AlexMennen comments on Constructive mathemathics and its dual - Less Wrong

13 Post author: MrMind 28 February 2013 05:21PM

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Comment author: AlexMennen 28 February 2013 08:25:12PM 2 points [-]

ECQ and TND are both theorems (or axioms) of classical logic. They are in fact equivalent.

Nitpick: any 2 theorems are equivalent to each other, so that's redundant. Perhaps you meant to say that they are in fact intuitively related to each other.

Comment author: MrMind 28 February 2013 10:22:23PM 0 points [-]

Maybe a better phrasing could have been: if you assume one as an axiom, the other follows as a theorem.

Comment author: AlexMennen 28 February 2013 11:36:51PM 3 points [-]

Yes, that's what equivalence means. And if X and Y are both theorems, then if you assume either as an axiom, then the other must follow as a theorem.

Comment author: Benja 01 March 2013 02:46:12AM *  1 point [-]

It's standard to say that the axiom of choice is equivalent to the well-ordering principle, which means that if you add either to ZF, the other follows and you get ZFC. As I'm reading this, the relevant paragraph in the OP reads,

Notice though: ECQ and TND are both theorems (or axioms) of classical logic. They are in fact equivalent. ECQ is espressed as , but under the DeMorgan laws, double negation elimination and commutativity of disjunction: , which is the TND.

So I interpret MrMind as giving the analogous statement with ZF -> (DeMorgan + DNE + commutativity of disjunction + possibly something else?), axiom of choice -> ECQ, well-ordering principle -> TND, and ZFC -> "classical logic" ... which is probably meant to mean classical propositional logic in this context?

I haven't checked this claim, if indeed that was what was intended, though.

Comment author: Pfft 01 March 2013 07:41:41PM 2 points [-]

The problem is that double negation elimination is already equivalent to classical logic (i.e. adding double negation elimination to minimal logic gives the law of excluded middle as a theorem), so there is no "background theory" like ZF to work in.

Comment author: AlexMennen 01 March 2013 03:49:02AM 0 points [-]

Ah, okay. I had been interpreting him as saying that they are equivalent in classical logic. But yeah, they probably are equivalent under a weaker subset of classical logic in which neither is a theorem on its own.