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I am lying!

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abyaly:
A friend of mine who is proficient with boolean rings deduced that
"A: A -> B" is an member of the proposition set only in case B is universally true.
I've been convinced that curry's paradox isn't really a problem, but I don't think I could explain it well enough to convince you :-/

jknilinux:
So he agrees with your idea- that it's not a proposition. Are you saying Tarski is wrong when he expresses it as a proposition in the proofs I uploaded?

I realize that's kind of a fallacy- "the super smart guy said x, I don't know why, therefore x" - but anyway, I still think it's a proposition. I mean, Tarski's proofs seem to show that it is logically valid, if not even logically sound...

By the way, what do you mean by "I don't think I could explain it well enough to convince you"? I've been convinced of a new idea before....

Numsgil:
I looked at the proof.  I still say that because you take the statement "this statement is not true" as an axiom (or assume it's true, same difference), and you arrive at a contradiction, you essentially formed a proof by contradiction.  Which means the initial statement is false and can't act as an axiom in the first place.  It also doesn't matter that if you take the statement as false you arrive at another contradiction.  It just means that neither it nor its negation are true.  And if you use contraposition, you can conclude that neither are false.

Which just demonstrates that it's possible for a statement to be neither true nor false.  This isn't a breakdown of logic.  Logic is designed to work in a discreet world of only pure truth and pure falsehood.  So long as your axioms are discreet (completely true or completely false), you can use logic, and any proofs you do will be discrete (purely true or false).  But you can't use logic in a non discreet problem space, which is the problem here.  The proof was flawed before you even started.  You need to use something more like Fuzzy logic.

abyaly:

--- Quote from: jknilinux ---So he agrees with your idea- that it's not a proposition. Are you saying Tarski is wrong when he expresses it as a proposition in the proofs I uploaded?
--- End quote ---
Tarksi's proof is about formal languages. The first two premises are true about our language, and Tarski showed that applying them to a formal language leads to a contradiction (it seems he later gave up the first premise). I have no problem with Tarski's proof. But if you try to apply the proof in general, the error will be in the first step.


--- Quote from: jknilinux ---By the way, what do you mean by "I don't think I could explain it well enough to convince you"? I've been convinced of a new idea before....
--- End quote ---
I suppose I could try, but I don't know how many of the terms will be familiar.
He used the idea that a boolean algebra is isomorphic to a boolean ring.
He assumed that
A = "A -> B"
was a member of the boolean algebra, and formulated what I assume was an equivalent statement in the ring.
A = A + AB + 1
+ in this case is symmetric difference. Multiplication is AND, 1 is universally true (multiplicative identity), and 0 is universally false (additive identity).
He then preformed the following operations:
0 = AB + 1   (left cancellation)
1 = AB + 1 + 1 (add 1 on the right)
1 = AB (boolean rings are characteristic 2, so 1 + 1 = 0)
From there, we get that A = B = 1.
So saying
A = "A -> B"
is the same as saying
A = B = 1
So the statement "A = 'A -> B'" being valid is equivalent to B being universally true. In boolean logic, at least.

jknilinux:
Okay, I never learned boolean algebra before, but now that I sat down and reasoned through what you said, it makes sense. Boolean algebra isn't all that different at all from predicate logic. Anyway, the proof seems valid, but when you say it is equivalent to B being universally true, aren't you just saying the paradox all over again? You seem to be admitting that B must be true, even though we haven't even defined what it is. So are you saying the paradox is correct?

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