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Jaime Sevilla Molina
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A Sketch of Good Communication
Jaime Sevilla Molina7y100

I appreciate that in the example it just so happens that the person assigning a lower probability ends up assigning a higher probability that the other person at the beginning, because it is not intuitive that this can happen but actually very reasonable. Good post!

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Toward a New Technical Explanation of Technical Explanation
Jaime Sevilla Molina7y220

This is rather random, but I really appreciate the work made by the moderators when explaining their reasons for curating an article. Keep this up please!

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First Order Linear Equations
Jaime Sevilla Molina8y*10

Might as well do it! No promises though.

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An Introductory Guide To Modern Logic
Jaime Sevilla Molina9y*20

My fault, it should be \ulcorner.

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Rices Theorem
Jaime Sevilla Molina9y*10

The problem I have in mind is deciding whether the Halting problem is equivalent to any statement of the form "You cannot decide membership for S", where S is a non-trivial set of computable functions.

Clearly the argument exposed above shows that the Halting problem implies any of these statements, but does the reverse implication hold directly? In my proof of how Rice implies Halting I am handpicking an S. Can we make do without the handpicking?

In other words, given a Halting oracle, can we make a Rice oracle for an arbitrary S?

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Euclids Lemma On Prime Numbers
Jaime Sevilla Molina9y*10

Out of curiosity, is there any reason you are avoiding calling this lemma by its traditional name, Euclid's lemma?

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A primer on provability logic
Jaime Sevilla Molina9yΩ240

Generalized fixed point theorem:

Suppose that Ai(p1,...,pn) are n modal sentences such that Ai is modalized in pn (possibly containing sentence letters other than pjs).

Then there exists H1,...,Hn in which no pj appears such that GL⊢∧i≤n{⊡(pi↔Ai(p1,...,pn)}↔∧i≤n{⊡(pi↔Hi)}.


We will prove it by induction.

For the base step, we know by the fixed point theorem that there is H such that GL⊢⊡(p1↔Ai(p1,...,pn))↔⊡(p1↔H(p2,...,pn))

Now suppose that for j we have H1,...,Hj such that GL⊢∧i≤j{⊡(pi↔Ai(p1,...,pn)}↔∧i≤j{⊡(pi↔Hi(pj+1,...,pn))}.

By the second substitution theorem, GL⊢⊡(A↔B)→[F(A)↔F(B)]. Therefore we have that GL⊢⊡(pi↔Hi(pj+1,...,pn)→[⊡(pj+1↔Aj+1(p1,...,pn))↔⊡(pj+1↔Aj+1(p1,...,pi−1,Hi(pj+1,...,pn),pi+1,...,pn))].

If we iterate the replacements, we finally end up with GL⊢∧i≤j{⊡(pi↔Ai(p1,...,pn)}→⊡(pj+1↔Aj+1(H1,...,Hj,pj+1,...,pn)).

Again by the fixed point theorem, there is H′j+1 such that GL⊢⊡(pj+1↔Aj+1(H1,...,Hj,pj+1,...,pn))↔⊡[pj+1↔H′j+1(pj+2,...,pn)].

But as before, by the second substitution theorem, GL⊢⊡[pj+1↔H′j+1(pj+2,...,pn)]→[⊡(pi↔Hi(pj+1,...,pn))↔⊡(pi↔Hi(H′j+1,...,pn)).

Let H′i stand for Hi(H′j+1,...,pn), and by combining the previous lines we find that GL⊢∧i≤j+1{⊡(pi↔Ai(p1,...,pn)}→∧i≤j+1{⊡(pi↔H′i(pj+2,...,pn))}.

By Goldfarb's lemma, we do not need to check the other direction, so GL⊢∧i≤j+1{⊡(pi↔Ai(p1,...,pn)}↔∧i≤j+1{⊡(pi↔H′i(pj+2,...,pn))} and the proof is finished □


An immediate consequence of the theorem is that for those fixed points Hi and every Ai, GL⊢Hi↔Ai(H1,...,Hn).

Indeed, since GL is closed under substitution, we can make the change pi for Hi in the theorem to get that GL⊢∧i≤n{⊡(Hi↔Ai(H1,...,Hn)}↔∧i≤n{⊡(Hi↔Hi)}.

Since the righthand side is trivially a theorem of GL, we get the desired result.


One remark: the proof is wholly constructive. You can iterate the construction of fixed point following the procedure implied by the construction of the H′i to compute fixed points.

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Free Group
Jaime Sevilla Molina9y*20

Pedantic remark: Aren't you missing the identity of free groups in your intuitive construction?

We have the ρx and the ρx−1. Where is ρϵ?

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99 LDT X 1 CDT Oneshot PD Tournament As Arguable Counterexample To LDT Doing Better Than CDT
Jaime Sevilla Molina9y*10

I do not think that the "Me facing off against 99 CDT agents and 1 LDT agent, versus an LDT agent facing 99 LDT agents and 1 CDT agent" is fair either.

The thing that confuses me is that you are changing the universe in which you put the agents to compete.

To my understanding, the universe should be something of the form "{blank} against a CDT agent and 100 LDT agents in a one-shot prisoners dilemma tournament", and then you fill the "blank" with agents and compare their scores.

If you are using different universe templates for different agents then you are violating extensionality, and I hardly can consider that a fair test.

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99 LDT X 1 CDT Oneshot PD Tournament As Arguable Counterexample To LDT Doing Better Than CDT
Jaime Sevilla Molina9y*10

I fail to see how this setup is not fair - but more importantly, I fail to see how LDT is losing in this situation. If the payoff matrix is CC:2/2, CD:0/3, DD: 1/1, then if LDT cooperates in every round it will get 99⋅2=198 utilons, while if it defected then it gets 100 utilons.

Thus LDT wins 198 utilons in this situation, while a CDT agent in his shoes would win 100 utilons by defecting each round.

The situation changes if the payoff becomes Having a higher score that CDT: 1, while Having an equal or lower score than CDT: 0.

Then the game is clearly rigged, as there is no deterministic strategy that LDT could follow that would lead to a win. But neither could CDT win if it was pitted in the same situation.

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Normal system of provability logic
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First order linear equations
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First order linear equations
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Fixed point theorem of provability logic
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