Perplexed comments on Making Reasoning Obviously Locally Correct - Less Wrong

19 Post author: JGWeissman 12 March 2011 07:41PM

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Comment author: Perplexed 12 March 2011 09:35:07PM 5 points [-]

Amusingly, try substituting x=y=0. You find a different flawed step than the one Eliezer found.

Comment author: cousin_it 13 March 2011 02:06:33PM *  3 points [-]

Nice. So here we have two testcases that catch different bugs, and you can never prove the absence of bugs by adding more testcases. But testing still works well enough in practice that people rarely write fully formal proofs for software, or for math theorems for that matter :-)

Comment author: Snowyowl 13 March 2011 03:59:43PM 0 points [-]

Actually, x=y=0 still catches the same flaw, it just catches another one at the same time.

Comment author: Perplexed 13 March 2011 04:27:20PM 1 point [-]

Our disagreement seems to derive from my use of the words "different flawed step" and your use of "same flaw". Eliezer suggested substituting 1 for x and y in :

(x+y)(x-y) = y(x-y)

x+y = y

yielding

(1+1)(1-1) = 1(1-1) (true)

1+1 = 1 (false)

Thus, since a true equation was transformed into a false one, the step must have been flawed.

Under my suggestion, we have:

(0+0)(0-0) = 0(0-0) (true)

0+0 = 0 (true)

So, under Eliezer's suggested criterion (turning true to false) this is not a flawed step, though if you look carefully enough, you can still notice the flaw - a division by zero.

Comment author: timtyler 13 March 2011 06:41:16PM *  4 points [-]

So, under Eliezer's suggested criterion (turning true to false) this is not a flawed step, though if you look carefully enough, you can still notice the flaw - a division by zero.

Hmm. A failure to identify a flawed step doesn't mean that the step isn't flawed.

A true statement turning into a false one does show that you manipulated it badly - but a true statement staying true doesn't show that you manipulated it well.