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Proof.
Pick an arbitrary world
,
and let
(
) =
![]()
![]()
n = 1
![]()
i 1, i 2, . . . , i nN
i n ( . . . (
i 2 (
i 1 (
) ) )
that is, (
)
is the set of all worlds that are reachable from
. Clearly, for each i
N,
i(
)
(
),
which shows that
is a coarsening of the partitions
i, i
N.
Hence
(
)
(
),
as
is the finest common coarsening of the
i's.
We need to show that (
)
(
)
to complete the proof. To do this, it suffices to show that for any sequence i 1, i 2, . . . , i n
N
( 1 ) i n ( . . . (
i 2 (
i 1(
) ) )
We will prove ( 1 ) by induction on n. By definition, i(
)
(
)
for each i
N, proving ( 1 ) for n = 1. Suppose now that ( 1 ) obtains for n = k, and for a given i
N, let
*
i( A )
where A =
i k ( . . . (
i 2 (
i 1 (
) ) ).
By induction hypothesis, A
(
).
Since
i( A )
states that i 1 thinks that i 2 thinks that . . . i k thinks that i thinks that
* is possible, A and
i(
*)
must overlap, that is,
i(
* )
A
.
If
*
(
),
then
i(
* )
(
),
which implies that
is not a common coarsening of the
i's, a contradiction. Hence
*
(
),
and since i was chosen arbitrarily from N, this shows that ( 1 ) obtains for n = k + 1.
Peter Vanderschraaf peterv@MAIL1.ANDREW.CMU.EDU |