as an English word), we have A=A and A != not-A
("true" and "not true" are different concepts in English).

However, consider B="in a million years", as in the sentence "When will
I finish this project? In a million years!". Now, B ("in a million years") and
"not B" ("not in a million years") can mean exactly the same thing in
many contexts. For casual English (as opposed to Geology research papers),
B = not-B.

I know this sounds really silly, but so does (sorry), the original question...
Unless I missed something really important...

======== The "logic" way to understand S ============
Normally, when you formulate the axioms of logics, you should have the
following axiom:

  For each truth value X (either TRUE or FALSE), you have the following
        1.      X=X
        2.      X=not not X
  (you can call these axioms, or part of the definitions of = and not).

This then extends to any "form", or function, that has an arbitrary number
of variables and produces a single truth value:
        f=f
        f=not not f

[f=g should be understood as "f(x1,...,xn)=g(x1,...xn) for any choice of
truth values x1..xn. Also, not f is a function defined as (not f)(x1..xn)
= not (f(x1...xn)).]

Example functions like that are CNF (conjunctive normal form) expressions,
and their "not" can be expressed as a CNF again using De-Morgan's law, by
the way.

So your statement S(X) ("X is X, X is not not X") is true for any CNF
expression, be it called A or B, and so your question "does it imply"
deserves a "yes" answer in the context of CNF expressions.


-- 
Nadav Har'El                        |    Thursday, Oct 25 2001, 9 Heshvan 5762
[EMAIL PROTECTED]             |-----------------------------------------
Phone: +972-53-245868, ICQ 13349191 |If Windows is the answer, you didn't
http://nadav.harel.org.il           |understand the question.

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