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VHanson comments on The Cartoon Guide to Löb's Theorem - Less Wrong

12 Post author: Eliezer_Yudkowsky 17 August 2008 08:35PM

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Comment author: VHanson 21 January 2014 02:37:26AM 0 points [-]

I did study stuff like this a LONG time ago, so I respect your trying to work this out from 'common sense'. The way I see it, the key to the puzzle is the truth-value of Y, not 'whether or not X is actually true'. By working out the truth-tables for the various implications, the statement "((◻C)->C)->C" has a False truth-value when both (◻C) and C are False, i.e. if C is both unprovable and false the statement is false. Even though the 'material implication' "(X->Y)->Y implies (not X)->Y" is a tautology (because when the 1st part is false the 2nd part is true & vice-versa) that does not guarantee the truth of the 2nd part alone. In fact, it is intuitively unsurprising that if C is false, the premise that 'C is provable' is also false (of course such intuition is logically unreliable, but it feels nice to me when it happens to be confirmed in a truth table). What may seem counterintuitive is that this is the only case for which the premise is True, but the encompassing implication is False. That's because a 'logical implication' is equivalent to stating that 'either the premise (1st part) is False or the conclusion (2nd part) is True (or both)'. So, for the entire statement "((◻C)->C)->C" to be True when C is False, means "((◻C)->C)" must be False. "((◻C)->C)" is only False when "(◻C)" is True and "C" is False. Here's where the anti-intuitive feature of material implication causes brain-freeze - that with a false premise any implication is assigned a True truth-value. But that does NOT mean that such an implication somehow forces the conclusion to be true! It only affirms that for any implication to be true, it must be the case that "IF the premise is true then the conclusion is true." If the premise is False, the conclusion may be True or False. If the premise is True and the conclusion is False, then the implication itself is False!

The translation of " (not ◻C)->C" as "all statements which lack proofs are true" is a ringer. I'd say the implication "If it is not the case that C is provable, then C is True" is equivalent to saying "Either C is provable or C is True or both." Both of these recognize that the case for which "C is provable" and "C is False" makes the implication itself False. The mistranslation erroneously eliminates consideration of the case in which C is both unprovable an False, which is (not coincidentally, I suspect) the one case for which the grand formulation "((◻C)->C)->C" is False.

Comment author: Ender 05 February 2014 11:04:20PM *  0 points [-]

Jeepers. I haven't thought about this problem for a long time. Thanks.

The answer that occurs to me for the original puzzle is that Yudkowsky never proved (◻(2 = 1) -> (2 = 1)). I don't know it that is actually the answer, but I really need to go do other work and stop thinking about this problem.