Larks comments on To signal effectively, use a non-human, non-stoppable enforcer - Less Wrong

31 Post author: Clippy 22 May 2010 10:03PM

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Comment author: Tyrrell_McAllister 22 May 2010 11:47:43PM 5 points [-]

I would cooperate with you if and only if (you would cooperate with me if and only if I would cooperate with you).

This is logically equivalent to, and hence carries no more information or persuasive power than

You would cooperate with me.

This may be checked with the following truth-table:

Let P = I would cooperate with you.

Let Q = You would cooperate with me.

Then we have

P <=> (Q <=> P)
T T T T T
T F F F T
F T T F F
F F F T F
Comment author: Larks 22 May 2010 11:58:52PM *  0 points [-]

I assume you mean,

P, <=>, (Q <=> P)

to be the headers of your truth table.

But even then the truth tables for (P iff ( Q iff P) ) and ( P iff Q) are different - consider the case where 'you' will co-operate with me no matter what. If I'm running ( P iff Q), I'll cooperate; if I'm running (P iff ( Q iff P) ), I'll defect.

Edit: formatting trouble.

Comment author: Tyrrell_McAllister 23 May 2010 12:07:45AM *  1 point [-]

I assume you mean . . .

No, I am giving the truth-table for P <=> (Q <=> P) in a compact form. It's constructed by first assigning truth-values to the first occurrence of "P" and the first occurrence of "Q". The second occurrence of "P" gets the same truth-value as the first occurrence in every case. Then you compute the truth-values for the inner-most logical operation, which is the second occurrence of "<=>". This produces the fourth column of truth values. Finally, you compute the truth-values for the outer-most logical operation, which is the first occurrence of "<=>".

Hence, the second column of truth-values gives the truth-values of P <=> (Q <=> P) in all possible cases. In particular, that column matches the third column. Since the third column contains the truth-values assigned to Q, this proves that P <=> (Q <=> P) and Q are logically equivalent.

ETA: You edited your comment. Those are indeed the correct headers, so my correction above no longer applies.

But even then the truth tables for (P iff ( Q iff P) ) and ( P iff Q) are different - consider the case where 'you' will co-operate with me no matter what. If I'm running ( P iff Q), I'll cooperate; if I'm running (P iff ( Q iff P) ), I'll defect.

Yes, the truth-table for P <=> (Q <=> P) is different from the truth-table for P <=> Q. But those aren't the propositions that I'm saying are equivalent. I'm saying that to assert P <=> (Q <=> P) is logically equivalent to asserting Q all by itself. In other words, to implement the belief that P <=> (Q <=> P) is functionally the same as implementing the belief that Q. This means that the belief that Clippy recommends signaling is logically equivalent to an unconditional belief that you will cooperate with me.

One can't help but suspect that Clippy is trying to sneak into us a belief that it will always cooperate with us ;).

Comment author: Larks 23 May 2010 10:40:00AM 1 point [-]

ETA: You edited your comment. Those are indeed the correct headers, so my correction above no longer applies.

Sorry for the confusion. I understand now; the extra space between two of the columns confused me.

However, I suspect we need a stronger logic to represent this properly. If Q always defects, no matter what, "you would cooperate with me if ... I ... cooperate with you" is false, but is given true in the propositional interpretation.

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