Tuesday, January 28, 2014

Liveblogging the Veritas Forum

21:30 And we're done. I think two things kinda say it all: more of the audience questions were appropriate for a preacher than for a scientist; and the book table on the lobby was packed with every desperate attempt to make theism intellectually respectable published in the last twenty years. Veritas indeed.

21:25 I made my way to the mike line and asked about personhood being an emergent phenomenon. In particular, since she had gone out of her way to dismiss the very idea in less than a sentence, my question is, do you have the scientific basis for such a rejection? Result: total deflection. I mean, yes, it's also interesting that other humans ascribe person-ness to dumb routines, which is why the Turing Test is less powerful than it first appears, but answer the damn question.

20:58 Did she just approvingly mention Killing Jesus? As in, Bill Fucking O'Reilly?

20:50 Oh, and apparently the Bible instructs us not to self-promote. Or at least not to do the research you already did last year.

20:45  Now Picard and Spickard are sitting down for Q&A. His first questions are about religion.

20:37 ...And she came from an "atheist faith" to a Christian faith. Lovely.

20:36 Ah, now we see the reveal. Imagine an alien race which stumbles across inductions for building a radio. Lo and behold, it's more than its components: there's music too! And it's an emergent phenomenon! Personhood is analogous to the music "emerging" from a radio.

20:29 We've now seen two going on three multimedia clips; yet for some reason the products didn't bother to figure out how to connect the laptop's sound to the PA.

20:13 Picard opens by polling the room about their beliefs on whether human beings are "just" machines. She says she used to. Very few in the audience raise their hands. (The room, by the way, is FULL.)

20:11 A student brings the room to order. Picard's research sounds well worth doing.

The guest of honor: Rosalind Picard of MIT.

The local host: Anderson Spickard. The Third. Captain, I detect no warning signs of incipient douchebaggery.

Wednesday, January 22, 2014

Elbow grease

Random thought of the morning:

It is a fact that there exist electric cookware appliances which are completely submersible and dishwasher-safe. (We have such a beast -- a fondue pot that Ms. Heel-Filcher got for Christmas a couple of years back.)

In view of this, it is time and past time for there to exist a dishwasher-safe miniature grill. Get on that, George Foreman.

Saturday, January 18, 2014

Proof by Contradiction

I spent one morning at the JMMs this week in a session on the Introduction-To-Proofs course (aka the Bridge course). I enjoyed the session; it contained some good techniques to include, some good warnings of things to avoid, etc. Such a course always includes learning about proof by contradiction.

Now, I have a bit of a thing about proof by contradiction. It's not that it's invalid; it's that, at least in classical logic, you can always transform a proof by contradiction into a proof by contraposition; in any case, while I'm not a constructivist and think that intuitionistic logic is mostly a curiosity, it's nice to have a proof which provides an implicit algorithm, when such a thing exists.

So today on the train I was puzzling over a little bit of boolean-algebra arithmetic which Tom Jech uses in his exposition of forcing. (I'm supervising an undergrad who wants to learn forcing, and the boolean-valued-model approach is so much nicer than the poset-only approach.) I tried to prove it forwards six ways to Sunday, and nothing seemed to work. So, remembering what we all tell students in Bridge courses, I turned it around and tried to prove it by contradiction, and that took all of six lines. I still don't see a direct proof that's not an artificial translation of the contradiction/contraposition result; if you know one, put it in the comments!

Theorem: Recall that in boolean algebras, the implication operation \( v_1 \Rightarrow v_2 \) is defined as \( v_2 \lor \neg v_1 \). If \(x,y,z \) belong to a boolean algebra \( \mathbf{B} \) and \( x \land y = x \land z \), then \( x \leq ( y \Leftrightarrow z) \).

Proof: Suppose otherwise. Then we can find elements \( x, y, z \) in a boolean algebra \( \mathbf{B} \) so that
\[ x > y \Rightarrow z = x \land ( z \lor (\neg y)) \]
Now observe that
\[ x \land ( z \lor (\neg y)) = (x \land z) \lor (x \land \neg y) = (x \land z) \lor (x \land y) \lor (x \land \neg y) = (x \land z) \lor x \geq x \]
Overall, \( x > x \), a contradiction.

Sunday, January 5, 2014

Forcing Fraisse

Maybe it's just because it's three in the morning, but this little throwaway exercise in some lecture notes (pdf) of David Marker made my night:

Two models \( \mathcal{M} \) and \( \mathcal{N} \) are \( \infty \)-back-and-forth equivalent iff there is a forcing extension of the universe in which they are isomorphic.
Mind. Blown. And the proof is obvious.