(CYHB13) Definability of Truth in Probabilistic Logic

Posted: 2016-08-19 , Modified: 2016-08-26

Tags: ai safety, logic

Intro

Undefinability of truth: no expressive and consistent language can contain its own truth predicate, satisfying \[\forall \ph: True(\ce{\ph}) \iff \ph.\]

Responses:

  1. Work with meta-languages.
  2. Accept some sentences are neither true nor false.

“Then we may continue to define True by the strong strong reflection property that the truth values of \(\ph\) and \(True(\ce{\ph})\) are the same for every \(\ph\). (?)

Unsatisfying.

Possible over prob. logic!

Fix a theory \(T\) over a language \(L\). (We will let also let \(L\) denote the set of sentences.)

Definition: A probability function \(\Pj:L\to [0,1]\) is coherent if there is a probability measure \(\mu\) over models of \(L\) such that \[\Pj(\ph) = \mu(\set{\mathcal M}{\mathcal M \vDash \ph}).\]

Theorem: \(\Pj\) is coherent iff all these hold.

  1. For each \(\ph,\psi\), \(\Pj(\ph)=\Pj(\ph\wedge \psi)+\Pj(\ph\wedge \neg \psi)\).
  2. For each tautology (something that is true in all models) \(\Pj(\ph)=1\).
  3. For each contradiction \(\ph\), \(\Pj(\ph)=0\).

Proof. Define \(T_{i+1}\) iteratively by choosing the first statement independent of \(T_i\), and including it with probability \(\Pj(\ph_j|T_i)\) and its negation otherwise. Let \(T=\bigcup_i T_i\). (Consistent by compactness.)

Axiom 1 implies \(\Pj(\ph|T_i)\) is a martingale. It eventually stabilizes at 0 or 1; the martingale property implies \(T\vdash \ph\) with probability \(\Pj(\ph|T_0)\).

Reflection

This reflection principle doesn’t work: \[\forall \ph\in L', \forall a,b\in \Q: a<\Pj(\ph)<b\iff \Pj(a<\Pj(\ce{\ph})<b)=1\] because we can construct \(G\iff \Pj(\ce{G})<1\), \[\Pj(G)<1\iff \Pj(\Pj(\ce{G})<1)=1 \iff \Pj(G)=1.\]

Instead imagine \(\Pj\) having arbitrarily precise information about \(\Pj\). Say \(\Pj\) is reflectively consistent if (these are equivalent) \[\begin{align} \forall \ph\in L',\forall a,b\in \Q: (a<\Pj(\ph)<b) &\implies \Pj(a<\Pj(\ce{\phi})<b) &= 1\\ \forall \ph\in L',\forall a,b\in \Q: (a\le \Pj(\ph)\le b) &\Leftarrow \Pj(a\le \Pj(\ce{\phi})\le b) &= 1 \end{align}\]

Theorem (Consistency of reflection): Let \(T\) be a consistent theory over \(L\) where \((\Q,+)\) can be embedded, and \(L'\) be the extension of \(L\) by \(\Pj\). There is a coherent probability function \(\Pj\) over \(L'\) assigning probability 1 to \(T\) and satisfying the reflection principle.

Proof.

Theorem (Kakutani fixed point): wikipedia

  1. Let \(S\subeq \R^n\) be non-empty compact convex, \(\ph:S\to 2^S\) such that \(\ph(x)\) is non-empty and convex for all \(x\in S\), and \(\ph\) has closed graph. Then \(\ph\) has a fixed point (\(x\in \ph(x)\)).
  2. \(\ph:X\to 2^Y\)is upper hemicontinuous if for every open \(W\subeq Y\), \(\set{x}{\ph(x)\subeq W}\) is open in \(X\). If \(X,Y\) are topological vector spaces, \(\ph:X\to 2^Y\), \(Y\) convex, then \(\ph\) is Kakutani if it is upper hemicontinuous and \(\ph(x)\) is non-empty, compact and convex. If \(S\) is non-empty, compact, convex subset of Hausdorff locally convex topological vector space, and \(\ph:S\to 2^S\) is Kakutani, then it has a fixed point.

(Why do we care that \(R_{\Pj'}\) is countable?)

“we don’t want P to assert that P is reflectively consistent, we want it to actually be reflectively consistent.”

our work shows that the obstructions presented by the liar’s paradox can be overcome by tolerating an infinitesimal error, and that Tarski’s result on the undefinability of truth is in some sense an artifact of the infinite precision demanded by reasoning about complete certainty.

(Don’t understand the last page: revisit.)