mamert comments on The Parable of the Dagger - Less Wrong

53 Post author: Eliezer_Yudkowsky 01 February 2008 08:53PM

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Comment author: HalFinney 01 February 2008 10:55:47PM 6 points [-]

I'm having some trouble with the logic here. I wonder if the parable got a bit garbled.

"You see," the jester said, "let us hypothesize that the first inscription is the true one."

The first inscription says, "Either this box contains an angry frog, or the box with a false inscription contains an angry frog, but not both." Now we are hypothesizing that this is the true one. Therefore "the box with a false inscription" means "the second box". So, "Either the 1st box contains an angry frog, or the 2nd box contains an angry frog, but not both".

The jester goes on, "Then suppose the first box contains an angry frog."

So we know (by assumption) that the 1st clause of the inscription is true, the 1st box contains an angry frog. Since "not both" clauses are true, it means the 2nd clause is false, and so the 2nd box does not contain an angry frog - it must contain gold.

But the jester claims that this is a contradiction: "Then the other box would contain gold and this would contradict the first inscription which we hypothesized to be true." For this to be a contradiction, the 1st inscription would have had to say that the 2nd box should contain an angry frog, but we just saw that it doesn't say that.

I can't make much progress with the 2nd inscription either. I'm getting pretty confused now!

Comment author: mamert 14 October 2015 10:46:37AM *  1 point [-]

Bx is true if box x has gold, false if frog. one contains frog, other gold -> B1 == ~B2. only one inscription is true -> Bf == ~Bt

We know:

B2 && Bf || Bt && B1 (I1)

B2 && Bt || B1 && Bt (I2)

Bt == B1 && Bf == B2 && I1 && ~I2 || Bf == B1 && Bt == B2 && ~I1 && I2 # only one inscription is true

From this:

((B2 && B2 || B1 && B1) && ~(B2 && B1 || B1 && B1)) || (~(B2 && B1 || B2 && B1) && (B2 && B2 || B1 && B2))

((B2 || B1) && ~(false || B1)) || (~(false || false) && (B2 || false))

(true && (true && B2)) || ((true && true) && B2)

B2 || B2

B2 # so, Box 2 contains gold