Viliam_Bur comments on Logical Pinpointing - Less Wrong

62 Post author: Eliezer_Yudkowsky 02 November 2012 03:33PM

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Comment author: Viliam_Bur 01 November 2012 08:50:17PM 11 points [-]

You just say: 'For every relation R that works exactly like addition, the following statement S is true about that relation.' It would look like, '∀ relations R: (∀x∀y∀z: R(x, 0, x) ∧ (R(x, y, z)→R(x, Sy, Sz))) → S)', where S says whatever you meant to say about +, using the token R.

The expression '(∀x∀y∀z: R(x, 0, x) ∧ (R(x, y, z)→R(x, Sy, Sz)))' is true for addition, but also for many other relations, such as a '∀x∀y∀z: R(x, y, z)' relation.

Comment author: moshez 01 November 2012 10:20:40PM 3 points [-]

I'm not sure that adding the conjunction (R(x,y,z)&R(x,y,w)->z=w) would have made things clearer...I thought it was obvious the hypothetical mathematician was just explaining what kind of steps you need to "taboo addition"

Comment author: Viliam_Bur 02 November 2012 09:43:41AM *  16 points [-]

Yes, the educational goal of that paragraph is to "taboo addition". Nonetheless, the tabooing should be done correctly. If it is too difficult to do, then it is Eliezer's problem for choosing a difficult example to illustrate a concept.

This may sound like nitpicking, but this website has a goal is to teach people rationality skills, as opposed to "guessing the teacher's password". The article spends five screens explaining why details are so important when defining the concept of a "number", and the reader is supposed to understand it. So it's unfortunate if that explanation is followed by another example, which accidentally gets the similar details wrong. My objections against the wrong formula are very similar to the in-story mathematician's objections to the definitions of "number"; the definition is too wide.

Your suggestion: '∀x∀y∀z∀w: R(x, 0, x) ∧ (R(x, y, z)↔R(x, Sy, Sz)) ∧ ((R(x, y, z)∧R(x, y, w))→z=w)'

My alternative: '∀x∀y∀z: (R(x, 0, z)↔(x=z)) ∧ (R(x, y, z)↔R(x, Sy, Sz)) ∧ (R(x, y, z)↔R(Sx, y, Sz))'.

Both seem correct, and anyone knows a shorter (or a more legible) way to express it, please contribute.

Comment author: Kindly 02 November 2012 02:31:47PM 16 points [-]

Shorter (but not necessarily more legible): ∀x∀y∀z: (R(x, 0, z)↔(x=z)) ∧ (R(x, Sy, z)↔R(Sx, y, z)).

Comment author: Eliezer_Yudkowsky 03 November 2012 05:32:04PM 8 points [-]

Done!

Comment author: Viliam_Bur 02 November 2012 03:40:52PM 3 points [-]

Perfect!

Comment author: Benja 02 November 2012 07:49:52PM *  3 points [-]

Both seem correct, and anyone knows a shorter (or a more legible) way to express it, please contribute.

The version in the article now, ∀x∀y∀z: R(x, 0, x) ∧ (R(x, y, z)↔R(x, Sy, Sz)), is better than before, but it leaves open the possibility that R(0,0,7) as well as R(0,0,0). One more possibility is:

"Not in second-order logic, which can quantify over functions as well as properties. (...) It would look like, '∀ functions f: ((∀x∀y: f(x, 0) = x ∧ f(x, Sy) = Sf(x, y)) → Q)' (...)"

(I guess I'm not entirely in favor of this version -- ETA: compared to Kindly's fix -- because quantifying over relations surely seems like a smaller step from quantifying over properties than does quantifying over functions, if you're new to this, but still thought it might be worth pointing out in a comment.)