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Proof.
Since
is a coarsening of
i for
each i ∈ N,
Ki((ω)).
Hence,
K1N((ω) ), and since by definition
Ki((ω)) =
{ ω |
i(ω)
⊆
(ω)}
=
(ω),
K1N((ω)) = ∩
i ∈ NKi((ω)) = (ω)
Applying the recursive definition of mutual knowledge, for any m ≥ 1,
KmN((ω)) = ∩
i ∈ NKi(Km−1N((ω)) = ∩
i ∈ NKi((ω)) = (ω)
so, since ω ∈ (ω), by definition we have ω ∈ K*N((ω)).
Peter Vanderschraaf peterv@cyrus.andrew.cmu.edu | Giacomo Sillari Carnegie Mellon University gsillari@andrew.cmu.edu |