on 20-Apr-2017 (Thu)

In AGM-theory of belief revision, the expansion T + ϕ of a theory T with a sentence ϕ is defined as T + ϕ := {ψ : T ∪ {ϕ} \(\vdash\) ψ}.

(*1) T ∗ ϕ is a theory;
(*2) ϕ ∈ T ∗ ϕ;
(*3-4) if \(\vdash\) ϕ then T ∗ ϕ = T ;
(*5) T ∗ ϕ = ⊥ iff \(\vdash\) ¬ϕ ;
(*6) if \(\vdash\) ϕ ↔ ψ then T ∗ ϕ = T ∗ ψ;
(*7-8) if ¬ψ \(\notin\) T ∗ ϕ then T ∗ (ϕ ∧ ψ) = (T ∗ ϕ) + ψ

(*5e) T ∗ ϕ = ⊥ iff T \(\vdash\) K¬ϕ (i.e. iff (K¬ϕ) ∈ T )

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TIP: Switching between Hiragana input mode and Direct Input mode through the language bar is tedious. Instead you can switch by pressing Alt-Tilde (the key below ESC on your keyboard). So, if you need to type Japanese, press Alt-Tilde and start typing. When you are done press Alt-Tilde again to switch back to English. To achieve the same result when in Japanese mode, press Control - ~.

Hope this answers your question

So we formulate our revised multi-agent (epistemic) AGM pos tulates, by giving, for every agent a ∈ A: a family T a ⊆ P(BKL) of sets of sentences in the language BKL, called a-theories, and a belief revision operator ∗ a : T a × BKL → T a , taking pairs of a-theories and BKL-sentences into new a-theories; and requiring them to satisfy the following conditions: (T1) ⊥ ∈ T a (where ⊥ := BKL is the inconsistent theory, containing all the sentences in BKL); (T2) every T ∈ T a is deductively closed, w.r.t. the complete proof system of BKL; (T3) for every ϕ ∈ BKL and every T ∈ T a , we have either K a ϕ ∈ T or (¬K a ϕ) ∈ T ; (T4) all the above postulates of epistemic AGM, in which we label with agent names both the knowledge K a and the revision ∗ a operators. Observe that it is not necessary to require an introspective condition corresponding to (T3) for belief, since this follows from the above conditions, given the axioms of BKL. Indeed, one can easily prove that for every ϕ ∈ BKL and every T ∈ T a , we have either B a ϕ ∈ T or (¬B a ϕ) ∈ T

each S-theory T ⊆ S gives rise to a syntactic theory th(T ) = {φ ∈ BKL : t |= S φ for all t ∈ T }.